Note that the result is equal to area of circle of radius a (pi × a^2) multiplied by the length of the torus centerline (2 × pi × b). Engineers dealing with piping know that the volume of a bent pipe equals the volume of the straight pipe you started with if the centerline length is unchanged, implying that the bending process has to stretch the extreme max radius of the bend and compress the min radius. This integration demonstrates that fact.
16:14 pi^2 instead of pi / good for exercising triple integrals, however path is more direct by integrating an elementary volume of (pi * a ^ 2 * b * d_theta) with theta between 0 and 2*pi
This is a pedagogic exercise in integral calculus. If we were interested more in the result than the method, we could recall the Pappus and Guldinas Rules which allow easy calculations of surface area and volume of solids. The rules are easily applied to solids of revolution, of which the torus is a classic example. Volume is obtained as the product of the area of the rotating plane and the length of the distance travelled by the centroid of that area. The surface area is the product of the length of the perimeter of the rotating plane and the distance travelled by its centroid. [ Try calculating the surface area of a torus using calculus for a bit of extra fun. Alternatively find the shape of intersections of a torus and various planes. Some of these are counterintuitive ! ]
Making the life simpler: this is the volume of a cylinder where the height is equal to the length of the circumference of radius b and the area of the base is the circle of radius a. No need calculus tondo that. :)
My math. instinct told me: Damn, that's easy to solve without a 3 x - integral, without calculus at all. The volume of the torus is just the slice area (pi x a^2) times the circumference of the outer b-circle (2pi x b), in other words: A(torus) = 2pi b x pi x a^2 = 2 pi^2 b a^2. What I did - is just to consider the torus geometrically as a "circle curved" cylinder.
I solved it by using a "toral" coordinate system that invented just right now, and I'm sure I'm not the first to do that. In my coordinate system phi defines a cross section of the torus, and on that cross section, the location of each point is defined by a polar coordinate system (r, theta). Take dA = dr (r d_theta) on the cross section. dV will be equal to dA * R d_phi, where R is the distance between the desired point and the z axis, define by R = b - r cos(theta). Now, the rest is easy. The volume is the triple integral of dV over r from 0 to 1, and theta and phi from 0 to 2pi. The integral of the cos(theta) term will be zero, which simplifies the problem a lot. The final answer is essentially 2pi * (2pi * b a^2 /2) = 2 pi^2 b a^2. BTW, there's a typo at 16:10. it should be pi^2.
A bit off base but: do you think that there is something deep(er) hidden in the torus/toroidal structure? I had a strong vision once, and gut feeling there is, I know for a fact there are some deep established relationships of it in mathematics and physics but I think its only scratching the surface. For instance it is interesting it is the shape of a tokamak and other magnetic confiment/phenomena. Or maybe i'm seeing things.. . . just have a feeling there is some new insight to unlock
Can’t we just use the washer method and create a circle with the radius we want, distance from the origin where point a is the closest to the origin on the x axis, and then make a point b being furthest from the x axis, then we use the formula (x-h)^2+(y-k)^2 = r^2, solve for x so we get the circle in terms of only y and then integrate
You actually don't even need calculus for this. Just take the torus and cut it on one of the circles of radius a. Then you can stretch it out to just be a cylinder of height 2pi*b (the circumference of the bigger circle) and radius a. Then the volume of the cylinder is pi*a^2(2pi*b) = 2pi^2*a^2b. Still a cool exercise to do with multivariable calculus, though.
What you say happens to be true, but without calculus or geometrical proof, how could you merely assume that you can take circle area times centerline length? When revolving circle around z-axis the outer part of circle sweepes out excess area while inner part sweeps out less. These exactly cancel out, but it takes a calculus or geometrical proof to prove it.
As Robert Pelini pointed out, you cannot assume that you do not change the volume at bending. An intuitive proof may look like: take the torus from above (view down the z-axis), cut the torus "like a cake" in n pieces, rearrange all pieces to a pile by turning every second piece by 180 degree (swap big and small side of the "cake piece") and put it on top of the last piece => for large n this is a good approximation of a cylinder with r=a and height=(2Pi)b="the circle r=b". Note, that is the torus not "bend" but "rearranged" to a cylinder, only the outer shell and the center are "snake-ish" but that's not an issue for n -> inf. "... and that's a good place to stop."
ye but lets say you wanna know how much bigger the outer half of the donut is than the inner. Then that'd be pretty gangsta to pull up in a sec with integration!
I sliced the bagel from top to bottom and got no clue whatcha talking which means theres waiting tons of fun stuff yet to be explored. guess thats the beauty of math:)
Note that the result is equal to area of circle of radius a (pi × a^2) multiplied by the length of the torus centerline (2 × pi × b). Engineers dealing with piping know that the volume of a bent pipe equals the volume of the straight pipe you started with if the centerline length is unchanged, implying that the bending process has to stretch the extreme max radius of the bend and compress the min radius. This integration demonstrates that fact.
16:14 pi^2 instead of pi /
good for exercising triple integrals, however path is more direct by integrating an elementary volume of (pi * a ^ 2 * b * d_theta) with theta between 0 and 2*pi
This is a pedagogic exercise in integral calculus.
If we were interested more in the result than the method, we could recall the Pappus and Guldinas Rules which allow easy calculations of surface area and volume of solids. The rules are easily applied to solids of revolution, of which the torus is a classic example. Volume is obtained as the product of the area of the rotating plane and the length of the distance travelled by the centroid of that area. The surface area is the product of the length of the perimeter of the rotating plane and the distance travelled by its centroid. [ Try calculating the surface area of a torus using calculus for a bit of extra fun. Alternatively find the shape of intersections of a torus and various planes. Some of these are counterintuitive ! ]
Thank you so much, it is very useful for my project at my university!!
THIS VIDEO SO USEFUL FINALY SOMEONE HAS ANSWERS I NEED YESSSSS U ARE SO EPIC MATH MAN I CAN CALCULATE VOLUME OF MODIFIED TORUS NOW
i couldn't use solid of revolution cause it was unequal but ur method is epic and does not require that
I once heard of a rule that the volume is the area times the length of the way of the center of gravity.
That gives: pi a a 2 pi b.
I think this is easier by considering a circle of radius a centrred on (b,0) and revolving it around the y-axis, that way you only need one integral.
Thank you buff math teacher guy I love you.
i learned it as a rotational body, but great to see your calculation
Making the life simpler: this is the volume of a cylinder where the height is equal to the length of the circumference of radius b and the area of the base is the circle of radius a.
No need calculus tondo that. :)
My math. instinct told me: Damn, that's easy to solve without a 3 x - integral, without calculus at all.
The volume of the torus is just the slice area (pi x a^2) times the circumference of the outer b-circle (2pi x b),
in other words:
A(torus) = 2pi b x pi x a^2 =
2 pi^2 b a^2.
What I did - is just to consider the torus geometrically as a "circle curved" cylinder.
I solved it by using a "toral" coordinate system that invented just right now, and I'm sure I'm not the first to do that. In my coordinate system phi defines a cross section of the torus, and on that cross section, the location of each point is defined by a polar coordinate system (r, theta). Take dA = dr (r d_theta) on the cross section. dV will be equal to dA * R d_phi, where R is the distance between the desired point and the z axis, define by R = b - r cos(theta). Now, the rest is easy. The volume is the triple integral of dV over r from 0 to 1, and theta and phi from 0 to 2pi. The integral of the cos(theta) term will be zero, which simplifies the problem a lot. The final answer is essentially 2pi * (2pi * b a^2 /2) = 2 pi^2 b a^2. BTW, there's a typo at 16:10. it should be pi^2.
A bit off base but: do you think that there is something deep(er) hidden in the torus/toroidal structure? I had a strong vision once, and gut feeling there is, I know for a fact there are some deep established relationships of it in mathematics and physics but I think its only scratching the surface. For instance it is interesting it is the shape of a tokamak and other magnetic confiment/phenomena. Or maybe i'm seeing things.. . . just have a feeling there is some new insight to unlock
or you think in a thorus as a cylinder where the hight is the circunference with radius b and the base is the circle with radius a
Can’t we just use the washer method and create a circle with the radius we want, distance from the origin where point a is the closest to the origin on the x axis, and then make a point b being furthest from the x axis, then we use the formula (x-h)^2+(y-k)^2 = r^2, solve for x so we get the circle in terms of only y and then integrate
You actually don't even need calculus for this. Just take the torus and cut it on one of the circles of radius a. Then you can stretch it out to just be a cylinder of height 2pi*b (the circumference of the bigger circle) and radius a. Then the volume of the cylinder is pi*a^2(2pi*b) = 2pi^2*a^2b.
Still a cool exercise to do with multivariable calculus, though.
What I was thinking
This is known as the pappus guldinus theorem
What you say happens to be true, but without calculus or geometrical proof, how could you merely assume that you can take circle area times centerline length? When revolving circle around z-axis the outer part of circle sweepes out excess area while inner part sweeps out less. These exactly cancel out, but it takes a calculus or geometrical proof to prove it.
As Robert Pelini pointed out, you cannot assume that you do not change the volume at bending.
An intuitive proof may look like: take the torus from above (view down the z-axis), cut the torus "like a cake" in n pieces, rearrange all pieces to a pile by turning every second piece by 180 degree (swap big and small side of the "cake piece") and put it on top of the last piece => for large n this is a good approximation of a cylinder with r=a and height=(2Pi)b="the circle r=b". Note, that is the torus not "bend" but "rearranged" to a cylinder, only the outer shell and the center are "snake-ish" but that's not an issue for n -> inf.
"... and that's a good place to stop."
ye but lets say you wanna know how much bigger the outer half of the donut is than the inner. Then that'd be pretty gangsta to pull up in a sec with integration!
What is the function r supposed to be, or what can it be?
Shouldn’t it be 2*pi^2*a^2*b ?
2pi b a^2, right?
The answer is 2pi^2(a^2b)
"Penn Pineapple, Michael Penn."
okay, good. Nice job
genius
yeah
I sliced the bagel from top to bottom and got no clue whatcha talking which means theres waiting tons of fun stuff yet to be explored. guess thats the beauty of math:)
Pi power 2 not Pi
But what's the temperature of a torus given the current price of bananas?
It took me a month to calculate it, but it's blue.
12:48 YOU WILL BE EQUAL TO A...to a...?...to a donut?
Next up: The surface area of a torus
probably could use cylindric idea
Wouldn't that be 2 pi r x 2 pi R?
What
Man’s got some pythons