Surface Area with Parametric Equations — Question 6

Question 6

Problem

Rotate $x=e^t$, $y=e^{-t}$, $0\le t\le\ln2$, about the $x$-axis. Write the exact surface-area integral in simplest form.

Question 6 – Solution

Solution

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

2. The curve lies above the $x$-axis because $y=e^{-t}>0$, so the radius of rotation is $y=e^{-t}$.

3. Multiply the radius by $ds$ and simplify: $$\begin{aligned} y\,ds &=e^{-t}\sqrt{e^{2t}+e^{-2t}}\,dt\\ &=\sqrt{e^{-2t}(e^{2t}+e^{-2t})}\,dt\\ &=\sqrt{1+e^{-4t}}\,dt. \end{aligned}$$

4. Hence the requested exact integral is $$\boxed{S=2\pi\int_0^{\ln2}\sqrt{1+e^{-4t}}\,dt}.$$