# What is the surface area of the solid created by revolving f(x) = 2x^2-x+5 , x in [1,2] around the x axis?

Jan 8, 2018

$S A \setminus \approx 181.265$

#### Explanation:

The surface area of the solid created by revolving a positive function, $f \left(x\right)$, for $x \in \left[a , b\right]$ around the x-axis is always

$\setminus {\int}_{a}^{b} 2 \pi f \left(x\right) \sqrt{1 + f ' {\left(x\right)}^{2}} \mathrm{dx}$

The derivation of this is fairly simple. It involves the surface area of a frustrum which is $2 \pi r l$. I would highly recommend looking up the derivation, as it make this formula make a lot more sense. This link is pretty good, http://tutorial.math.lamar.edu/Classes/CalcII/SurfaceArea.aspx

Anyways, in the case of this problem, the first thing to notice is that $f \left(x\right)$ is negative over the interval indicated.

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

$= \setminus {\int}_{1}^{2} 2 \pi \left(2 {x}^{2} - 2 x + 5\right) \sqrt{1 + {\left(4 x - 2\right)}^{2}} \mathrm{dx}$

at this point I would use a calculator to find the value of this integral which would end up being the answer.

$S A \setminus \approx 181.265$