As always, yet another EXCELLENT presentation simplifying a complex topic ... I've been struggling to get my mind around surface integrals, and this is one of the best videos on this subject ... Calculating the surface area of a sphere is big bonus ...
I did a simpler method. I used the surface area integral 2pi integral(r(x) * sqrt(1 + derivative(x)^2) from -r to r. The equation of a circle is x^2 + y^2 = r^2. In terms of x it is y = sqrt(r^2 - x^2) .When you do the derivative and the algebra, the sqrt(r^2 - x^2) cancels, leaving you with sqrt(r^2). Since sqrt(r^2) is a constant r, put it in the front so you get 2pir(integral(1)dx.) from -r to r. Then you solve the integral which is 2pir(x) from r to -r. Plugging in r and -r it is 2pi(r)^2 - (- 2pi(r)^2), add both and you get 4pi(r)^2
I would like to ask a complicated question with a possibly trivial solution. What if you cut up a convex 3d volume by cutting up the surface and slicing to the origin (convex here being defined as "the surface crosses each Ray passing through the origin exactly once") and sorted the infinitesimally small pyramids in order from shortest around -x and longest around +x? Are there any guarantees about this shape? I don't think it would, but I also wouldn't be too surprised if it had to be a spheroid
@@drpeyam the same way the limit of the Riemann sum is the area I think. In all fairness just taking the derivative doesn't work but the principle of adding differential shell volumes to get the total volume remains the same I believe
@@drpeyam It's interesting, even though I don't have knowledge on subjects "above" calculus 2, can you do a video proving that the derivative of the "volume" of a n-hypersphere(n is the dimension) is always its "surface area" So in a in a short future I can enjoy the proof? (:
Yes, you can. That's the usual and simpler way to find the surface area of a solid of revolution. However, the purpose of the video was to show a method for computing the surface of *any* kind function, not only the symmetric ones. He applied it to a case where you can find the answer via both methods
I really love your videos about different kinds of integrals! Thank you Peyam
As always, yet another EXCELLENT presentation simplifying a complex topic ... I've been struggling to get my mind around surface integrals, and this is one of the best videos on this subject ... Calculating the surface area of a sphere is big bonus ...
My new life motto:
@5:02 We're not scared - We're mathematicians!
Wow, seeing that circle surface area derived logically is crazy! This is some powerful mind blowing stuff
I don't understand these videos yet, but I know I'm going to love them some day
Can you please make videos on Change of Basis? I find this topic very confusing, and your amazing at explaining? (i have an exam next week )
I remember doing this exact problem in Calc. 3, man what a time.
I did a simpler method. I used the surface area integral 2pi integral(r(x) * sqrt(1 + derivative(x)^2) from -r to r. The equation of a circle is x^2 + y^2 = r^2. In terms of x it is y = sqrt(r^2 - x^2) .When you do the derivative and the algebra, the sqrt(r^2 - x^2) cancels, leaving you with sqrt(r^2). Since sqrt(r^2) is a constant r, put it in the front so you get 2pir(integral(1)dx.) from -r to r. Then you solve the integral which is 2pir(x) from r to -r. Plugging in r and -r it is 2pi(r)^2 - (- 2pi(r)^2), add both and you get 4pi(r)^2
I would like to ask a complicated question with a possibly trivial solution. What if you cut up a convex 3d volume by cutting up the surface and slicing to the origin (convex here being defined as "the surface crosses each Ray passing through the origin exactly once") and sorted the infinitesimally small pyramids in order from shortest around -x and longest around +x? Are there any guarantees about this shape? I don't think it would, but I also wouldn't be too surprised if it had to be a spheroid
If I understood your question, that's the idea behind finding the volume in polar coordinates
Brilliant! Your channel is better than bprp!
Me: ... (loading)
Me: Prof. Steve, come here! I have something to show you!
I would like you to do a derivation of the spherical coordinate equations.
Coming in 10 days
Thank uou sir
Take the derivative of the volume :O
Yeah, but how do you know that’s true?
@@drpeyam the same way the limit of the Riemann sum is the area I think. In all fairness just taking the derivative doesn't work but the principle of adding differential shell volumes to get the total volume remains the same I believe
I know it’s intuitively true, but we want to show it’s rigorously true! In any case, there will be a video on that!
@@drpeyam It's interesting, even though I don't have knowledge on subjects "above" calculus 2, can you do a video proving that the derivative of the "volume" of a n-hypersphere(n is the dimension) is always its "surface area"
So in a in a short future I can enjoy the proof? (:
Absolutely, it’s on my to-do list! There’s an elegant proof for that
15:37 lmao
Why not take some common factors out of the determinant first, then you can find it much less pain and, when you find the magnitude, it's much easier?
Haha, 2 reasons: 1) It takes the magic out of it, and 2) Usually multivariable students haven’t taken linear algebra, so they don’t know that property
Can't you just use arc length and rotate it 360?
You mean around 2pi radians?
Yes, you can. That's the usual and simpler way to find the surface area of a solid of revolution. However, the purpose of the video was to show a method for computing the surface of *any* kind function, not only the symmetric ones. He applied it to a case where you can find the answer via both methods
Derive spherical coordinates please
Coming in 10 days
I'm Sorry, I'm just a physicist. But I'm also not scared..
Area of a donut please!!!
Coming on around January 21
Like surface area, or just area of an annulus. Cause that's pretty easy.
@@drpeyam and also a video for a general ellipsoid 😁
First