Surface Area with Parametric Equations — Question 2

Question 2

Problem

The upper semicircle $x=R\cos t$, $y=R\sin t$, $0\le t\le\pi$, rotates about the $x$-axis. Recover the sphere’s area.

Question 2 – Solution

Solution

1. Differentiate: $$x'(t)=-R\sin t, \qquad y'(t)=R\cos t.$$ Thus $$ds=\sqrt{R^2\sin^2t+R^2\cos^2t}\,dt=R\,dt.$$

2. On $0\le t\le\pi$, the generating semicircle lies above the $x$-axis, so its radius of rotation is $$y=R\sin t\ge0.$$ This semicircle generates the sphere exactly once.

3. Substitute into $S=2\pi\int y\,ds$: $$\begin{aligned} S&=2\pi\int_0^\pi(R\sin t)(R\,dt)\\ &=2\pi R^2\left[-\cos t\right]_0^\pi\\ &=2\pi R^2(2)=\boxed{4\pi R^2}. \end{aligned}$$