Triple Integrals in Spherical Coordinates — Question 4

Question 4 – Solution

Strategy. Write $z=\rho\cos\phi$ and use the upper-hemisphere bounds $0\le\phi\le\pi/2$.

Step 1: Geometry and integral

$$\begin{align*} I&=\int_0^{2\pi}\int_0^{\pi/2}\int_0^3 (\rho\cos\phi)\rho^2\sin\phi\,d\rho\,d\phi\,d\theta\\ &=(2\pi)\left[\frac{\sin^2\phi}{2}\right]_0^{\pi/2} \left[\frac{\rho^4}{4}\right]_0^3 =\boxed{\frac{81\pi}{4}}. \end{align*}$$

Step 2: Average height The hemisphere volume is $V=\frac 12\frac{4\pi}{3}(3^3)=18\pi$. Thus $$\boxed{z_{\mathrm{avg}}=\frac{I}{V}=\frac 98}.$$

Verification Every point has $0\le z\le 3$, and $9/8$ lies in that interval. The value is below the mid-height $3/2$ because cross-sectional area is greatest near $z=0$.