Triple Integrals in Spherical Coordinates — Question 2

Question 2 – Solution

Strategy. In a meridian plane, the cone fixes the polar angle while the sphere fixes the radial endpoint.

Step 1: Angular boundary Since $z=\rho\cos\phi$ and $\sqrt{x^2+y^2}=\rho\sin\phi$, $$\rho\cos\phi=\sqrt 3\rho\sin\phi \quad\Longrightarrow\quad\tan\phi=\frac 1{\sqrt 3}.$$ The region above the cone therefore has $0\le\phi\le\pi/6$.

Step 2: Evaluate $$\begin{align*} V&=\int_0^{2\pi}\int_0^{\pi/6}\int_0^4 \rho^2\sin\phi\,d\rho\,d\phi\,d\theta\\ &=2\pi\left(1-\frac{\sqrt 3}{2}\right)\frac{64}{3} =\boxed{\frac{64\pi}{3}(2-\sqrt 3)}. \end{align*}$$

Verification The angular factor $1-\sqrt 3/2$ is between $0$ and $1$, so this positive sector is smaller than the upper half-ball and therefore smaller than the entire ball of volume $256\pi/3$.