# What is the surface area of the solid created by revolving f(x) = e^(x)/2 , x in [2,7] around the x axis?

Nov 3, 2016

The area is =pi/8((4(ln(e^(14)+4)+e^7)+e^7sqrt(e^(14)+4)))-pi/8(4(ln(e^(4)+4)+e^2)+e^2sqrt(e^(4)+4)))

#### Explanation:

The volume of a small disc is $\mathrm{dS} = \pi y \mathrm{dr}$
and $\mathrm{dr} = \sqrt{{\left(\mathrm{dx}\right)}^{2} + {\left(\mathrm{dy}\right)}^{2}}$
$\mathrm{dS} = \pi y \sqrt{{\left(\mathrm{dx}\right)}^{2} + {\left(\mathrm{dy}\right)}^{2}} = \pi y \sqrt{1 + {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2}} \mathrm{dx}$
So $y = {e}^{x} / 2$$\implies$$\frac{\mathrm{dy}}{\mathrm{dx}} = {e}^{x} / 2$
so $\mathrm{dS} = \pi {e}^{x} / 2 \cdot \sqrt{1 + {e}^{2 x} / 4} \mathrm{dx}$
$\mathrm{dS} = \pi {e}^{x} / 4 \cdot \sqrt{4 + {e}^{2 x}} \mathrm{dx}$
so $S = \frac{\pi}{4} \int {e}^{x} \sqrt{4 + {e}^{2 x}} \mathrm{dx}$
let $u = {e}^{x}$ $\implies$$\mathrm{du} = {e}^{x} \mathrm{dx}$
$S = \frac{\pi}{4} \cdot \int \sqrt{4 + {u}^{2}} \mathrm{du}$
$u = 2 \tan v$$\implies$$\mathrm{du} = {\sec}^{2} u \left(\mathrm{dv}\right)$
and $4 {\tan}^{2} v + 4 = 4 {\sec}^{2} v$
$S = 4 \frac{\pi}{4} \int {\sec}^{3} v \mathrm{dv} = \pi \left(\frac{1}{2} \int \sec v \mathrm{dv} + \sec v \tan \frac{v}{2}\right)$
$\pi \left(\left(\ln \frac{\tan v + \sec v}{2}\right) + \sec v \tan \frac{v}{2}\right)$
S=pi/8(4(ln(e^(2x)+4)+e^x)+e^xsqrt(e^(2x)+4)))_2^7
=pi/8((4(ln(e^(14)+4)+e^7)+e^7sqrt(e^(14)+4)))-pi/8(4(ln(e^(4)+4)+e^2)+e^2sqrt(e^(4)+4)))