Find volume of body given by $x^2+y^2+z^2 \leq a^2$ and $x^2+y^2 \geq a(a-2z)$, $a>0$. Is this answer right? What are mistakes in my solution? It appears to be an upside-down 'bowl', cut out of a sphere by a paraboloid. Moving to cylindrical coordinates,
\begin{align}
4 \int_0^{\pi/2}\ d\phi \int_{0}^{a}\   r \ dr \int_{(a^2-r^2)/2a}^{\sqrt{a^2-r^2}} \ dz
&=
4 \int_0^{\pi/2}\ d\phi \int_{0}^{a}\ \left( \frac{a^2-r^2}{2}-\frac{(a^2-r^2)^2}{2a\cdot\ 2a\cdot2} \right) r \ dr 
\\ \ \\
&=
2 \int_0^{\pi/2}\ d\phi \int_{0}^{a}\ \left( \frac{a^2-r^2}{1}-\frac{(a^2-r^2)^2}{4a^2} \right) r \ dr 
\\ \ \\
&=
\int_0^{\pi/2}\ d\phi \int_{0}^{a}\ \left( \frac{3a^2}{2}r-r^3-\frac{r^5}{2a^2} \right)\ dr 
\\ \ \\
&=
\int_0^{\pi/2}\   \frac{3a^2}{2}\cdot \frac{a^2}{2}-\frac{a^4}{4}-\frac{1}{2a^2}\cdot\frac{a^6}{6} \ d\phi
\\ \ \\
&=
\frac{5\pi a^4}{24}
\end{align}
Was this correct? If not, where?
I'd also love to see alternative ways of doing this.

*

*Your setup is the canonical one, with the exception of taking $0\leq \phi\leq \pi/2$ and then multiplying by $4$: you could have just integrated from $0$ to $2\pi$. 

*You are assuming that $a>0$, which is probably the case but you don't say. 

*From the first to the second line: instead of $\sqrt{a^2-r^2}-\frac{a^2-r^2}{2a}$ you wrote $\frac{a^2-r^2}2-\frac{(a^2-r^2)^2}{2a\cdot2a\cdot2}$. I can't imagine why. That's where things derail. 
What you should have done: 
\begin{align}
\int_0^{2\pi}\int_0^a \int_{\frac{a^2-r^2}{2a}}^{\sqrt{a^2-r^2}}\,1\,dz\,r\,dr\,d\theta
&=\int_0^{2\pi}\int_0^a\left({\sqrt{a^2-r^2}}-{\frac{a^2-r^2}{2a}}\right)\,r\,dr\,d\theta\\ \ \\
&={2\pi}\int_0^a\left({\sqrt{a^2-r^2}}-{\frac{a^2-r^2}{2a}}\right)\,r\,dr\\ \ \\
&={2\pi}\left.\vphantom{\int}\left(-\frac23(a^2-r^2)^{3/2}+\frac{(a^2-r^2)^2}{4a}\right)\right|_0^a\\\ \ \\
&=2\pi\,\left(\frac13(a^2)^{3/2}-\frac{(a^2)^2}{8a}\right)\\ \ \\
&=\frac{5\pi a^3}{12}
\end{align}
As a way to confirm the result, if you   calculate the bottom half, the limits of the inner integral would have been
$$
\int_{\sqrt{a^2-r^2}}^{\frac{a^2-r^2}{2a}},
$$
and a very similar calculation as above yields $\frac{11\pi a^3}{12}$. Not surprisingly, if you add both volumes you get 
$$
\frac{5\pi a^3}{12}+\frac{11\pi a^3}{12}=\frac{16\pi a^3}{12}=\frac{4\pi a^3}3, 
$$
the volume of the whole interior of the sphere. 
