Surface Area with Parametric Equations — Question 3

Question 3 – Solution

Solution

1. The interval $0\le t\le2\pi$ traces the generating circle once. Differentiate: $$x'(t)=-\sin t, \qquad y'(t)=\cos t.$$ Hence $$ds=\sqrt{\sin^2t+\cos^2t}\,dt=dt.$$

2. The circle lies between $x=2$ and $x=4$, entirely to the right of the $y$-axis. Therefore, the radius of rotation is the nonnegative distance $$x=3+\cos t.$$

3. Apply $S=2\pi\int x\,ds$: $$\begin{aligned} S&=2\pi\int_0^{2\pi}(3+\cos t)\,dt\\ &=2\pi\left[3t+\sin t\right]_0^{2\pi}\\ &=2\pi(6\pi)=\boxed{12\pi^2}. \end{aligned}$$