I was working on finding the volume of the intersection of two cylinders $$ x^2 + y^2 = a^2 \quad \text{and} \quad y^2 + z^2 = a^2. $$ I am well aware of the volume being $16a^3/3$ and I know how to compute it.
However, when I compute it using polar coordinates, the following happens: $$ \begin{align*} &\phantom{{}={}}\int ^{2\pi} _{0} { \int ^{a} _{0} { \sqrt{a^2 - r^2 \sin^2 \theta} \, r } \, dr } \, d\theta \\ &= -\frac{1}{2} \int ^{2\pi} _{0} { \int ^{a^2 \cos ^2 \theta} _{0} { u ^{1/2} \cdot \frac{r}{r \sin ^2 \theta} } \, du } \, d\theta \tag{$u = a^2 - r^2 \sin ^2 \theta$} \\ &= -\frac{1}{2} \int ^{2\pi} _{0} { \frac{1}{\sin ^2 \theta} \left[ \frac{2}{3} u^\frac{3}{2} \right] ^{a^2 \cos ^2 \theta} _{a^2} } \, d\theta \\ &= -\frac{a^3}{3} \int ^{2\pi} _{0} { \frac{\cos^3 \theta - 1}{\sin ^2 \theta} } \, d\theta \\ &= -\frac{a^3}{3} \int ^{2\pi} _{0} { \frac{\cos \theta (1 - \sin ^2 \theta)}{\sin ^2 \theta} - \csc ^2 \theta } \, d\theta \\ &= -\frac{a^3}{3} \int ^{2\pi} _{0} { \cot \theta \csc \theta - \cos \theta - \csc ^2 \theta } \, d\theta \\ &= - \frac{a^3}{3} \left[ -\csc \theta - \sin \theta + \cot \theta \right] ^{2\pi} _0. \end{align*} $$
The integration domain of $\theta$ includes $\pi$, which $$ -\csc \theta - \sin \theta + \cot \theta $$ diverges. Despite this, one can get the correct value by excluding $\pi$ in the integration.
My question is that I do not understand what the change of variable makes the integral divergent. Geometrically, the volume is bounded by the cylinder so it could not be divergent. A possible reason I can think of is that by letting $u = a^2 - r^2 \sin \theta$, the area element becomes $\csc^2 \theta$, which is very large when $\theta$ is small. However, this does not answer why there is only a discontinuity at $\theta = \pi$, not $\theta = 0$.
Please contribute if you have any thought. Any help is appreciated.
It seems like you made the mistake of assuming $\sqrt{x^2} = x$. The correct simplification after plugging in your bounds should have been $$(a^2\cos^2\theta)^{\frac{3}{2}} = a^3 \left|\cos\theta\right|^3$$ You can check that for this version of the integrand $$\lim_{\theta\to n\pi}\frac{1-\left|\cos\theta\right|^3}{\sin^2\theta} = \lim_{\theta\to n\pi}\frac{1-|\cos\theta|}{(\theta-n\pi)^2}\cdot\frac{(\theta-n\pi)^2}{\sin^2\theta}\cdot\left(1+|\cos\theta|+\cos^2\theta\right) = \frac{1}{2}\cdot 1 \cdot 3 = \frac{3}{2}$$ i.e. the singularities in the integrand are nonexistent.