 Shirley Peck

2023-03-22

What is the surface area of the solid created by revolving $f\left(x\right)={e}^{2-x},x\in \left[1,2\right]$ around the x axis? The surface area for a revolution around the x axis is given by:
$S=2\pi {\int }_{a}^{b}f\left(x\right)\sqrt{1+{\left(\frac{dy}{dx}\right)}^{2}}dx$
(which is basically a projection of the circumference along the function f(x) whose arc length you could have found.)
In this case, ${\left(\frac{dy}{dx}\right)}^{2}$ is given by:
${\left(\frac{dy}{dx}\right)}^{2}={\left(-{e}^{2-x}\right)}^{2}$
And then we have:
$S=2\pi {\int }_{1}^{2}{e}^{2-x}\sqrt{1+{\left(-{e}^{2-x}\right)}^{2}}dx$
First, let $u=-{e}^{2-x}$. Therefore, $du={e}^{2-x}dx$ and:
$S=2\pi \int \sqrt{1+{u}^{2}}du$
where we omit the integral bounds for now. Then we can see it looks like the form $\sqrt{{a}^{2}+{x}^{2}}$, then let $u=\mathrm{tan}\theta$ to get $du={\mathrm{sec}}^{2}\theta d\theta$. Therefore:
$S=2\pi \int \sqrt{1+{\mathrm{tan}}^{2}\theta }{\mathrm{sec}}^{2}\theta d\theta$
$=2\pi \int {\mathrm{sec}}^{3}\theta d\theta$
And you should have written down the following integral in class:
$=2\pi \stackrel{\int {\mathrm{sec}}^{3}\theta d\theta }{\stackrel{⏞}{\left[\frac{1}{2}\left(\mathrm{sec}\theta \mathrm{tan}\theta +\mathrm{ln}|\mathrm{sec}\theta +\mathrm{tan}\theta |\right)\right]}}$
$=\pi \mathrm{sec}\theta \mathrm{tan}\theta +\pi \mathrm{ln}|\mathrm{sec}\theta +\mathrm{tan}\theta |$
Next, back-substitute.
$⇒\pi u\sqrt{1+{u}^{2}}+\pi \mathrm{ln}|\sqrt{1+{u}^{2}}+u|$
$=\pi u\sqrt{1+{\left(-{e}^{2-x}\right)}^{2}}+\pi \mathrm{ln}|\sqrt{1+{\left(-{e}^{2-x}\right)}^{2}}+\left(-{e}^{2-x}\right)|$
$=-\pi {e}^{2-x}\sqrt{1+{e}^{4-2x}}+\pi \mathrm{ln}|\sqrt{1+{e}^{4-2x}}-{e}^{2-x}|$
As it turns out, $\sqrt{1+{e}^{4-2x}}-{e}^{2-x}\ge 0$, so we can remove the absolute values.
$=-\pi {e}^{2-x}\sqrt{1+{e}^{4-2x}}+\pi \mathrm{ln}\left(\sqrt{1+{e}^{4-2x}}-{e}^{2-x}\right)$
Thus, we evaluate from 1 to 2:
$⇒\left[-\pi {e}^{2-2}\sqrt{1+{e}^{4-2\cdot 2}}+\pi \mathrm{ln}\left(\sqrt{1+{e}^{4-2\cdot 2}}-{e}^{2-2}\right)\right]-\left[-\pi {e}^{2-1}\sqrt{1+{e}^{4-2\cdot 1}}+\pi \mathrm{ln}\left(\sqrt{1+{e}^{4-2\cdot 1}}-{e}^{2-1}\right)\right]$
$=\left[-\pi \sqrt{2}+\pi \mathrm{ln}\left(\sqrt{2}-1\right)\right]-\left[-\pi e\sqrt{1+{e}^{2}}+\pi \mathrm{ln}\left(\sqrt{1+{e}^{2}}-e\right)\right]$
$=-\pi \sqrt{2}+\pi \mathrm{ln}\left(\sqrt{2}-1\right)+\pi e\sqrt{1+{e}^{2}}-\pi \mathrm{ln}\left(\sqrt{1+{e}^{2}}-e\right)$
$22.943$

Do you have a similar question?