<s>[INST] Find the volume between the sphere $x ^2 + y^2 + z^2 = 4$ and the plane $z = 1$ Suppose $y \geq 3$. I want to compute the volume between the sphere $x ^2 + (y − 2)^2 + z^2 = 4$
and the plane $y = 3$.
So I move left the sphere and and the plan, and rotate it counterclockwise. I got the new sphere and the new plan:
Suppose $z \geq 1$. Then compute the volume between $x ^2 + y^2 + z^2 = 4$ and the plan $z = 1$.
Here is my attempt using spherical coordinates:
$$\int_{0}^{2 \pi} \int_{0}^{\frac{\pi}{6}} \int_1^2 \rho^2 \sin \phi dp d \phi d\theta  = 2\pi (1- \frac{\sqrt{3}}{2})\frac{1}{3}$$ I am supposed to get $\frac{5 \pi}{3}$. Where am I wrong?
 [/INST] OP's attempt can be rectified after correction the following two mistakes:

*

*At different $\phi$ values, the $\rho$ does not always vary from 1 to 2. By drawing a diagram, we can check that at angle $\phi$, the range of $\rho$ is from $\frac{1}{\cos \phi}$ to 2.

*$\phi$ is supposed to range from 0 to $\pi/3$ (calculated from $\cos^{-1} (\frac{1}{2})$; again a diagram will help)

Thus, the correct integral to compute is
$$\int\limits_{0}^{2\pi} d\theta  \int\limits_{0}^{\pi/3}\int\limits_{\frac{1}{\cos\phi}}^{2} \rho^2\sin\phi \ d\rho\ d\phi$$
$$ = 2\pi \int\limits_{0}^{\pi/3} \left.\left(\frac{1}{3}\rho^3\sin\phi\right)\right\rvert_{\frac{1}{\cos\phi}}^{2} \  d\phi$$
$$= 2\pi \int\limits_{0}^{\pi/3} \frac{8}{3}\sin\phi-\frac{1}{3}\tan\phi\sec^2\phi\  d\phi$$
$$= 2\pi \left.\left(-\frac{8}{3}\cos\phi-\frac{1}{6}\tan^2\phi\right)\right\rvert_{0}^{\pi/3}$$
$$ = \frac{5\pi}{3}$$
</s>