Surface Integrals — Question 3

Question 3 – Solution

Strategy. Parametrize by angle and height. Rotational symmetry immediately locates two coordinates of the center of mass.

Step 1: Geometry $$\mathbf r(\theta,z)=\langle 2\cos\theta,2\sin\theta,z\rangle, \quad 0\le\theta\le 2\pi,\quad 0\le z\le 3.$$ Since $\lVert\mathbf r_\theta\times\mathbf r_z\rVert=2$, $$\boxed{dS=2\,d\theta\,dz}.$$

Step 2: Mass $$\begin{align*} M&=\int_0^{2\pi}\int_0^3(1+z)2\,dz\,d\theta\\ &=4\pi\left[z+\frac{z^2}{2}\right]_0^3=\boxed{30\pi}. \end{align*}$$

Step 3: Center of mass Symmetry gives $\bar x=\bar y=0$. The first moment about the $xy$-plane is $$\begin{align*} M_{xy}&=\int_0^{2\pi}\int_0^3 z(1+z)2\,dz\,d\theta\\ &=4\pi\left[\frac{z^2}{2}+\frac{z^3}{3}\right]_0^3=54\pi. \end{align*}$$ Thus $$\boxed{(\bar x,\bar y,\bar z)=\left(0,0,\frac 95\right)}.$$

Verification Because density increases with height, $\bar z=1.8$ is correctly above the geometric midpoint $1.5$ and remains inside $0\le z\le 3$.