Triple Integrals in Cylindrical Coordinates — Question 5

Question 5 – Solution

Strategy. The annular cylinder becomes a rectangular box in $(r,\theta,z)$, while the density becomes $r+z$.

Step 1: Mass The bounds are $0\le\theta\le 2\pi$, $1\le r\le 2$, and $0\le z\le 3$. Therefore $$\begin{align*} M&=\int_0^{2\pi}\int_1^2\int_0^3(r+z)r\,dz\,dr\,d\theta\\ &=2\pi\int_1^2\left(3r^2+\frac 92r\right)dr =2\pi\left[r^3+\frac 94r^2\right]_1^2 =\boxed{\frac{55\pi}{2}}. \end{align*}$$

Step 2: Average density The volume is $\pi(2^2-1^2)(3)=9\pi$. Hence $$\boxed{\rho_{\mathrm{avg}}=\frac{M}{V}=\frac{55}{18}}.$$

Verification On the solid, $1\le r+z\le 5$, and $55/18\approx 3.06$ lies strictly in that range. The units of $M/V$ are density units, as required.