# What is the surface area of the solid created by revolving f(x) = x^2-3x+2 , x in [3,4] around the x axis?

Feb 25, 2017

$S A = 103 \setminus u n i {t}^{2}$ (3sf)

#### Explanation:

The surface area of a solid created by revolving $y = f \left(x\right)$ about $O x$ is given by:

$S A = 2 \pi \setminus {\int}_{x = \alpha}^{x = \beta} \setminus f \left(x\right) \setminus \sqrt{1 + {\left(f ' \left(x\right)\right)}^{2}} \setminus \mathrm{dx}$

So we have;

$\setminus \setminus \setminus \setminus \setminus \setminus f \left(x\right) = {x}^{2} - 3 x + 2$
$\therefore f ' \left(x\right) = 2 x - 3$

And so the Surface Area is;

$S A = 2 \pi \setminus {\int}_{3}^{4} \setminus \left({x}^{2} - 3 x + 2\right) \setminus \sqrt{1 + {\left(2 x - 3\right)}^{2}} \setminus \mathrm{dx}$
$\setminus \setminus \setminus \setminus \setminus = 2 \pi \setminus {\int}_{3}^{4} \setminus \left({x}^{2} - 3 x + 2\right) \setminus \sqrt{1 + \left(4 {x}^{2} - 12 x + 9\right)} \setminus \mathrm{dx}$
$\setminus \setminus \setminus \setminus \setminus = 2 \pi \setminus {\int}_{3}^{4} \setminus \left({x}^{2} - 3 x + 2\right) \setminus \sqrt{4 {x}^{2} - 12 x + 10} \setminus \mathrm{dx}$

Although the integral can be established it is quite complex,so I will just quote the result

$S A = \frac{\pi}{32} \left(5 {\sinh}^{-} 1 \left(3\right) - 45 \sqrt{10} - 5 {\sinh}^{-} 1 \left(5\right) + 235 \sqrt{26}\right)$
$\setminus \setminus \setminus \setminus \setminus = 103.426839 \ldots$#