Triple Integrals in Cylindrical Coordinates — Question 8

Question 8 – Solution

Strategy. Find the nonzero intersection radius, then integrate $z$ between the radial graphs.

Step 1: Geometry The surfaces meet where $r^2=2r$, so $r=0$ or $r=2$. On $0\le r\le 2$, the cone lies above the paraboloid.

Thus $0\le\theta\le 2\pi$, $0\le r\le 2$, and $r^2\le z\le 2r$.

Step 2: Integral of $z$ $$\begin{align*} I&=\int_0^{2\pi}\int_0^2\int_{r^2}^{2r}zr\,dz\,dr\,d\theta\\ &=2\pi\int_0^2\left(2r^3-\frac 12r^5\right)dr =\boxed{\frac{16\pi}{3}}. \end{align*}$$

Step 3: Average The volume is $$V=2\pi\int_0^2(2r-r^2)r\,dr=\frac{8\pi}{3}.$$ Therefore $\boxed{z_{\mathrm{avg}}=I/V=2}$. This lies between the solid’s minimum and maximum heights, $0$ and $4$.