4. (12 pts) Using spherical coordinates, rewrite the triple integral with all limits of integration, \[ \iiint_{E} z d V \] $E$ is the solid that lies below the sphere $x^{2}+y^{2}+z^{2}=4$ and above the cone $z=\sqrt{x^{2}+y^{2}}$. DO NOT EVALUATE the integral.
Solution
Step 1 ：Given the triple integral \(\iiint_{E} z d V\), where E is the solid that lies below the sphere \(x^{2}+y^{2}+z^{2}=4\) and above the cone \(z=\sqrt{x^{2}+y^{2}}\).
Step 2 ：We need to convert this integral from Cartesian coordinates to spherical coordinates. In spherical coordinates, \(x = \rho\sin\phi\cos\theta\), \(y = \rho\sin\phi\sin\theta\), and \(z = \rho\cos\phi\). The volume element \(dV\) in spherical coordinates is \(\rho^2\sin\phi d\rho d\phi d\theta\).
Step 3 ：The solid E is bounded by the sphere \(x^2 + y^2 + z^2 = 4\) and the cone \(z = \sqrt{x^2 + y^2}\). In spherical coordinates, the sphere becomes \(\rho = 2\) and the cone becomes \(\phi = \pi/4\).
Step 4 ：The limits of integration for \(\rho\) are from 0 to 2, for \(\phi\) are from 0 to \(\pi/4\), and for \(\theta\) are from 0 to \(2\pi\).
Step 5 ：The integrand in spherical coordinates is \(\rho^{3}\sin\phi\cos\phi\).
Step 6 ：Finally, the triple integral in spherical coordinates is \(\boxed{2\pi\int_{0}^{2}\int_{0}^{\pi/4} \rho^{3}\sin\phi\cos\phi\, d\phi\, d\rho}\)
