Alternately, I would've just transformed/rotated the x and y axes by an angle of 45° about the origin using the rotational matrix. Then using the standard method of computing volumes of revolution ,the problem becomes easy.
Yeah, it is actually the same thing (just less convoluted). The change of variable is (u,v)=1/sqrt(2) (y-x , x+y) and you get the curve by solving sqrt(2) (u+v)=(v-u)^2. Then you can apply the usual technique on the interval u in [0,sqrt(2)].
@@quintonpierre Thing is, it becomes multivariable when you let u = (y-x) and v = (y + x). Then you have to deal with Jacobians, which you did implicitly.
@@emanuellandeholm5657 things are nicer if you include a factor 1/sqrt(2) in the new variables. otherwise you need to think about scaling factors in the integral limit and differential
I used a rotation matrix to get a parametric equation for the curve: x=sqrt(2)/2*(t^2 + t) y=sqrt(2)/2*(t^2 - t) Then you can integrate pi*y(x)^2 over x from 0 to sqrt(2), which after making a u-substitution of x in terms of t turns into the integral over t from 0 to 1 (that’s the only t interval on which y is negative) of pi*sqrt(2)/2*(t^2 - t)^2*(sqrt(2)*t+sqrt(2)/2), which works out nicely to the pi/(30*sqrt(2)).
Much simpler method. The volume is an onion made of cones which slant height is h(x) = x - x^2. The radius is just h(x) / sqrt(2) (simple isocele 45° triangle) Thus, the surface of this cone is s(x) = pi * h(x)^2 / sqrt(2) =pi * (x - x^2) ^2 / sqrt(2) = pi * (x^4 - 2x^3 + x^2) / sqrt(2) The thickness of each cone is dx Then V = / s(x) dx in interval (0,1) V = pi * (F(1) - F(0)) / sqrt(2) with primitive F(x) = x^5 / 5 - x^4 / 2 + x^3/3 F(0) = 0 F(1) = 1/30 " 1/5 - 1/2 + 1/3 = 1/30 " V = pi / (30* sqrt(2) ) CQFD NOTE : Démonstration rapide de la surface d'un cône de longueur latérale L et de rayon R. Un cône déplié fait un secteur d'angle Alpha et sa surface est le ratio Alpha/ 2*pi du cercle de rayon L (pi*L^2) Sa surface est donc S = pi*L^2 * Alpha/ 2*pi = L^2* Alpha/ 2 La circonférence de la base du cône fait 2*pi*R mais aussi Alpha*L Donc Alpha = 2*pi*R /L donc Alpha/ 2 = pi*R /L Finalement S = L^2 * pi*R/L = pi*R *L la surface d'un cône de longueur latérale L et de rayon R vaut S = pi*R *L Conic onion peels characteristics : PRIMO : Each cone has a constant angle of 45° with its axis (base radius and height are equals) SECUNDO : dx is the thickness of the peel because x axis is perpendicular to the cone generating line Then s(x) is something simple proportional to the square of the slant height and slant height is also simple y(top of the conic peel ) - y(parabola matching lower point of the peel)
@@thomashoffmann8857 No that's where it changes the problem entirely to make it far easier to solve than it otherwise would have been. This error persisted throughout the video, and I'm not sure whether the answer that he got is incorrect or whether he just pulled it from mathematica without checking it through the calculations.
@@sweetcornwhiskey No, the value (4*xi+1-sqrt(8*xi+1))/2 us the value that makes the problem easier to solve. xi - (2*xi+1-sqrt(8*xi+1))/2 = (-1+sqrt(8*xi+1))/2 which removes the xi term so the two parts would not be the same.
Wow - this takes me back to the 80's. I was fortunate enough to have a high-school teacher who once worked for NACA (later became NASA) - named Roselle D'Orazio. She was a war-effort mathematician and worked on the X-1 too. A brilliant lady - a kind grandmotherly type who could run circles around most anyone with math. Most fun I ever had in a class. She did bring up such a topic, as extra out-of-interest work. She touched on Hamilton's work and quaternions, among other ways of addressing these types of problems. If the USA would remove the politics from teaching, and get back to attracting (and keeping) teachers like this with backgrounds like this who really love the subjects ... this country would be accelerating forward ...
@@UltraMaXAtAXX How backasswards is it that funding should depend on test scores. Low test scores should indicate students and teachers need _more_ funding not less.
My way for this kind of problems is to remember: The volume of a rotated area is the area times the distance the center of gravity traveled in this rotation. So I break it down to several steps. Step 1: Calculate the area of this [1/6]. Step 2: Calculate the volume of it if rotated by the x-axis [2*Pi/15]. Step 3: Calculate the volume if rotated by the y-axis [Pi/6]. Step 4: Use the numbers to calculate the coordinates of the center of gravity [1/2; 2/5]. Step 5: Calculate the distance between the center of gravity and the line of actual rotation [sqrt(2)/20]. Step 6: With these numbers, calculate the volume [Pi*sqrt(2)/60]. All done with simple integrations, and moreover: Without thinking.
+1 I was going to comment on the same idea: Calculate the center of gravity and use the Pappo-Guldin Theorem. Depending of the problem, calculating the center of gravity can be done in different way, for a parabola segment is known since Archimedes times, sometime symmetries can help, or there is a general formula in worse cases.
The point is that there is not ONE way to arrive at a correct answer. That the answer is correct because there is agreement of the solution INDEPENDENT of the method of its derivation. You can integrate disks or revolve curves, it's the same outcome. If you only ever confine yourself to revolving curves, you may not include integrating disks into your thinking, and will be subsequently stopped because there are no appropriate curves to revolve. Remember, everything looks like a nail when all you have is a hammer. But if you never learn to use a screwdriver, screws just appear as poor performing nails.
Can you explain how you calculated the centre of gravity in step 4 please? I don't understand what function you are rotating about the axes to get the volumes in parts 2 and 3
I spent like 2.5 hours on this and thought it was going to require a bunch of trig sub and whatnot... and then I realized I used 2*pi*r instead of pi*r^2 for the disc volume 😂 which meant I just had an extra square root in there unnecessarily complicating the heck out of my integral
There is a simpler way to do this problem since we have the line y=x which is a nice 45 degree angle. The trick is to take a small rectangle (width dx) and note that when you rotate it about y=x you get a lateral surface of a cone with volume: V_conearea = pi*r*L*dx. The slant height "L" is easy to calculate since it's the difference of the curves: L = x-x^2. Next, the radius "r" is a bit more complicated since that line is at an angle. However, since we know that for that 45 degree cone h = r = L/sqrt(2), we know that r = (x-x^2)/sqrt(2). Now integrate x from 0 to 1! V = integral(pi*r*L dx,0,1) = integral(pi*(x-x^2)^2/sqrt(2) dx,0,1) = pi/(30*sqrt(2)) I think this is a "cooler" way to do this problem since we don't need to do any rotation, or use washers or shells! Instead, we use lateral area of cones with thickness dx! You can do this "cone" trick with any angle but that just makes finding the cone radius "r" from the slant height "L" involve an arc trigonometric function. Have fun!
|\ |\ \ | \ \ | \ So the infinitesimal volume (V_conearea) is the volume between those 2 slanted lines rotated around the vertical line (vertically extruded lateral cone area), where the vertical line represents y=x.
I am a math teacher in Germany. First of all thanks a lot for the idea of this challenge. It is not necessarily very hard, yet a good training in arithmetics! Here are my two cents: The orthogonal distance from the line y=x to the parabola is: sqrt(2xi^2 + 6xi + 1 -2(2xi + 1)sqrt(0.25 + 2xi)) Finally the whole integration leads to the volume of V= 1/60 * sqrt(2) * pi. Hope that's the right answer, but it seems at least plausible... Thanks again for this great channel!
I guess the next challenge would be to find the volume of the solid of revolution formed by rotating the region bounded by y=x and y=x^2 about the parabola y=x^2.
what would it even mean to rotate sth about a curve? afaik that doesnt make sense. take a random point in the shaded area. can you tell me unambiguously where it would rotate to?
@@jakobunfried2669 Hmm... how about using the curvature of a point on the parabola to generate a circle of inversion and then invert the point on the line?
@@derendohoda3891 only 90% sure i understand what you mean, but assuming i do: a point on the line could then be on several different such circles and it would be unclear, where it should rotate to
@@jakobunfried2669 there's only one circle perpendicular to the curvature with all points equidistant from a particular point of the curve. The team problem we would run into is overlapping circles -- a naive integral would double count some volumes when rotating about a curve
This was just a wonderful video! I know I would have really struggled with that if a student in a Calc II class had asked it as a question and I didn't have any notes prepared. I thought for sure the square root integrals were going to need to be done with trig substitutions; it's always a little nice when they can just be done with u-substitutions (I guess you'd need squares under the radical for trig substitutions).
Had teachers cover this in '96 and after. They not only did y=x, but a lot of other values for the slope and we had to calculate by hand. I loved my TI-92 that would do this!
My way was "simple" but fraught with tiny mistakes: I wrote A+B = , where r = 1/sqrt(2). Then just write the integral of (pi*B^2)dA. This simplifies greatly because 2rA = x+x^2 and 2rB=x-x^2, resulting in an integral of powers of x from 0..1.
Rotating the old coordinate system X by 45° the coordinates for (x,x^2) in the new system X' become (x',y')=R*(x,y)=1/sqrt(2)*(x^2+x,x^2-x) where R=1/sqrt(2)*[ (1,-1) , (1,1) ] is the rotation matrix by -45° (rotating the coordinate system by 45° is equivalent to rotating the old coordinates by -45°). Hence y' becomes the radius of the rotating body and V=Pi*Int(y'^2 * dx',x=0..1)=Pi/2*Int( (x^2-x)^2 * d(x^2+x)/sqrt(2) , x=0..1) = Pi/(2*sqrt(2))*Int( (x^4 - 2x^3 + x^2)*(1+2x)*dx,x=0..1)=Pi*sqrt(2)/60.
i found it much more intuitive to do a double integral over all rings and then you can even let the curve be an arbitrary function f(x). the radius of the ring that intersects the shaded surface at (x, y) is (x - y)/sqrt(2). integrating the circumference of these rings over the shaded area (x from 0 to 1, y from f(x) to x) gives a general formula V = π/sqrt(2) \int_0^1 dx (x - f(x))^2 valid for a function with f(0)=0, f(1)=1, f(x)
I believe there is a more easy way to solve this problem. The distance between any point of the parabola and the line y=x in [0,1] is (by basic geometry courses) d(x)=(x-x*x)/sqrt(2). Because it is going to be rotate around y=x the length is sqrt(2)*(xi+1-xi), then the volume will be pi*integral[0,1] d2(x) sqrt(2) dx. Solving this integral the result is pi/30sqrt(2)
Thank you, Mr. Penn. This video just taught me how to do "u substitution". My calculus class (in 1977) didn't get this far. Hey, I learned something today! Thanks again!
this can be trivialized by a change of basis, but it's nice to see this alternative method. Also yeah, not sure what's up with calc teachers skipping this. My class didn't cover it, as far as I can remember.
Probably not that exciting, but this was definitely covered in my calc 2 course. I remember because it was one of the only problems I skipped on the final (not because I couldn't do it, but I recognized right away it would take too much of my time, because I wasn't as prepared for that problem as I was the rest of the material.)
At least one Mr. Turnbaugh in Bentonville, AR didn't skip it! Spent about 3 weeks on it, and like 4 people in the class ended up being comfortable with it. But we did it.
I think a combination of both. In general, surfaces of revolution are kind of skimmed over. They just give the integral formula without giving it too much thought, which results in not knowing how to do these, since there's not a standard formula for this problem.
As someone who has taught college calculus a few times I can tell you that the semester tends to be pretty packed. This would work well in something like calculus BC in high school where there's a fair amount of time and you can do some neat projects, but if you tried this in a college calculus class (probably calc 2, but some people do solids of rotation at the end of calc 1 in a semester system; I don't remember where everything goes in quarters) you'd be pressed for time and the students wouldn't get as much out of it as you might want. This would make for a cool extra credit project, though, and if you were looking for an undergrad paper this would probably be an interesting one. It does kind of get trivialized by rotation matrices, though. Another reason I would want to skip this is that I generally think of solids of rotation as a somewhat gentle introduction to applications of integration. They're easy to visualize - the slices for the Riemann sum are literal slices - rather than trying to "slice" time or some other abstract concept. It goes well alongside the classics like lifting objects with ropes of non-zero mass and pumping water out of various troughs and such. If you then make the algebra complicated it defeats the purpose of having a gentle introduction as the students get lost in the algebra and don't comprehend the cool stuff you're showing them.
@@skylardeslypere9909 It's funny, I never even remember the formulas myself because I always want to draw everything. I think my students want or expect the formula but I just keep drawing stuff; the formula is in the book, they can look it up if they want to. I'm just going to keep unrolling cylinders and stacking discs. I mean, who cares about the formula anyway? The whole point of having solids of revolution as part of the introduction to Riemann sums is to help gain intuition about what a Riemann sum really is by having a visualizable application, so why ever worry about the formula? That just skips all the visualizing and jumps to the answer, which defeats the purpose. Also, I'm pretty sure (though I haven't done it) that you could derive a general formula for rotations about lines not square to the axes by using a rotation matrix to rotate your axes, do all the calculations in the new coordinate system, and then another to rotate them back. My guess is that you can probably just run the existing formulas for discs and shells through those rotation matrices to get the formulas straight away. You could probably also do it with standard algebra (like Michael did in the video), but that feels like it would be harder. So I strongly suspect there's a reasonably-derived standard formula, but again, why bother if the whole point is to think about the problem and how Riemann sums work? Now that I think of it, this might make a good project in a linear algebra class when talking about either rotation matrices or change of basis, as this is a really good use of a change of basis matrix. I think I might steal this for exactly that purpose. Thanks for making me think on this topic Skylar, I think my students will really like this :-)
@@jgray2718 You mentioned that it gets trivialized by rotation matrices, and I was thinking something along those lines too while watching the video. Is it actually easier to find the volume of the surface of revolution around any line in the plane, if you apply a change of base? That is, translate your coordinate system so the origin is on that line, and then rotate it so it becomes one of the axes.
@@skylardeslypere9909 In general, yes it's probably easier. Applying a rotation matrix is easy, then you have a standard solid of revolution, which is also easy. The algebra Michael did seemed much harder to me. "Trivialized" was probably an overstatement on my part, but it's not difficult, just a little bit longer than the usual solid of revolution. This particular problem was also chosen specifically to have "easy" algebra since it was rotated around y = x. I'd think a rotation matrix would be the way to go for sure if you wanted to rotate around y = 2x. God help you if you want a general formula for rotating about y = ax + b, though. I'd have to consider how to do that, but I think you're into affine algebra rather than linear algebra at that point, so maybe the approach Michael took in the video would actually be easier in that case. I don't even know what an affine rotation matrix looks like, now that I think of it. Probably a slide then a spin?
I normally teach this topic for honors classes, specially the revolution about the line y = mx. Revolution about y=mx+b can be little more involved. Thank you for doing these.
How is revolution around y=mx+b more involved? You just move everything down by b and you're back at y=mx, don't you? Well, technically, that one simple thing makes it a little more involved. But you know what I mean.
alternative solution: generate weight field which is 0 at the revolving axis and linearly goes to 1pi when it reaches distance 1 from the revolving axis, integrate over x. y equals the integral of the weight field beween the revolving axis and other function
Using Archimedes’ geometric proof of the parabolic quadrature, the relative height and width of each triangle is known and the position of each center can also be calculated. So the centroid can be calculated as the weighted sum of the centers. Then take the foot/elevation, r, of the centroid from y=x. This is 1/(10*2^(1/2)). Multiply 2πr by 1/6 (per Pappus) an lid the volume obtains.
i need to write, that the perpendicular point xl,yl on y=x is the solution of 2 linear equations. after this is done, r(x)=sqr((x-xl)^2+(y-yl)^2). then there is the guldin rule or a for next loop to add parts with different r(x) together. finally, a part of the volume is dv=pi*r(x)^2*dl l=sqr(xs^2+ys^2) so dl=l/n (n=number after "to" in a for next loop)
Let Sₖ be the set containing 2 and 5 and the first _k_ primes that end in 7. For example, S₃ = {2, 5, 7, 17, 37}. Define a _k-Ruff_ number to be one that is not divisible by any element in Sₖ. If Nₖ is the product of the numbers in Sₖ then define _F(k)_ to be the sum of all _k-Ruff_ numbers less than Nₖ that have the last digit 7. You are given _F(3)_ = 76101452. Find _F(97),_ give your answer modulo 1,000,000,007.
hi! the difficulty would be greater if the axis y=x was not in the xy plane. but so, the rotated points are, if xl and yl are perpendicular points: xr=x+(xl-x)*cos(w) yr=y+(yl-y)*cos(w) zr=sqr((x-xl)^2+(y-yl)^2)*sin(w) have fun using ba$ic instead of calculus!
The distance from a point (x,x^2) up to the point (x,x) is L=x^2-x. The shortest distance from (x,x^2) to the line x=y is thus r=(x^2-x)/sqrt(2) since there is a 45-45-90 triangle involved. Rotating the vertical line segment from (x,x^2) to (x,x) around the x=y line gives a cone with slant height L and radius r. The volume can then be found from the surface area of a cone: pi r L, i.e. \int_0^1 pi /sqrt(2)* (x-x^2)^2 = pi/ (30 *sqrt(2)). Disks, shells, washers... and cones!
Great idea! Only the distance between the point of the parabola (x,x^2) and the axis of rotation y=x is equal to L=(x-x^2)/√2, 0≤ x≤ 1. When the parabola rotates, this segment forms a cross-section of the figure perpendicular to the axis of rotation - a circle with an area of π L^2. If dl is an element of the axis of the geometric body , then dl=√2dx. V=∫(from 0 to 1)π*[(x-x^2)^2/2]√2dx=π√2/60.
We calculate the volume of the geometric body as the sum of the volumes of elementary cylinders of differentially small height. In this sense, the cone has nothing to do with it.
@@Vladimir_Pavlov You are right of course, r=(x-x^2)/sqrt(2), since (x^2-x)/sqrt(2) is negative. I think what you call L was r for me. Splitting the volume up into disks as you suggest is very good and simpler. I do like the idea of splitting it into cones though!
It should be noted that this approach to solving the problem makes it easy to get a result if the axis of rotation is an arbitrary straight line passing through the origin, y=tga *x, a≠π/2.)
I used a series of cone shaped shells each having slant height of x-x^2 and radius sqrt(2)/2 of that. Integrated 0 to 1 dx. I got the same answer. First I started the same way as in the video but stopped short of calculating that diagonal distance. Then I tried rotating axes but couldn't figure that bit out.
Same, although I think the cones have the same radius and height of r(x)=(x-x^2)/sqrt(2), we want the shells so I did Vshell(x)=pi/3*r(x)^3-pi/3*(r(x)-sqrt(2)dx)^3 I.e. volume outer cone minus volume of inner cone. Note you need sqrt(2)dx because the cone height lies on the diagonal line rather than aligned with the x axis. After cancelling and dropping terms with dx^2 or dx^3, you get Vshell(x)=pi/sqrt(2)(x-x^2)^2 dx and when you integrate this from 0 to 1 you get pi/(30*sqrt(2))
I used a slightly different approach, I set P(x) as the projection of the point A = (x, f(x)) onto the axis (the line x=y). P(x) can be found as P(x)=(t,t) setting t=x-c=f(x)+c. This solves as c= ½(x-x²), so that P(x)= (½(x+x²), ½(x+x²)). Now let s(x) be the distance of P(x) from the origin: s(x) = ½√2 (x+x²), and let r(x) be the distance from P to A: A-P = (x- ½(x+x²) , x²- ½(x+x²)) = (½(x-x²) , ½(x²-x)) and r(x)=|A-P|= ½√2(x-x²). We now find ∫ πr²ds = ∫ πr² • ds/dx dx = ∫ π ½(x-x²)² • ½√2 (1+2x) dx = ¼π√2 ∫ (x-x²)² •(1+2x) dx = ¼π√2 ∫ x^2 -3x^4 + 2x^5 dx = ¼π√2 [1/3 x^3 - 3/5 x^5 + 1/3 x^6] Because x runs from 0 to 1, the last expression must be evaluated at the boundaries 0 and 1 this gives a volume of π√2/60
If you just subtract x from y = x^2 then you the object bounded above by the axis. This is equivalent and EASIER than the rotation matrix. This problem is simply rotation y = x^2 - x about the x-axis between 0 and 1.
If for a parabola y=x^2 the axis of rotation is a straight line y=tga *x (a ≠π /2), then the required volume of the body of rotation will be equal to V=(π/30) *cosa *(tga)^5. At a = 45 degrees, we get an answer for a this case.
I applied a different method, in principle it should be more general. Let us assume to have a straight line t, with equation y=mx+q in the plane, and a function y=f(x) to rotate w.r.t. to the direction defined by the straight line by an angle of 2pi. Then, if we take a general point P=(s,f(s)) on the graph of the function f, the distance between point P and the straight line is given by the classical formula d(P,t)=|ms+q-f(s)|/sqrt(1+m^2). So the volume of the rotation of the graph of bounded by the interval [a,b] is simply the sum of all the circles of radius d(P,t) all along the interval [a,b]. Obviously the sum becomes the following integral \int_a^b \pi*d(P(s),t)^2ds In the case proposed in this video, a=0, b=1, f(x)=x^2, m=1, q=0 and the integral becomes \int_0^1 \pi/2*(s-s^2)^2ds=\pi/2*(x^3/3-x^4/2+x^5/5)|_0^1=\pi/2*(1/3-1/2+1/5)=\pi/2*1/30=\pi/60 The result is not the same, I do not have the square root of 2 but just 2 at the denominator. I hope you enjoy this solution.
Very cool approach and easier than in the video. I would like to point out that until the line "sum becomes the following integral" everything is correct. The problem comes after that. If u take an area of πr² then u need to multiply the differential element perpendicular to the given area. So multiplying "ds" simply gives the wrong solution. Differential element should be along the line ms+q acc. to notations u have taken which then gives the differential element as ds.√(1+m²). (From pythagoras theorem directly; taking a triangle of ds base and dy =mds as altitude) The entire integral then becomes Integration from 0 to 1[π(s-s²)²/√2 ds] I think this will give the correct value now.
Instead of relying on that magic, you can avoid the sqrt altogether by not solving for x in terms of xi. Instead turn x^2+x=2xi into: xi=x(x+1)/2 dxi=(x+1/2)dx and substitute accordingly so x is integration variable (and bounds remain at 0 and 1). After much reduction, end up with: sqrt(2)pi/4 integral from 0 to 1 dx [2x^5 - 3x^4 + x^2] which is very easy to integrate.
To clarify, the "much reduction" part involves many but easy reductions since we are simplifying sums and multiplications of small polynomials (no radicals).
When I took AP Calc in High School, we were supposed to come up with problems and put together our own "final examn". My group put this problem on it and our teacher wasn't able to solve it 😉
This situation happened to me on accident when our prof made an error in the homework. Me and another guy couldn't solve it so we asked, but he just fixed the error instead of telling us how to solve it.
Both r and w are obviously hypotenuses in right, isosceles triangles (you already established their slopes of 1 and -1). That would save a lot of writing. Just take one of the legs, multiply by √2 and call it a day.
Going about it the hard way. Set the slope as your x value and calculate the area under the curve. Convoluted methods take too much time. Good exercise for fun.
"I think you can work out a general formula for this.." Yep, and it's really easy: the volume is the integral of the cross sectional area d h. Go forth and calculate. :)
Rotation of coordinate systems was covered on one my senior year math courses leading to my electrical engineering degree. Course was called "Multidimensional Calculus". The material in the course I found to be quite helpful in my employment for a defense contractor in missile and weapons systems targeting/engagement. (Been retired for over 30 years, and still remember some of that stuff)
You will be hard pressed to find a better mathematician than an EE grad. I earned my BSEE studying computer chip design. Graduated in the tech bubble burst. No jobs anywhere, world-wide. Ended up doing statistics for the insurance industry.
Hi Michael, there is an easier way to solve this: if OP represents the segment from the origin to P(x,y=x^2), and t is the angle between OP and x-axis, then radius of revolution is r=OP*sin(pi/4-t)=sqrt(2)/2*OP*(cost-sint)=sqrt(2)/2(x-y). Then V=integral_0^1 pi*[(sqrt(2)/2(x-x^2))]^2*sqrt(2)dx = pi*sqrt(2)/60
This is actually the standard situation (covered for example in section 5.1 in Stewart's Calculus) of a solid whose volume is computed as the integral of the cross-section area. Volumes of solids obtained by rotating a region bounded by graphs of functions about the x-axis (V=\int_a^b \pi f(x)^2 dx) is a special case of it.
@@adrianignacioirias I don't have the 4th edition but I guess you are talking about the Discover project "Rotation about a slant line" in the chapter about Applications of Integration. You don't need this for the example of the video. And, if you read the whole chapter, you'd better use the Theorem of Pappus to compute the volume.
I see comments regarding forward rotation of the axes. … I am not impressed by the counterintuitive nature of that idea. Rotate the curves instead as you suggest. Det(J) or det(g) =1 so there is no need to do any heroic math. The standard parametric statement of the volume integral definition includes the derivative of x with respect to a parameter t. y’^2(dx’/dx)dx can all be expressed in terms of the original variable x, so the bounds of integration are unchanged.
You could also use Pappus, you only need to find the area of the curve, his center of gravity and the distince of that center to the axis of rotation (y=x in this case). I had these problem (but with a different curve) in my Calc 2 final exam
I am sure there are others who have commented that the rotation of axis method would be simpler to implement, but this is also a good video to demonstrate the use of the Riemann Sum. I wonder how this can be solved using Lebesgue Integral...
I used Lebesgue integration, and it felt much neater to me! It worked by splitting the volume into hollow cylinders with the central axis being y=x. The radius r cylinder would have width dr and volume 2 pi r len(r) dr, where len(r) is the length of the cylinder of radius r. To find len(r), we just need to find the points such that y = x^2 and y = x - r/sqrt(2) (since that is the parallel line to y=x at distance r from it). The x values of the points are the solutions to x^2 - x + r/sqrt(2) = 0. The difference between the x values is sqrt(b^2 - 4ac), which in this case is sqrt(1 - 2sqrt(2)r). The distance between the points is then sqrt(2)sqrt(1-2sqrt(2)r) (because the slope of the line is 1). So our answer is the integral of 2sqrt(2) pi r sqrt(1-2sqrt(2)r) dr, where r goes from 0 to... 1/2sqrt(2), since that's where len(r) becomes 0. Substitute t=1-2sqrt(2)r, so r = (1-t)/2sqrt(2) and dr = -dt/2sqrt(2). Now we're solving integral of 2sqrt(2) pi (1-t)/2sqrt(2) sqrt(t) (-dt/2sqrt(2)), which is: -pi/2sqrt(2) times integral of sqrt(t)-tsqrt(t) dt, where t goes from 1 to 0. The final integral to solve is super easy, and you get -pi/2sqrt(2) times [2/5 - 2/3], which is 2pi/15sqrt(2), which is different from the answer given in the video. :o I have no idea where I went wrong. I had a silly calculation mistake in my first attempt and it magically matched the video answer with that mistake.
What about taking a surface of revolution around a curve? The radii of the circles would be the perpendicular distance between the two curves. Maybe that's a silly thing to do, but it would make for an interesting problem.
I solved this way differently (desmos demo: desmos com/calculator/fci6ffmtdx) by parametrizing the distance between the two functions with a line orthogonal two x=y. I will call the intercepts k_i. First I considered that the function for the orthogonal line must be: -x + 2x_0 = y Then I found the intercepts with the two other functions at x_0. For x=y its almost by definition: k_1 = (x_0, x_0) and for x^2=y its a little more complicated, but basically you can solve the equation -x + 2x_0 = x^2 and you get: k_2 = ((sqrt(1+8*x)-1)/2,((sqrt(1+8*x)-1)/2)^2) Which looks a bit terrifying, but doesn't cause any major problems down the way. Then I considered that the radius of the volume, r, at x_0 must be given by the function: r(x) = ||k_1 - k_2|| Where we are a bit lucky because actually both terms in the norm are equal (not intuitive in my opinion). However that simplifies the equation A LOT: r(x) = sqrt(2) * ((sqrt(1+8x)-1)/2 - x) Which we can use to calculate the area of a circle-slice: c(x) = r(x)^2 * pi = (sqrt(2) * ((sqrt(1+8x)-1)/2 - x))^2 * pi = 2 * pi * ((sqrt(1+8x)-1)/2 - x)^2 Then we can construct a integral using the formula for the area of a circle-slice to calculate the volume: V = int_0^1{c(x) dx} = int_0^1{2pi * ((sqrt(1+8x)-1)/2 - x) ^2 dx} = pi/60 ~ 0.52 So, I must say I am pretty confused. Weirdly this is "off" from your result by a factor of exactly sqrt(2), so that is probably not a coincidence, however I for the life of me cannot fathom it right now.
Man, I did this a much simpler way. Rather than finding the distance from the point (x_i,x_i) to the corresponding point on the parabola, I found the distance from the the point (x_i, x_i^2) to the corresponding point on the line. This quickly gave pi sqrt 2 ∫ ((x - x^2)/sqrt 2)^2 dx
I did the same, but differentiated with respect to t. Like you, I get w = √2 dx But when solving for slope from (x, x) to (t, t²) = −1, I solve for x in terms of t instead of t in terms of x (x−t²)/(x−t) = −1 xᵢ− t² = −x + t 2x = t² + t x = 1/2 (t² + t) Differentiting both sides we get: dx = 1/2 (2t + 1) dt Note that points (t, t²) go from (0,0) to (1,1), so t is between 0 and 1 Now we calculare r², where r is distance between (x, x) = (1/2 (t² + t), 1/2 (t² + t)) and (t, t²) r² = [1/2 (t² + t) − t]² + [1/2 (t² + t) − t²]² = [1/2 (t² − t)]² + [1/2 (t − t²)]² = 1/4 (t² − t)² + 1/4 (t − t²)² = 1/2 (t⁴ − 2t³ + t²) Now we can calculate volume of a disk with radius r and width w v = πr²w = π · 1/2 (t⁴ − 2t³ + t²) · √2 dx = π · 1/2 (t⁴ − 2t³ + t²) · √2/2 (2t + 1) dt Integrating over all disks, we get volume of rotation V = π/(2√2) ∫ [0, 1] (t⁴ − 2t³ + t²) (2t + 1) dt = π/(2√2) ∫ [0, 1] (2t⁵ − 3t⁴ + t²) dt = π/(2√2) (t⁶/3 − 3t⁵/5 + t³/3) | [0, 1] = π/(2√2) (1/3 − 3/5 + 1/3) = π/(2√2) (1/15) V = π/(30√2)
Could you not have simply subtracted x from x^2 (or vise versa) and done a horizontal rotation on that? (of course that rotation would be from 0 to sqrt(2) since that's the length of the line)
Had a lot more fun finding the length of the line perpendicular to the rotation axis between its intersections with the curves, and getting area with pi r^2 (squaring the square root), and integrating the resulting simple polynomial over x.
im guessing starting out by rotating the picture so that y=x is flat would have made most of the steps turn out simpler, but it would have been harder to get info about the transformed parabola. but i guess having a rotated parabola defining the curve is better than having a rotating axis defining the solid, and probably involves a lot less writing. maybe thats why they dont do slanted examples, because you can transform a slanted example into a standard one at the start, albeit it would be good to know how that process plays out.
The area of the depicted figure is A = \int_0^1 x dx - \int_0^1 x^2 dx = 1/6 A rectangle with width a and length sqrt(2) with the same area: a * sqrt(2) = 1/6 => a = 1/(6*sqrt(2)) Rotating this rectangle gives the solution: pi*a^2*sqrt(2) = pi * sqrt(2)/72 =pi/(36*sqrt(2)) However Michael obtained pi/(30*sqrt(2)). Can someone point me to my error?
I was thinking all the time what may be wrong with such an easy approach and I found the explanation: Rotating figures with the same cross-section do not lead to bodies with the same volumina, e.g. a rectangle A_r=a*h_r has the same area as a right-triangle A_t=½*a*h_t if h_r=½*h_t. Now I introduce r=h_t that the formulas are looking more familiar: A_r=a*½*r, A_t=½*a*r. If I rotate the rectangle around a, I obtain a cylinder with V_cy=¼*pi*a*r^2 and if I rotate the right triangle around a, I obtain a cone with V_co=1/3*pi*a*r^2
Math arguments should not be interrupted by ads. An ad at the beginning is enough for youtube and Mr Penn to be well compensated. We taxpayers, who paid for the development of the interned receive no royalties at all for the risk of our taxes. We should not have to tolerate the greed of youtube. The is part of the corruption of our government that permits such greed.
Teachers 'skip' this because it's obviously just a co-ordinate change away from regular revolution. What would be more interesting is if you rotated every point on the 'slanted' curve around the X axis, but the centre of rotation was given by the other curve.
That's because they don't throw the prolate sphere aka football. I was a post-doc in physics at UPenn and have also taught calculus (my favorite was teaching Calculus-3).
@@williamwolfs4819 It's a shame because Pappus' centroid theorem is easily understood earlier, and it is FAR more general than solids of revolution so restricting it to that is a disservice to the theorem.
21:06 My German math teacher in those situations 20 years ago: *Deduction of points!! Pi/(30sqrt2) what? Potatoes? Pancakes? It's "volume units"!* (which very often downgraded me from a "1" ("very good", "A") to a "2" ("good", "B+"). Different times though. Today, grading in German secondary schools is less strict and less pedantic.
mmmm imagine as an extra advanced problem if the line being rotated around was not one of the boundaries (i.e. not y = x ) , but it would have to be parallel to y = x to stop it intersecting the region. Consider rotating around y = x + 1 ( or y = x + c for c not equal to 0 ).
mmm actually, it doesn't even need to be constrained to have the same slope as y = x. It just needs to be such that it does not intersect the region between y = x and y = x^2.
00:06 That skipping must have started recently. 🤔I did many of these from the Swokowski (?) textbook in 1993 at LSU. And the book was purple. 😁👍🏻 (I also think that I have taught these problems in a class or two, but that was many moons ago.)
You might be complicating stuff, just rotate everything to the y=0 axis and integrate as usual, notice that this is a rotation by 45° and we know the matrix for that
Of course I am supposing you can use the formula for volumes generated by a curve rotated about the y=0 axis, otherwise just use the same logic to prove it
Wouldn't it be easier to rotate everything by -45 degrees, i.e. go from (x, y) to (u, v) where u = (x+y)/sqrt(2), v = (x-y)/sqrt(2) (maybe need to change some plus to minus or vice versa), express the line y = x^2 in terms of u and v, get it in a form v = f(u) on the interval from 0 to 1, and then calculate ∫πf(u)^2 du from 0 to 1?
When rotating the coordinate system by 45 degrees , it is necessary to calculate π*∫[from 0 to√2 ]{(u+1/√2)-√(√8 u+1/2)}^2 du =√2/60 When the coordinate system rotates by (-45)degrees will have to calculate the same integral. Note that you will have to integrate from 0 to √2.
Couldn't you also take the area of the 45/45/90 triangle between 0 and f(x) = g(x), and then just integrate y=x^2 within the same bounds? You'd Have to fiddle with it a bit; like you might have to take the complement of the integral within the quadrant before you rotated it around the linear function, but i'm just curious. It's too late for me to tackle right now the full "slanted revolution" technique: i'm sleepy, but i don"t see why my kludge wouldn"t work in this case at least.
Le volume étant simplement connexe et la symétrie xy, pourquoi s'embêter à prendre des rectangles d'intégration penchés, les verticaux auraient aussi bien fonctionnés dans ce cas précis. Un exemple plus compliqué avec un sinus "rapide" borné par un sinus x lent" par rapport à y=1/2x aurait montré la généralité de la méthode présentée
At 3:09 the interval [0,1] is sudivided into n+1 subintervals, not n subintervals. Maybe he had in mind to dtart with x1 as his first point, not x0. Anyway is is an error.
Alternately, I would've just transformed/rotated the x and y axes by an angle of 45° about the origin using the rotational matrix. Then using the standard method of computing volumes of revolution ,the problem becomes easy.
That's a ingenious idea. This makes sense👌
That was my first thought. That parabola is still a "nice" function in a rotated coordinate system (x' = y - x, y' = y + x). Trivial substitution.
Yeah, it is actually the same thing (just less convoluted). The change of variable is (u,v)=1/sqrt(2) (y-x , x+y) and you get the curve by solving sqrt(2) (u+v)=(v-u)^2. Then you can apply the usual technique on the interval u in [0,sqrt(2)].
@@quintonpierre Thing is, it becomes multivariable when you let u = (y-x) and v = (y + x). Then you have to deal with Jacobians, which you did implicitly.
@@emanuellandeholm5657 things are nicer if you include a factor 1/sqrt(2) in the new variables. otherwise you need to think about scaling factors in the integral limit and differential
I used a rotation matrix to get a parametric equation for the curve:
x=sqrt(2)/2*(t^2 + t)
y=sqrt(2)/2*(t^2 - t)
Then you can integrate pi*y(x)^2 over x from 0 to sqrt(2), which after making a u-substitution of x in terms of t turns into the integral over t from 0 to 1 (that’s the only t interval on which y is negative) of pi*sqrt(2)/2*(t^2 - t)^2*(sqrt(2)*t+sqrt(2)/2), which works out nicely to the pi/(30*sqrt(2)).
Much simpler method. The volume is an onion made of cones which slant height is h(x) = x - x^2.
The radius is just h(x) / sqrt(2) (simple isocele 45° triangle)
Thus, the surface of this cone is s(x) = pi * h(x)^2 / sqrt(2) =pi * (x - x^2) ^2 / sqrt(2) = pi * (x^4 - 2x^3 + x^2) / sqrt(2)
The thickness of each cone is dx
Then V = / s(x) dx in interval (0,1)
V = pi * (F(1) - F(0)) / sqrt(2) with primitive F(x) = x^5 / 5 - x^4 / 2 + x^3/3
F(0) = 0 F(1) = 1/30 " 1/5 - 1/2 + 1/3 = 1/30 "
V = pi / (30* sqrt(2) ) CQFD
NOTE : Démonstration rapide de la surface d'un cône de longueur latérale L et de rayon R.
Un cône déplié fait un secteur d'angle Alpha et sa surface est le ratio Alpha/ 2*pi du cercle de rayon L (pi*L^2)
Sa surface est donc S = pi*L^2 * Alpha/ 2*pi = L^2* Alpha/ 2
La circonférence de la base du cône fait 2*pi*R mais aussi Alpha*L
Donc Alpha = 2*pi*R /L donc Alpha/ 2 = pi*R /L
Finalement S = L^2 * pi*R/L = pi*R *L
la surface d'un cône de longueur latérale L et de rayon R vaut S = pi*R *L
Conic onion peels characteristics :
PRIMO : Each cone has a constant angle of 45° with its axis (base radius and height are equals)
SECUNDO : dx is the thickness of the peel because x axis is perpendicular to the cone generating line
Then s(x) is something simple proportional to the square of the slant height
and slant height is also simple
y(top of the conic peel ) - y(parabola matching lower point of the peel)
Pretty much any other method is simpler than his
it is a very satisfying feeling to just find rotated coordinates and ending up at pretty much exactly the same integral and hence final value
8:40 the y-coordinate should have been (4*xi+1-sqrt(8*xi+1))/2
Yes i noticed the same thing
It's compensated around 13:20 when Delta - y is calculated for the distance (?)
@@thomashoffmann8857 No that's where it changes the problem entirely to make it far easier to solve than it otherwise would have been. This error persisted throughout the video, and I'm not sure whether the answer that he got is incorrect or whether he just pulled it from mathematica without checking it through the calculations.
@@sweetcornwhiskey No, the value (4*xi+1-sqrt(8*xi+1))/2 us the value that makes the problem easier to solve.
xi - (2*xi+1-sqrt(8*xi+1))/2 = (-1+sqrt(8*xi+1))/2 which removes the xi term so the two parts would not be the same.
Yeah that honestly bothered me a lot lol.
Wow - this takes me back to the 80's. I was fortunate enough to have a high-school teacher who once worked for NACA (later became NASA) - named Roselle D'Orazio. She was a war-effort mathematician and worked on the X-1 too. A brilliant lady - a kind grandmotherly type who could run circles around most anyone with math. Most fun I ever had in a class. She did bring up such a topic, as extra out-of-interest work. She touched on Hamilton's work and quaternions, among other ways of addressing these types of problems. If the USA would remove the politics from teaching, and get back to attracting (and keeping) teachers like this with backgrounds like this who really love the subjects ... this country would be accelerating forward ...
Except districts are worried about test scores. I love my subject but my district may have a hard time keeping me.
@@UltraMaXAtAXX How backasswards is it that funding should depend on test scores. Low test scores should indicate students and teachers need _more_ funding not less.
My way for this kind of problems is to remember: The volume of a rotated area is the area times the distance the center of gravity traveled in this rotation. So I break it down to several steps. Step 1: Calculate the area of this [1/6]. Step 2: Calculate the volume of it if rotated by the x-axis [2*Pi/15]. Step 3: Calculate the volume if rotated by the y-axis [Pi/6]. Step 4: Use the numbers to calculate the coordinates of the center of gravity [1/2; 2/5]. Step 5: Calculate the distance between the center of gravity and the line of actual rotation [sqrt(2)/20]. Step 6: With these numbers, calculate the volume [Pi*sqrt(2)/60].
All done with simple integrations, and moreover: Without thinking.
Pappus theorem.
+1 I was going to comment on the same idea: Calculate the center of gravity and use the Pappo-Guldin Theorem. Depending of the problem, calculating the center of gravity can be done in different way, for a parabola segment is known since Archimedes times, sometime symmetries can help, or there is a general formula in worse cases.
The point is that there is not ONE way to arrive at a correct answer. That the answer is correct because there is agreement of the solution INDEPENDENT of the method of its derivation. You can integrate disks or revolve curves, it's the same outcome. If you only ever confine yourself to revolving curves, you may not include integrating disks into your thinking, and will be subsequently stopped because there are no appropriate curves to revolve. Remember, everything looks like a nail when all you have is a hammer. But if you never learn to use a screwdriver, screws just appear as poor performing nails.
@@abenedict85 For this, I prefer integrating cones. (Take the segment from x^2 to x and revolve it around the x=y line.)
Can you explain how you calculated the centre of gravity in step 4 please? I don't understand what function you are rotating about the axes to get the volumes in parts 2 and 3
I spent like 2.5 hours on this and thought it was going to require a bunch of trig sub and whatnot... and then I realized I used 2*pi*r instead of pi*r^2 for the disc volume 😂 which meant I just had an extra square root in there unnecessarily complicating the heck out of my integral
There is a simpler way to do this problem since we have the line y=x which is a nice 45 degree angle.
The trick is to take a small rectangle (width dx) and note that when you rotate it about y=x you get a lateral surface of a cone with volume: V_conearea = pi*r*L*dx. The slant height "L" is easy to calculate since it's the difference of the curves: L = x-x^2. Next, the radius "r" is a bit more complicated since that line is at an angle. However, since we know that for that 45 degree cone h = r = L/sqrt(2), we know that r = (x-x^2)/sqrt(2). Now integrate x from 0 to 1!
V = integral(pi*r*L dx,0,1) = integral(pi*(x-x^2)^2/sqrt(2) dx,0,1) = pi/(30*sqrt(2))
I think this is a "cooler" way to do this problem since we don't need to do any rotation, or use washers or shells! Instead, we use lateral area of cones with thickness dx!
You can do this "cone" trick with any angle but that just makes finding the cone radius "r" from the slant height "L" involve an arc trigonometric function.
Have fun!
|\
|\ \
| \ \
| \
So the infinitesimal volume (V_conearea) is the volume between those 2 slanted lines rotated around the vertical line (vertically extruded lateral cone area), where the vertical line represents y=x.
I am a math teacher in Germany. First of all thanks a lot for the idea of this challenge. It is not necessarily very hard, yet a good training in arithmetics!
Here are my two cents: The orthogonal distance from the line y=x to the parabola is: sqrt(2xi^2 + 6xi + 1 -2(2xi + 1)sqrt(0.25 + 2xi))
Finally the whole integration leads to the volume of V= 1/60 * sqrt(2) * pi.
Hope that's the right answer, but it seems at least plausible...
Thanks again for this great channel!
This is something new, this is the first video about this new idea I have ever seen in UA-cam. Thank you for this amazing video
I guess the next challenge would be to find the volume of the solid of revolution formed by rotating the region bounded by y=x and y=x^2 about the parabola y=x^2.
what would it even mean to rotate sth about a curve? afaik that doesnt make sense.
take a random point in the shaded area. can you tell me unambiguously where it would rotate to?
@@jakobunfried2669 Hmm... how about using the curvature of a point on the parabola to generate a circle of inversion and then invert the point on the line?
@@derendohoda3891 only 90% sure i understand what you mean, but assuming i do: a point on the line could then be on several different such circles and it would be unclear, where it should rotate to
@@jakobunfried2669 there's only one circle perpendicular to the curvature with all points equidistant from a particular point of the curve.
The team problem we would run into is overlapping circles -- a naive integral would double count some volumes when rotating about a curve
Maybe next year
This was just a wonderful video! I know I would have really struggled with that if a student in a Calc II class had asked it as a question and I didn't have any notes prepared. I thought for sure the square root integrals were going to need to be done with trig substitutions; it's always a little nice when they can just be done with u-substitutions (I guess you'd need squares under the radical for trig substitutions).
(P.S. Your "Mr. Rogers mannerisms" are really nice; I should try to emulate that more)
11:04 - should be sqrt(2*(deltaX)^2) = sqrt(2)*deltaX
I worked through this when I found it in the James Stewart Calculus book. What a great problem!
Had teachers cover this in '96 and after. They not only did y=x, but a lot of other values for the slope and we had to calculate by hand. I loved my TI-92 that would do this!
My way was "simple" but fraught with tiny mistakes: I wrote A+B = , where r = 1/sqrt(2). Then just write the integral of (pi*B^2)dA. This simplifies greatly because 2rA = x+x^2 and 2rB=x-x^2, resulting in an integral of powers of x from 0..1.
Rotating the old coordinate system X by 45° the coordinates for (x,x^2) in the new system X' become
(x',y')=R*(x,y)=1/sqrt(2)*(x^2+x,x^2-x)
where R=1/sqrt(2)*[ (1,-1) , (1,1) ] is the rotation matrix by -45° (rotating the coordinate system by 45° is equivalent to rotating the old coordinates by -45°).
Hence y' becomes the radius of the rotating body and
V=Pi*Int(y'^2 * dx',x=0..1)=Pi/2*Int( (x^2-x)^2 * d(x^2+x)/sqrt(2) , x=0..1) = Pi/(2*sqrt(2))*Int( (x^4 - 2x^3 + x^2)*(1+2x)*dx,x=0..1)=Pi*sqrt(2)/60.
i found it much more intuitive to do a double integral over all rings and then you can even let the curve be an arbitrary function f(x).
the radius of the ring that intersects the shaded surface at (x, y) is (x - y)/sqrt(2).
integrating the circumference of these rings over the shaded area (x from 0 to 1, y from f(x) to x) gives a general formula
V = π/sqrt(2) \int_0^1 dx (x - f(x))^2
valid for a function with f(0)=0, f(1)=1, f(x)
I believe there is a more easy way to solve this problem. The distance between any point of the parabola and the line y=x in [0,1] is (by basic geometry courses) d(x)=(x-x*x)/sqrt(2). Because it is going to be rotate around y=x the length is sqrt(2)*(xi+1-xi), then the volume will be pi*integral[0,1] d2(x) sqrt(2) dx. Solving this integral the result is pi/30sqrt(2)
Thank you, Mr. Penn. This video just taught me how to do "u substitution". My calculus class (in 1977) didn't get this far. Hey, I learned something today! Thanks again!
this can be trivialized by a change of basis, but it's nice to see this alternative method.
Also yeah, not sure what's up with calc teachers skipping this. My class didn't cover it, as far as I can remember.
Probably not that exciting, but this was definitely covered in my calc 2 course. I remember because it was one of the only problems I skipped on the final (not because I couldn't do it, but I recognized right away it would take too much of my time, because I wasn't as prepared for that problem as I was the rest of the material.)
At least one Mr. Turnbaugh in Bentonville, AR didn't skip it!
Spent about 3 weeks on it, and like 4 people in the class ended up being comfortable with it. But we did it.
0:03 Is it skipped because they don’t have enough time in the semester or becaue it’s too hard? 🤔
21:07 Good Place To Stop
I think a combination of both. In general, surfaces of revolution are kind of skimmed over. They just give the integral formula without giving it too much thought, which results in not knowing how to do these, since there's not a standard formula for this problem.
As someone who has taught college calculus a few times I can tell you that the semester tends to be pretty packed. This would work well in something like calculus BC in high school where there's a fair amount of time and you can do some neat projects, but if you tried this in a college calculus class (probably calc 2, but some people do solids of rotation at the end of calc 1 in a semester system; I don't remember where everything goes in quarters) you'd be pressed for time and the students wouldn't get as much out of it as you might want. This would make for a cool extra credit project, though, and if you were looking for an undergrad paper this would probably be an interesting one. It does kind of get trivialized by rotation matrices, though.
Another reason I would want to skip this is that I generally think of solids of rotation as a somewhat gentle introduction to applications of integration. They're easy to visualize - the slices for the Riemann sum are literal slices - rather than trying to "slice" time or some other abstract concept. It goes well alongside the classics like lifting objects with ropes of non-zero mass and pumping water out of various troughs and such. If you then make the algebra complicated it defeats the purpose of having a gentle introduction as the students get lost in the algebra and don't comprehend the cool stuff you're showing them.
@@skylardeslypere9909 It's funny, I never even remember the formulas myself because I always want to draw everything. I think my students want or expect the formula but I just keep drawing stuff; the formula is in the book, they can look it up if they want to. I'm just going to keep unrolling cylinders and stacking discs. I mean, who cares about the formula anyway? The whole point of having solids of revolution as part of the introduction to Riemann sums is to help gain intuition about what a Riemann sum really is by having a visualizable application, so why ever worry about the formula? That just skips all the visualizing and jumps to the answer, which defeats the purpose.
Also, I'm pretty sure (though I haven't done it) that you could derive a general formula for rotations about lines not square to the axes by using a rotation matrix to rotate your axes, do all the calculations in the new coordinate system, and then another to rotate them back. My guess is that you can probably just run the existing formulas for discs and shells through those rotation matrices to get the formulas straight away. You could probably also do it with standard algebra (like Michael did in the video), but that feels like it would be harder. So I strongly suspect there's a reasonably-derived standard formula, but again, why bother if the whole point is to think about the problem and how Riemann sums work?
Now that I think of it, this might make a good project in a linear algebra class when talking about either rotation matrices or change of basis, as this is a really good use of a change of basis matrix. I think I might steal this for exactly that purpose. Thanks for making me think on this topic Skylar, I think my students will really like this :-)
@@jgray2718 You mentioned that it gets trivialized by rotation matrices, and I was thinking something along those lines too while watching the video. Is it actually easier to find the volume of the surface of revolution around any line in the plane, if you apply a change of base? That is, translate your coordinate system so the origin is on that line, and then rotate it so it becomes one of the axes.
@@skylardeslypere9909 In general, yes it's probably easier. Applying a rotation matrix is easy, then you have a standard solid of revolution, which is also easy. The algebra Michael did seemed much harder to me. "Trivialized" was probably an overstatement on my part, but it's not difficult, just a little bit longer than the usual solid of revolution.
This particular problem was also chosen specifically to have "easy" algebra since it was rotated around y = x. I'd think a rotation matrix would be the way to go for sure if you wanted to rotate around y = 2x.
God help you if you want a general formula for rotating about y = ax + b, though. I'd have to consider how to do that, but I think you're into affine algebra rather than linear algebra at that point, so maybe the approach Michael took in the video would actually be easier in that case. I don't even know what an affine rotation matrix looks like, now that I think of it. Probably a slide then a spin?
I normally teach this topic for honors classes, specially the revolution about the line y = mx. Revolution about y=mx+b can be little more involved. Thank you for doing these.
How is revolution around y=mx+b more involved? You just move everything down by b and you're back at y=mx, don't you?
Well, technically, that one simple thing makes it a little more involved. But you know what I mean.
Why?
alternative solution: generate weight field which is 0 at the revolving axis and linearly goes to 1pi when it reaches distance 1 from the revolving axis, integrate over x. y equals the integral of the weight field beween the revolving axis and other function
Using Archimedes’ geometric proof of the parabolic quadrature, the relative height and width of each triangle is known and the position of each center can also be calculated.
So the centroid can be calculated as the weighted sum of the centers. Then take the foot/elevation, r, of the centroid from y=x. This is 1/(10*2^(1/2)). Multiply 2πr by 1/6 (per Pappus) an lid the volume obtains.
i need to write, that the perpendicular point xl,yl on y=x is the solution of 2 linear equations.
after this is done, r(x)=sqr((x-xl)^2+(y-yl)^2). then there is the
guldin rule
or
a for next loop to add parts with different r(x) together.
finally, a part of the volume is dv=pi*r(x)^2*dl
l=sqr(xs^2+ys^2) so dl=l/n (n=number after "to" in a for next loop)
I think I've done one similar to this before. The method I chose was rotate the curve pi/4, then use the normal method around the x-axis.
That how I would do it.
Without doing it, you would need to change the integration to 0 to sqrt(2) no?
Not entirely trivial to rotate the parabola. Not impossible. But I'm not sure it's easier than this video.
Let Sₖ be the set containing 2 and 5 and the first _k_ primes that end in 7. For example, S₃ = {2, 5, 7, 17, 37}.
Define a _k-Ruff_ number to be one that is not divisible by any element in Sₖ.
If Nₖ is the product of the numbers in Sₖ then define _F(k)_ to be the sum of all _k-Ruff_ numbers less than Nₖ that have the last digit 7. You are given _F(3)_ = 76101452.
Find _F(97),_ give your answer modulo 1,000,000,007.
hi! the difficulty would be greater if the axis y=x was not in the xy plane. but so, the rotated points
are, if xl and yl are perpendicular points:
xr=x+(xl-x)*cos(w)
yr=y+(yl-y)*cos(w)
zr=sqr((x-xl)^2+(y-yl)^2)*sin(w)
have fun using ba$ic instead of calculus!
The distance from a point (x,x^2) up to the point (x,x) is L=x^2-x. The shortest distance from (x,x^2) to the line x=y is thus r=(x^2-x)/sqrt(2) since there is a 45-45-90 triangle involved. Rotating the vertical line segment from (x,x^2) to (x,x) around the x=y line gives a cone with slant height L and radius r. The volume can then be found from the surface area of a cone: pi r L, i.e.
\int_0^1 pi /sqrt(2)* (x-x^2)^2 = pi/ (30 *sqrt(2)). Disks, shells, washers... and cones!
Genius
Great idea!
Only the distance between the point of the parabola (x,x^2) and the axis of rotation y=x is equal to
L=(x-x^2)/√2, 0≤ x≤ 1. When the parabola rotates, this segment forms a cross-section of the figure perpendicular to the axis of rotation - a circle with an area of π L^2.
If dl is an element of the axis of the geometric body , then dl=√2dx.
V=∫(from 0 to 1)π*[(x-x^2)^2/2]√2dx=π√2/60.
We calculate the volume of the geometric body as the sum of the volumes of elementary cylinders of differentially small height. In this sense, the cone has nothing to do with it.
@@Vladimir_Pavlov You are right of course, r=(x-x^2)/sqrt(2), since (x^2-x)/sqrt(2) is negative. I think what you call L was r for me. Splitting the volume up into disks as you suggest is very good and simpler. I do like the idea of splitting it into cones though!
It should be noted that this approach to solving the problem makes it easy to get a result if the axis of rotation is an arbitrary straight line passing through the origin, y=tga *x, a≠π/2.)
I used a series of cone shaped shells each having slant height of x-x^2 and radius sqrt(2)/2 of that. Integrated 0 to 1 dx. I got the same answer. First I started the same way as in the video but stopped short of calculating that diagonal distance. Then I tried rotating axes but couldn't figure that bit out.
Same, although I think the cones have the same radius and height of r(x)=(x-x^2)/sqrt(2), we want the shells so I did Vshell(x)=pi/3*r(x)^3-pi/3*(r(x)-sqrt(2)dx)^3 I.e. volume outer cone minus volume of inner cone. Note you need sqrt(2)dx because the cone height lies on the diagonal line rather than aligned with the x axis. After cancelling and dropping terms with dx^2 or dx^3, you get Vshell(x)=pi/sqrt(2)(x-x^2)^2 dx and when you integrate this from 0 to 1 you get pi/(30*sqrt(2))
8:42 the y coordinate should be =1/2* (4Xi+1+....
That's why the magic happens at 13:00...
Classic michael...
The right result shows that you made an even number of errors 😉
I used a slightly different approach, I set P(x) as the projection of the point A = (x, f(x)) onto the axis (the line x=y).
P(x) can be found as P(x)=(t,t) setting t=x-c=f(x)+c. This solves as c= ½(x-x²), so that
P(x)= (½(x+x²), ½(x+x²)).
Now let s(x) be the distance of P(x) from the origin: s(x) = ½√2 (x+x²),
and let r(x) be the distance from P to A:
A-P = (x- ½(x+x²) , x²- ½(x+x²)) = (½(x-x²) , ½(x²-x)) and r(x)=|A-P|= ½√2(x-x²).
We now find ∫ πr²ds = ∫ πr² • ds/dx dx = ∫ π ½(x-x²)² • ½√2 (1+2x) dx = ¼π√2 ∫ (x-x²)² •(1+2x) dx
= ¼π√2 ∫ x^2 -3x^4 + 2x^5 dx = ¼π√2 [1/3 x^3 - 3/5 x^5 + 1/3 x^6]
Because x runs from 0 to 1, the last expression must be evaluated at the boundaries 0 and 1 this gives a volume of π√2/60
If you just subtract x from y = x^2 then you the object bounded above by the axis. This is equivalent and EASIER than the rotation matrix. This problem is simply rotation y = x^2 - x about the x-axis between 0 and 1.
If for a parabola y=x^2 the axis of rotation is a straight line y=tga *x (a ≠π /2), then the required
volume of the body of rotation will be equal to V=(π/30) *cosa *(tga)^5.
At a = 45 degrees, we get an answer for a this case.
I applied a different method, in principle it should be more general. Let us assume to have a straight line t, with equation y=mx+q in the plane, and a function y=f(x) to rotate w.r.t. to the direction defined by the straight line by an angle of 2pi. Then, if we take a general point P=(s,f(s)) on the graph of the function f, the distance between point P and the straight line is given by the classical formula d(P,t)=|ms+q-f(s)|/sqrt(1+m^2). So the volume of the rotation of the graph of bounded by the interval [a,b] is simply the sum of all the circles of radius d(P,t) all along the interval [a,b]. Obviously the sum becomes the following integral
\int_a^b \pi*d(P(s),t)^2ds
In the case proposed in this video, a=0, b=1, f(x)=x^2, m=1, q=0 and the integral becomes
\int_0^1 \pi/2*(s-s^2)^2ds=\pi/2*(x^3/3-x^4/2+x^5/5)|_0^1=\pi/2*(1/3-1/2+1/5)=\pi/2*1/30=\pi/60
The result is not the same, I do not have the square root of 2 but just 2 at the denominator. I hope you enjoy this solution.
Very cool approach and easier than in the video. I would like to point out that until the line "sum becomes the following integral" everything is correct. The problem comes after that. If u take an area of πr² then u need to multiply the differential element perpendicular to the given area. So multiplying "ds" simply gives the wrong solution. Differential element should be along the line ms+q acc. to notations u have taken which then gives the differential element as ds.√(1+m²). (From pythagoras theorem directly; taking a triangle of ds base and dy =mds as altitude)
The entire integral then becomes
Integration from 0 to 1[π(s-s²)²/√2 ds]
I think this will give the correct value now.
@@Goku_is_my_idol Yes, you're right, very smart fix. Thank you
12:48 Actually the magic easily derives from the fact that the slope of the line QP being -1 has the absolute value 1, doesn't it?
Instead of relying on that magic, you can avoid the sqrt altogether by not solving for x in terms of xi. Instead turn x^2+x=2xi into:
xi=x(x+1)/2
dxi=(x+1/2)dx
and substitute accordingly so x is integration variable (and bounds remain at 0 and 1).
After much reduction, end up with:
sqrt(2)pi/4 integral from 0 to 1 dx [2x^5 - 3x^4 + x^2]
which is very easy to integrate.
To clarify, the "much reduction" part involves many but easy reductions since we are simplifying sums and multiplications of small polynomials (no radicals).
rotating gives y=x^2 --> f(x)=x+(1-sqrt(1+4x*sqrt(2)))/sqrt(2), now calc pi*integral (f^2)dx
Oh man, I JUST so happened to derive the formula for a 45º rotated parabola a year ago. I never would've thought that would come in handy
When I took AP Calc in High School, we were supposed to come up with problems and put together our own "final examn".
My group put this problem on it and our teacher wasn't able to solve it 😉
I covered this once. The class had a very hard time following.
This situation happened to me on accident when our prof made an error in the homework. Me and another guy couldn't solve it so we asked, but he just fixed the error instead of telling us how to solve it.
Both r and w are obviously hypotenuses in right, isosceles triangles (you already established their slopes of 1 and -1). That would save a lot of writing. Just take one of the legs, multiply by √2 and call it a day.
Going about it the hard way. Set the slope as your x value and calculate the area under the curve. Convoluted methods take too much time. Good exercise for fun.
"I think you can work out a general formula for this.." Yep, and it's really easy: the volume is the integral of the cross sectional area d h. Go forth and calculate. :)
i just got the parametric form of the rotated function and used the parametric version of the volume integral from 0 to 1 :D
Rotation of coordinate systems was covered on one my senior year math courses leading to my electrical engineering degree. Course was called "Multidimensional Calculus". The material in the course I found to be quite helpful in my employment for a defense contractor in missile and weapons systems targeting/engagement. (Been retired for over 30 years, and still remember some of that stuff)
You will be hard pressed to find a better mathematician than an EE grad. I earned my BSEE studying computer chip design. Graduated in the tech bubble burst. No jobs anywhere, world-wide. Ended up doing statistics for the insurance industry.
@@billyoung8118 A better mathematician than an EE grad is pretty easy to find if you've ever heard of a 'mathematician'.
Hi Michael, there is an easier way to solve this: if OP represents the segment from the origin to P(x,y=x^2), and t is the angle between OP and x-axis, then radius of revolution is r=OP*sin(pi/4-t)=sqrt(2)/2*OP*(cost-sint)=sqrt(2)/2(x-y). Then V=integral_0^1 pi*[(sqrt(2)/2(x-x^2))]^2*sqrt(2)dx = pi*sqrt(2)/60
This is actually the standard situation (covered for example in section 5.1 in Stewart's Calculus) of a solid whose volume is computed as the integral of the cross-section area.
Volumes of solids obtained by rotating a region bounded by graphs of functions about the x-axis (V=\int_a^b \pi f(x)^2 dx) is a special case of it.
No! The better example for method is in page 588 ( chapter 9 ) 4th Edition. (Discovery Project). Adrián Irias
@@adrianignacioirias I don't have the 4th edition but I guess you are talking about the Discover project "Rotation about a slant line" in the chapter about Applications of Integration. You don't need this for the example of the video. And, if you read the whole chapter, you'd better use the Theorem of Pappus to compute the volume.
Convoluted but elegant. Around 11.20 mark there is a slight error in computation related to W.
My calc teacher did this exact problem as an extra-curricular example, but we weren't tested on it.
I think you could rotate the two curves '-45 deg' and evaluate the regular way -- effectively, this is what resulted.
I see comments regarding forward rotation of the axes. … I am not impressed by the counterintuitive nature of that idea.
Rotate the curves instead as you suggest. Det(J) or det(g) =1 so there is no need to do any heroic math. The standard parametric statement of the volume integral definition includes the derivative of x with respect to a parameter t. y’^2(dx’/dx)dx can all be expressed in terms of the original variable x, so the bounds of integration are unchanged.
I always love pausing the video and trying this out first.
Nice! Never did this before! Thanks!
You could also use Pappus, you only need to find the area of the curve, his center of gravity and the distince of that center to the axis of rotation (y=x in this case). I had these problem (but with a different curve) in my Calc 2 final exam
I am sure there are others who have commented that the rotation of axis method would be simpler to implement, but this is also a good video to demonstrate the use of the Riemann Sum. I wonder how this can be solved using Lebesgue Integral...
I used Lebesgue integration, and it felt much neater to me!
It worked by splitting the volume into hollow cylinders with the central axis being y=x. The radius r cylinder would have width dr and volume 2 pi r len(r) dr, where len(r) is the length of the cylinder of radius r.
To find len(r), we just need to find the points such that y = x^2 and y = x - r/sqrt(2) (since that is the parallel line to y=x at distance r from it). The x values of the points are the solutions to x^2 - x + r/sqrt(2) = 0. The difference between the x values is sqrt(b^2 - 4ac), which in this case is sqrt(1 - 2sqrt(2)r). The distance between the points is then sqrt(2)sqrt(1-2sqrt(2)r) (because the slope of the line is 1).
So our answer is the integral of 2sqrt(2) pi r sqrt(1-2sqrt(2)r) dr, where r goes from 0 to... 1/2sqrt(2), since that's where len(r) becomes 0.
Substitute t=1-2sqrt(2)r, so r = (1-t)/2sqrt(2) and dr = -dt/2sqrt(2).
Now we're solving integral of 2sqrt(2) pi (1-t)/2sqrt(2) sqrt(t) (-dt/2sqrt(2)), which is:
-pi/2sqrt(2) times integral of sqrt(t)-tsqrt(t) dt, where t goes from 1 to 0.
The final integral to solve is super easy, and you get -pi/2sqrt(2) times [2/5 - 2/3], which is 2pi/15sqrt(2), which is different from the answer given in the video. :o I have no idea where I went wrong. I had a silly calculation mistake in my first attempt and it magically matched the video answer with that mistake.
What about taking a surface of revolution around a curve? The radii of the circles would be the perpendicular distance between the two curves. Maybe that's a silly thing to do, but it would make for an interesting problem.
I solved this way differently (desmos demo: desmos com/calculator/fci6ffmtdx) by parametrizing the distance between the two functions with a line orthogonal two x=y.
I will call the intercepts k_i. First I considered that the function for the orthogonal line must be:
-x + 2x_0 = y
Then I found the intercepts with the two other functions at x_0. For x=y its almost by definition:
k_1 = (x_0, x_0)
and for x^2=y its a little more complicated, but basically you can solve the equation -x + 2x_0 = x^2 and you get:
k_2 = ((sqrt(1+8*x)-1)/2,((sqrt(1+8*x)-1)/2)^2)
Which looks a bit terrifying, but doesn't cause any major problems down the way. Then I considered that the radius of the volume, r, at x_0 must be given by the function:
r(x) = ||k_1 - k_2||
Where we are a bit lucky because actually both terms in the norm are equal (not intuitive in my opinion). However that simplifies the equation A LOT:
r(x) = sqrt(2) * ((sqrt(1+8x)-1)/2 - x)
Which we can use to calculate the area of a circle-slice:
c(x) = r(x)^2 * pi = (sqrt(2) * ((sqrt(1+8x)-1)/2 - x))^2 * pi = 2 * pi * ((sqrt(1+8x)-1)/2 - x)^2
Then we can construct a integral using the formula for the area of a circle-slice to calculate the volume:
V = int_0^1{c(x) dx} = int_0^1{2pi * ((sqrt(1+8x)-1)/2 - x) ^2 dx} = pi/60 ~ 0.52
So, I must say I am pretty confused. Weirdly this is "off" from your result by a factor of exactly sqrt(2), so that is probably not a coincidence, however I for the life of me cannot fathom it right now.
Man, I did this a much simpler way. Rather than finding the distance from the point (x_i,x_i) to the corresponding point on the parabola, I found the distance from the the point (x_i, x_i^2) to the corresponding point on the line. This quickly gave pi sqrt 2 ∫ ((x - x^2)/sqrt 2)^2 dx
I did the same, but differentiated with respect to t.
Like you, I get
w = √2 dx
But when solving for slope from (x, x) to (t, t²) = −1, I solve for x in terms of t instead of t in terms of x
(x−t²)/(x−t) = −1
xᵢ− t² = −x + t
2x = t² + t
x = 1/2 (t² + t)
Differentiting both sides we get:
dx = 1/2 (2t + 1) dt
Note that points (t, t²) go from (0,0) to (1,1), so t is between 0 and 1
Now we calculare r², where r is distance between (x, x) = (1/2 (t² + t), 1/2 (t² + t)) and (t, t²)
r² = [1/2 (t² + t) − t]² + [1/2 (t² + t) − t²]²
= [1/2 (t² − t)]² + [1/2 (t − t²)]²
= 1/4 (t² − t)² + 1/4 (t − t²)²
= 1/2 (t⁴ − 2t³ + t²)
Now we can calculate volume of a disk with radius r and width w
v = πr²w = π · 1/2 (t⁴ − 2t³ + t²) · √2 dx
= π · 1/2 (t⁴ − 2t³ + t²) · √2/2 (2t + 1) dt
Integrating over all disks, we get volume of rotation
V = π/(2√2) ∫ [0, 1] (t⁴ − 2t³ + t²) (2t + 1) dt
= π/(2√2) ∫ [0, 1] (2t⁵ − 3t⁴ + t²) dt
= π/(2√2) (t⁶/3 − 3t⁵/5 + t³/3) | [0, 1]
= π/(2√2) (1/3 − 3/5 + 1/3)
= π/(2√2) (1/15)
V = π/(30√2)
Could you not have simply subtracted x from x^2 (or vise versa) and done a horizontal rotation on that? (of course that rotation would be from 0 to sqrt(2) since that's the length of the line)
Had a lot more fun finding the length of the line perpendicular to the rotation axis between its intersections with the curves, and getting area with pi r^2 (squaring the square root), and integrating the resulting simple polynomial over x.
Yes i solved it this way,it was nice
Me too, check out my solve!
Got to the same result using the integral of surface area of cones
General case with that method gives V=pi/sqrt(m^2+1) * integral[(mx+q-f(x))^2]
That's a revolutionary problem
im guessing starting out by rotating the picture so that y=x is flat would have made most of the steps turn out simpler, but it would have been harder to get info about the transformed parabola. but i guess having a rotated parabola defining the curve is better than having a rotating axis defining the solid, and probably involves a lot less writing. maybe thats why they dont do slanted examples, because you can transform a slanted example into a standard one at the start, albeit it would be good to know how that process plays out.
"Gnarly equations with a side of Pappus to go!"
"Coming right up!"
The area of the depicted figure is A = \int_0^1 x dx - \int_0^1 x^2 dx = 1/6
A rectangle with width a and length sqrt(2) with the same area: a * sqrt(2) = 1/6 => a = 1/(6*sqrt(2))
Rotating this rectangle gives the solution: pi*a^2*sqrt(2) = pi * sqrt(2)/72 =pi/(36*sqrt(2))
However Michael obtained pi/(30*sqrt(2)). Can someone point me to my error?
I was thinking all the time what may be wrong with such an easy approach and I found the explanation:
Rotating figures with the same cross-section do not lead to bodies with the same volumina, e.g.
a rectangle A_r=a*h_r has the same area as a right-triangle A_t=½*a*h_t if h_r=½*h_t. Now I introduce r=h_t that the formulas are looking more familiar: A_r=a*½*r, A_t=½*a*r.
If I rotate the rectangle around a, I obtain a cylinder with V_cy=¼*pi*a*r^2 and if I rotate the right triangle around a, I obtain a cone with V_co=1/3*pi*a*r^2
Math arguments should not be interrupted by ads. An ad at the beginning is enough for youtube and Mr Penn to be well compensated. We taxpayers, who paid for the development of the interned receive no royalties at all for the risk of our taxes. We should not have to tolerate the greed of youtube. The is part of the corruption of our government that permits such greed.
Teachers 'skip' this because it's obviously just a co-ordinate change away from regular revolution.
What would be more interesting is if you rotated every point on the 'slanted' curve around the X axis, but the centre of rotation was given by the other curve.
That's because they don't throw the prolate sphere aka football. I was a post-doc in physics at UPenn and have also taught calculus (my favorite was teaching Calculus-3).
Honestly an easier way is to use pappus's Centroid theorem.
Yes, but that's not something a second semester calculus student would have access to when first studying solids of revolution
@@williamwolfs4819 It's a shame because Pappus' centroid theorem is easily understood earlier, and it is FAR more general than solids of revolution so restricting it to that is a disservice to the theorem.
@@1ucasvb can you explain to me how to use it in this specific problem?
21:06 My German math teacher in those situations 20 years ago:
*Deduction of points!! Pi/(30sqrt2) what? Potatoes? Pancakes? It's "volume units"!*
(which very often downgraded me from a "1" ("very good", "A") to a "2" ("good", "B+").
Different times though. Today, grading in German secondary schools is less strict and less pedantic.
Lebesgue measure is inavariant under the action of isomtries, so I would rotate around e_1 instead
this was always left as an exercise to the readers in our courses.
mmmm imagine as an extra advanced problem if the line being rotated around was not one of the boundaries (i.e. not y = x ) , but it would have to be parallel to y = x to stop it intersecting the region.
Consider rotating around y = x + 1 ( or y = x + c for c not equal to 0 ).
mmm actually, it doesn't even need to be constrained to have the same slope as y = x. It just needs to be such that it does not intersect the region between y = x and y = x^2.
I think coordinate transformation is the easiest way to deal with this...but sure, there are other methods too.
problem would have been much easier if you used the inverse, and did things in terms of yi, yi+1 instead of xi, xi+1 (horizontal rieman)
Small mistake 8:40 it should be 4xi in the y coordinate of the blue point, not 2xi.
00:06 That skipping must have started recently. 🤔I did many of these from the Swokowski (?) textbook in 1993 at LSU. And the book was purple. 😁👍🏻 (I also think that I have taught these problems in a class or two, but that was many moons ago.)
You might be complicating stuff, just rotate everything to the y=0 axis and integrate as usual, notice that this is a rotation by 45° and we know the matrix for that
Of course I am supposing you can use the formula for volumes generated by a curve rotated about the y=0 axis, otherwise just use the same logic to prove it
11:10 Small mistake: sqrt(2 * delta x) does not equal sqrt(2) * delta x
It was delta x squared inside the square root, he just forgot
Your definition was that x0 = 0 ... xn = 1, but your sum had index i = 1 ... n
Very informative... Amazing content ... Tnx prof...
Wouldn't it be easier to rotate everything by -45 degrees, i.e. go from (x, y) to (u, v) where u = (x+y)/sqrt(2), v = (x-y)/sqrt(2) (maybe need to change some plus to minus or vice versa), express the line y = x^2 in terms of u and v, get it in a form v = f(u) on the interval from 0 to 1, and then calculate ∫πf(u)^2 du from 0 to 1?
When rotating the coordinate system by 45 degrees , it is necessary to calculate
π*∫[from 0 to√2 ]{(u+1/√2)-√(√8 u+1/2)}^2 du =√2/60
When the coordinate system rotates by (-45)degrees will have to calculate the same integral.
Note that you will have to integrate from 0 to √2.
@@Vladimir_Pavlov right, thank you
Have you tried rotating the axes? Maybe you get a complicated equation for the parabola.
Because it just standard revolution volume with extra geometric steps. ;)
I would have partitioned the solid into cones and integrated over the lateral areas of all the cones
I did learn this in calc 2...and the exact example.
Substitute teachers:
I don’t like to substitute u
It looks like the radius formula has wrong sign: it is negative if x = 1/2.
Math is the best, when well explained like this.
I've thought about this and realized it would not be too hard to find the center of mass, then use Pappus's theorem.
Couldn't you also take the area of the 45/45/90 triangle between 0 and f(x) = g(x), and then just integrate y=x^2 within the same bounds? You'd Have to fiddle with it a bit; like you might have to take the complement of the integral within the quadrant before you rotated it around the linear function, but i'm just curious. It's too late for me to tackle right now the full "slanted revolution" technique: i'm sleepy, but i don"t see why my kludge wouldn"t work in this case at least.
Le volume étant simplement connexe et la symétrie xy, pourquoi s'embêter à prendre des rectangles d'intégration penchés, les verticaux auraient aussi bien fonctionnés dans ce cas précis.
Un exemple plus compliqué avec un sinus "rapide" borné par un sinus x lent" par rapport à y=1/2x aurait montré la généralité de la méthode présentée
At 3:09 the interval [0,1] is sudivided into n+1 subintervals, not n subintervals. Maybe he had in mind to dtart with x1 as his first point, not x0. Anyway is is an error.
True in my case just completed my graduation never in my calculus class teacher used slanted evolution idea to explain a topic
4:25 aw man you're doing it the hard way
y = x - x^2 with rotation => simpler solve..
Cool derivation. It was a bit frustrating repeatedly hearing "disc" instead of "cylinder".