Question

Set up a triple integral in spherical coordinates to compute the volume of the region inside the sphere x² + y² + z² = 4 and above the plane z = 1​

Answer 1

The plane meets the sphere in the cylinder

$z=1 \implies x^2+y^2+1^2 = 4 \implies x^2+y^2 = 3 = \left(\sqrt3\right)^2$

In spherical coordinates, the plane has equation

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

and since

$\begin{cases}x = \rho\cos(\theta)\sin(\phi) \\ y=\rho\sin(\theta)\sin(\phi)\end{cases} \implies x^2+y^2 = \rho^2 \sin^2(\phi)$

the cylinder has equation

$\rho^2 \sin^2(\phi) = 3 \implies \rho = \sqrt3 \, \csc(\phi)$

The plane and cylinder thus meet when

$\sec(\phi) = \sqrt3\,\csc(\phi) \implies \tan(\phi) = \sqrt3 \implies \phi = \dfrac\pi3$

So, the volume of the region is given by

$\displaystyle \boxed{\int_0^{2\pi} \int_0^{\pi/3} \int_{\sec(\phi)}^2 \rho^2 \sin(\phi) \, d\rho \, d\phi \, d\theta}$