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?

There are two mistakes in your work. At the intersection of the sphere and the plane,

$z = 2 \cos\phi = 1 \implies \phi = \pi/3$

Also the lower bound of $\rho$ is defined by the plane

$z = \rho\cos\phi = 1 \implies \rho = \sec\phi$

So the integral should be,

$ \displaystyle \int_{0}^{2 \pi} \int_0^{\pi/3} \int_{\sec\phi}^2 \rho^2 \sin\phi ~ d\rho ~ d\phi ~d\theta = \frac{5\pi}{3}$

Also note that,

$x^2 + (y-2)^2 + z^2 = 4 \implies x^2 + y^2 + z^2 = 4 y$

So using $x = \rho \cos\theta \sin\phi, z = \rho\sin\theta\sin\phi, y = \rho\cos\phi$, we have

$\rho = 4 \cos\phi$

$y = \rho \cos\phi = 3 \implies \rho = 3 \sec\phi$

At intersection of the sphere and the plane,

$\rho = 4 \cos\phi = 3 \sec\phi \implies \cos\phi = \frac{\sqrt3}{2}$

So, $\phi = \pi/6$

So the integral can also be written as,

$ \displaystyle \int_{0}^{2 \pi} \int_0^{\pi/6} \int_{3\sec\phi}^{4\cos\phi} \rho^2 \sin\phi ~ d\rho ~ d\phi ~d\theta = \frac{5\pi}{3}$