# What is the surface area of the solid created by revolving f(x) = 6x^2-3x+22 , x in [2,3] around the x axis?

Jun 21, 2016

$\frac{14086}{5} \pi$

#### Explanation:

around the x axis, the volume of revolution is

$V = \pi {\int}_{x = 2}^{3} {y}^{2} \setminus \mathrm{dx}$

that looks horrendous and might be made simpler by a shift in axis. completing the square should ultimately make the algebra a little bit easier to bear

so $y \left(x\right) = 6 \left({x}^{2} - \frac{1}{2} x + \frac{22}{6}\right)$
$= 6 \left({\left(x - \frac{1}{4}\right)}^{2} - \frac{1}{16} + \frac{22}{6}\right)$
$= 6 \left({\left(x - \frac{1}{4}\right)}^{2} + \frac{173}{48}\right)$
so $y \left(u\right) = 6 {u}^{2} + \frac{173}{8}$ where $u = x - \frac{1}{4} , \mathrm{du} = \mathrm{dx}$

and the integral becomes

$\pi {\int}_{u = \frac{7}{4}}^{\frac{11}{4}} {y}^{2} \setminus \mathrm{du}$
$= \pi {\int}_{u = \frac{7}{4}}^{\frac{11}{4}} {\left(6 {u}^{2} + \frac{173}{8}\right)}^{2} \setminus \mathrm{du}$
$= \pi {\int}_{u = \frac{7}{4}}^{\frac{11}{4}} 36 {u}^{4} + \frac{519}{2} {u}^{2} + \frac{29929}{64} \setminus \mathrm{du}$
$= \pi {\left(\frac{36}{5} {u}^{5} + \frac{519}{6} {u}^{3} + \frac{29929}{64} u\right)}_{u = \frac{7}{4}}^{\frac{11}{4}}$
$= \frac{14086}{5} \pi$