# What is the surface area of the solid created by revolving f(x) = 3x, x in [2,5] around the x axis?

May 24, 2016

$A = 63 \sqrt{10} \pi$

#### Explanation:

$A = 2 \pi {\int}_{2}^{5} f \left(x\right) \cdot {\sqrt{1 + \frac{d f \left(x\right)}{d x}}}^{2} \cdot d x$

$f \left(x\right) = 3 x$

$\frac{d}{d x} f \left(x\right) = 3$

$A = 2 \pi {\int}_{2}^{5} 3 x \cdot \sqrt{1 + {3}^{2}} \cdot d x$

$A = 2 \pi {\int}_{2}^{5} 3 x \sqrt{10} \cdot d x$

$A = 6 \sqrt{10} \pi {\int}_{2}^{5} x \cdot d x$

$A = 6 \sqrt{10} {\left[\frac{1}{2} {x}^{2}\right]}_{2}^{5}$

$A = 6 \sqrt{10} \pi \left[\left(\frac{1}{2} \cdot {5}^{2} - \frac{1}{2} \cdot {2}^{2}\right)\right]$

$A = 6 \sqrt{10} \pi \left[\frac{25}{2} - \frac{4}{2}\right]$

$A = 6 \sqrt{10} \pi \left[\frac{21}{2}\right]$

$A = 63 \sqrt{10} \pi$