Double Integral in Polar Coordinates Symmetry Issue

[wadawalnut]

Homework Statement

Find the volume of the solid lying inside both the sphere x^2 + y^2 + z^2 = 4a^2 and the cylinder x^2 + y^2 = 2ay above the xy plane.

Homework Equations

Polar coordinates:
r^2 = x^2 + y^2
x = r\cos(\theta)
y = r\sin(\theta)

The Attempt at a Solution

So I tried this problem and got it almost perfectly correct. My result was: V = \frac{8\pi a^3}{3}, and the right answer is V = \frac{8\pi a^3}{3} - \frac{32 a^3}{9}, so I'm just missing one term. The first thing that crossed my mind was that one of my integrals evaluated to 0, so obviously that's where I went wrong. I checked the solution and noticed the only difference from my work was that the book solution used symmetry and integrated \theta from 0 to \frac{\pi}{2}, and I just integrated from 0 to \pi. This is what I did:
x^2 + y^2 = 2ay describes a cylinder with a circle base that is centered at (0,a) and has radius a. Therefore, this cylinder lies entirely in the first and second quadrants, where \theta ranges from (0,\pi). Using cylindrical coordinates, the sphere has the equation z = \sqrt{4a^2 - r^2} and the cylinder has the equation r^2 = 2ar\sin(\theta), or r = 2a\sin(\theta).
V = \int_{0}^{\pi}\int_{0}^{2a\sin(\theta)}\sqrt{4a^2 - r^2}rdrd\theta
Making the substitution u = 4a^2 - r^2, du = -2rdr
V = \frac{-1}{2} \int_{0}^{\pi}\int_{r = 0}^{r = 2a\sin(\theta)}\sqrt{u}dud\theta
V = \frac{-1}{3} \int_{0}^{\pi}(4a^2 - r^2)^{3/2}\bigg|_{0}^{2a\sin(\theta)}d\theta
V = \frac{-1}{3} \int_{0}^{\pi}\left[(4a^2(1 - sin^2(\theta))^{3/2} - (4a^2)^{3/2}\right]d\theta
V = \frac{-1}{3} \int_{0}^{\pi}\left[(4a^2)^{3/2}(cos^2(\theta))^{3/2} - 8a^3\right]d\theta
V = \frac{-8a^3}{3} \int_{0}^{\pi}cos^3(\theta)d\theta + \frac{8\pi a^3}{3}
Now, the remaining integral evaluates to 0 because it will be a function of \sin(\theta) where \theta \in (0,\pi).

I see that if I were to use symmetry like it was done in the book (\theta \in (0,\frac{\pi}{2})) and multiply everything by 2, my integral should return the right answer. I just don't understand why what I did was wrong.

 

[andrewkirk]

Always bear in mind the possibility that the book answer may be wrong.

The integration over $\theta$ only happens in your last step. So the two methods should be the same up to there. Prior to that integration, your formula is (simplifying a bit):

$$
V = \frac{8a^3}{3} \int_{0}^{\pi}\left[(1-\cos^3\theta\right]d\theta
$$
Theirs should be
$$V = 2\cdot \frac{8a^3}{3} \int_{0}^{\frac{\pi}{2}}\left[(1-\cos^3\theta\right]d\theta$$

Does their formula match that?

By the way, note that the integrand is not symmetric around $\theta=\frac{\pi}{2}$. It is not the case that the integral over $[0,\pi]$ is twice the integral over $[0,\frac{\pi}{2}]$.

 

[LCKurtz]

V = \frac{-1}{3} \int_{0}^{\pi}\left[(4a^2)^{3/2}(cos^2(\theta))^{3/2} - 8a^3\right]d\theta
V = \frac{-8a^3}{3} \int_{0}^{\pi}cos^3(\theta)d\theta + \frac{8\pi a^3}{3}
Now, the remaining integral evaluates to 0 because it will be a function of \sin(\theta) where \theta \in (0,\pi).

I see that if I were to use symmetry like it was done in the book (\theta \in (0,\frac{\pi}{2})) and multiply everything by 2, my integral should return the right answer. I just don't understand why what I did was wrong.

Note that $$(\cos^2\theta)^{\frac 3 2}= \left(\sqrt{\cos^2\theta}\right)^3 = |\cos\theta|^3\ne \cos^3\theta$$

 

[wadawalnut]

Note that $$(\cos^2\theta)^{\frac 3 2}= \left(\sqrt{\cos^2\theta}\right)^3 = |\cos\theta|^3\ne \cos^3\theta$$

Ahhhh that's why...
Forgot about those sneaky absolute values. So I would have to split it up into an integral from 0 to pi/2 and a negative one from pi/2 to pi, right?
That explains why the book was so quick to use symmetry, otherwise the symmetry wouldn't have been all that helpful.
Thanks a lot for the help.