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so the question I was trying to answer was, find the volume of the solid bounded above by the ellipsoid of revolution

$$b^2 x^2 + b^2 y^2 + a^2 z^2 = a^2 b^2,$$

below by the x-y plane

and on the sides by the cylinder $$x^2 + y^2 - ay = 0$$

I converted this into polar coordinates and got two possible double integrals to get my answer. The second one got me the correct answer, the first one got me the wrong answer. Could someone please help me figure out why the first integral calculates the wrong answer? It's been driving me mad for a day now!

Here is the one that gave me an incorrect answer (http://imgur.com/8V7EJhQ)

and Here is the one that gave me the correct answer. (http://imgur.com/6VpxfsK)

In the first one I tried to exploit the symmetry, the volume is the volume above the first quadrant in the xy plane multiplied by two.

Help would be very much appreciated. I guess I'll need to rack up some reputation points and put a bounty on this.. Not getting anywhere by myself

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  • $\begingroup$ I think that your "wrong" answer is actually correct. $\endgroup$ – Martín-Blas Pérez Pinilla Jun 7 '17 at 9:01
  • $\begingroup$ If the first method is correct then I'm doubly confused as to why the second one isn't! I'm pretty sure the second one is correct, I've assumed even less and it matches the answer in the back of my textbook $\endgroup$ – Ravi Jun 7 '17 at 9:09
  • $\begingroup$ Never trust blindly in textbook answers. $\endgroup$ – Martín-Blas Pérez Pinilla Jun 7 '17 at 9:16
  • $\begingroup$ Sure but the fact that I've found a solid method that matches the textbook answer surely indicates that the textbook answer is correct and my first one is wrong? $\endgroup$ – Ravi Jun 7 '17 at 9:18
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    $\begingroup$ Problems like this are already know: See mathoverflow.net/questions/11517/computer-algebra-errors. $\endgroup$ – Martín-Blas Pérez Pinilla Jun 11 '17 at 21:13
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Possible problem: doing the inner integral: $$\int_0^{a\sin\theta}br\sqrt{1 - r^2/a^2}\,dr = -\frac{b}{3a}(a^2 - r^2)^{3/2}\Big\vert_{r = 0}^{r = a\sin\theta} = \frac{a^2b}3(1 - |\cos^3\theta|)$$ and the $|\dot{}|$ is essential.

EDIT: The symmetry is true. For $\theta\in[0,\pi/2]$ we have $|\cos\theta| = \cos\theta$ and: $$ \int_0^\pi\frac{a^2b}3(1 - |\cos^3\theta|)\,d\theta = 2\int_0^{\pi/2}\frac{a^2b}3(1 - |\cos^3\theta|)\,d\theta = 2\int_0^{\pi/2}\frac{a^2b}3(1 - \cos^3\theta)\,d\theta\ne \int_0^{\pi}\frac{a^2b}3(1 - \cos^3\theta)\,d\theta. $$

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  • $\begingroup$ Thanks for the reply, I don't think the issue is with calculation of the integral itself, I've double checked that with symbolab and wolframalpha. Maybe I've misunderstood what you've said? $\endgroup$ – Ravi Jun 7 '17 at 9:07
  • $\begingroup$ The values of your integrals are correct. Only one is the volume. You can check that with the $|\dot{}|$ you have $\int_0^\pi = 2 \int_0^{\pi/2}$. $\endgroup$ – Martín-Blas Pérez Pinilla Jun 7 '17 at 9:23
  • $\begingroup$ I don't really understand, I have worked through the integral making sure im using |cos^3(theta)| and not cos^3(theta), in neither case do I get your equation $\endgroup$ – Ravi Jun 7 '17 at 9:27
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    $\begingroup$ @Ravi, see my edit. $\endgroup$ – Martín-Blas Pérez Pinilla Jun 7 '17 at 9:49
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    $\begingroup$ @Ravi, the term will always be positive because is the positive square root of $(\cdots)^3$. $\endgroup$ – Martín-Blas Pérez Pinilla Jun 7 '17 at 19:42
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If you look to the cylinder equation you'll see that it is shifted about $\frac{2}{a}$ in the $y$-axis, so there is no symmetry, but if it was centered in the origin, like the ellipsoid, the first method would work.

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    $\begingroup$ but there is symmetry about the y-axis for that equation because $$(-x)^2 + y^2 - ay = x^2 + y^2 - ay = 0$$ $\endgroup$ – Ravi Jun 5 '17 at 2:51
  • $\begingroup$ There is symmetry about the y axis but not x, hence integration on a quarter turn doesn't work. $\endgroup$ – Yves Daoust Jun 7 '17 at 10:11

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