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.