<s>[INST] How to find $\theta$ bounds when calculating the volume enclosed between sphere $x^2+y^2+z^2=4a^2$ and cylinder $x^2+(y-a)^2=a^2$? 
I need to find $\theta$ bounds when calculating the volume enclosed between sphere $x^2+y^2+z^2=4a^2$ and cylinder $x^2+(y-a)^2=a^2$. The final answer must be that volume=$\frac{48\pi-64}{9}$.

Here's an illustration:

Let $$V=\int\int\int_Bf(\theta,r,z)dV$$
Because of the symmetry of volume relative of the $y$ axis we can calculate the volume of the upper half then multiply it by 2.
Then:
$$
0\le z\le \sqrt{4a^2-x^2-y^2} \le \sqrt{4a^2-r^2}
$$
Because our the given surfaces don't have their center at the origin we need to adjust the $r$. We can retrieve it from the cylinder equation, $r=2a\sin\theta$. So:
$$
0\le r\le 2a\sin\theta
$$
Finally, the projection of the surface onto the $xy$ plane is a circle. So I think that $0\le \theta \le 2\pi$. 
But I don't get the correct answer with this angle range.
If we evaluate the triple integral we'll get to this expression in the end:
$$
E=-\frac{8a^3}{3}(\frac{1}{3}\cos^2\theta \sin\theta+\frac{2}{3}\sin\theta-\theta)
$$
For example:
$$
\int_0^{2\pi}E=\frac{16\pi}{3}a^3
$$
which is not the answer I should've got.
However if I choose the following bounds for $\theta$:
$$
\int_0^{\frac{\pi}{2}}E=\frac{12\pi-16}{9}a^3
$$
The last bounds give the volume of the quarter of the original volume so if we multiply it by $4$ we get the desired answer:
$$
\frac{12\pi-16}{9}a^3\cdot 4=\frac{48\pi-64}{9}
$$
Please explain why my bounds are not good.
 [/INST] Step 2 is incorrect because both limits vanish $E$  at $0, 2\pi$ . Steps 3 and 4 are correct evaluating between limits $ \int _{\pi/2}^0 $ and multiplying by $4.$
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