# What is the surface area of the solid created by revolving f(x)=2-x over x in [2,3] around the x-axis?

Apr 17, 2016

$A = - \sqrt{2} \pi$

#### Explanation:

$A = 2 \pi {\int}_{2}^{3} f \left(x\right) \sqrt{1 + {\left(\frac{d}{d x} f \left(x\right)\right)}^{2}} d x$

$\frac{d}{d x} f \left(x\right) = - 1$

${\left(\frac{d}{d x} f \left(x\right)\right)}^{2} = 1$

$A = 2 \pi {\int}_{2}^{3} \left(2 - x\right) \sqrt{1 + 1} d x$

$A = 2 \pi {\int}_{2}^{3} \left(2 - x\right) \sqrt{2} \cdot d x$

$A = 2 \sqrt{2} \pi {\int}_{2}^{3} \left(2 - x\right) d x$

$A = 2 \sqrt{2} \pi | 2 x - {x}^{2} / 2 {|}_{2}^{3}$

$A = 2 \sqrt{2} \pi \left[\left(2 \cdot 3 - {3}^{2} / 2\right) - \left(2 \cdot 2 - {2}^{2} / 2\right)\right]$

$A = 2 \sqrt{2} \pi \left[\left(6 - \frac{9}{2}\right) - \left(4 - 2\right)\right]$

$A = 2 \sqrt{2} \pi \left[\frac{3}{2} - 2\right]$

$A = 2 \sqrt{2} \pi \left(- \frac{1}{2}\right)$

$A = - \sqrt{2} \pi$