# What is the surface area of the solid created by revolving f(x) =e^(2-x) , x in [1,2] around the x axis?

Jun 16, 2017

I got $22.943$, and so does Wolfram Alpha here.

The surface area for a revolution around the $x$ axis is given by:

$S = 2 \pi {\int}_{a}^{b} f \left(x\right) \sqrt{1 + {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2}} \mathrm{dx}$

(which is basically a projection of the circumference along the function $f \left(x\right)$ whose arc length you could have found.)

In this case, ${\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2}$ is given by:

${\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2} = {\left(- {e}^{2 - x}\right)}^{2}$

And so we have:

$S = 2 \pi {\int}_{1}^{2} {e}^{2 - x} \sqrt{1 + {\left(- {e}^{2 - x}\right)}^{2}} \mathrm{dx}$

First, let $u = - {e}^{2 - x}$. Thus, $\mathrm{du} = {e}^{2 - x} \mathrm{dx}$ and:

$S = 2 \pi \int \sqrt{1 + {u}^{2}} \mathrm{du}$

where we omit the integral bounds for now. Then we can see it looks like the form $\sqrt{{a}^{2} + {x}^{2}}$, so let $u = \tan \theta$ to get $\mathrm{du} = {\sec}^{2} \theta d \theta$. Thus:

$S = 2 \pi \int \sqrt{1 + {\tan}^{2} \theta} {\sec}^{2} \theta d \theta$

$= 2 \pi \int {\sec}^{3} \theta d \theta$

And you should have written down this integral in class to be:

$= 2 \pi {\overbrace{\left[\frac{1}{2} \left(\sec \theta \tan \theta + \ln | \sec \theta + \tan \theta |\right)\right]}}^{\int {\sec}^{3} \theta d \theta}$

$= \pi \sec \theta \tan \theta + \pi \ln | \sec \theta + \tan \theta |$

(if you didn't, then you may have unknowingly asked a difficult question!)

Now, back-substitute.

$\implies \pi u \sqrt{1 + {u}^{2}} + \pi \ln | \sqrt{1 + {u}^{2}} + u |$

$= \pi u \sqrt{1 + {\left(- {e}^{2 - x}\right)}^{2}} + \pi \ln | \sqrt{1 + {\left(- {e}^{2 - x}\right)}^{2}} + \left(- {e}^{2 - x}\right) |$

$= - \pi {e}^{2 - x} \sqrt{1 + {e}^{4 - 2 x}} + \pi \ln | \sqrt{1 + {e}^{4 - 2 x}} - {e}^{2 - x} |$

As it turns out, $\sqrt{1 + {e}^{4 - 2 x}} - {e}^{2 - x} \ge 0$, so we can remove the absolute values.

$= - \pi {e}^{2 - x} \sqrt{1 + {e}^{4 - 2 x}} + \pi \ln \left(\sqrt{1 + {e}^{4 - 2 x}} - {e}^{2 - x}\right)$

Now, we evaluate from $1$ to $2$:

$\implies \left[- \pi {e}^{2 - 2} \sqrt{1 + {e}^{4 - 2 \cdot 2}} + \pi \ln \left(\sqrt{1 + {e}^{4 - 2 \cdot 2}} - {e}^{2 - 2}\right)\right] - \left[- \pi {e}^{2 - 1} \sqrt{1 + {e}^{4 - 2 \cdot 1}} + \pi \ln \left(\sqrt{1 + {e}^{4 - 2 \cdot 1}} - {e}^{2 - 1}\right)\right]$

$= \left[- \pi \sqrt{2} + \pi \ln \left(\sqrt{2} - 1\right)\right] - \left[- \pi e \sqrt{1 + {e}^{2}} + \pi \ln \left(\sqrt{1 + {e}^{2}} - e\right)\right]$

$= - \pi \sqrt{2} + \pi \ln \left(\sqrt{2} - 1\right) + \pi e \sqrt{1 + {e}^{2}} - \pi \ln \left(\sqrt{1 + {e}^{2}} - e\right)$

$\approx$ $\textcolor{b l u e}{22.943}$