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

Aug 26, 2017

$S \approx 4.46$

#### Explanation:

The first step is to check to make sure that $f \left(x\right)$ is defined on the interval $\left[1 , 2\right]$. The only discontinuity of the the function is at $x = 0$, so we do not have to concern ourselves with this.

The surface area of a curve on $\left[a , b\right]$ rotated around the x-axis is defined by $S = 2 \pi {\int}_{a}^{b} f \left(x\right) \sqrt{1 + {\left(f ' \left(x\right)\right)}^{2}} \mathrm{dx}$. The derivative of $f \left(x\right)$ is $f ' \left(x\right) = - \frac{2}{x} ^ 3$. This means that ${\left(f ' \left(x\right)\right)}^{2} = {\left(- \frac{2}{x} ^ 3\right)}^{2} = \frac{4}{x} ^ 6$.

$S = 2 \pi {\int}_{1}^{2} \frac{1}{x} ^ 2 \sqrt{1 + \frac{4}{x} ^ 6} \mathrm{dx}$

$S = 2 \pi {\int}_{1}^{2} \frac{1}{x} ^ 2 \sqrt{\frac{{x}^{6} + 4}{x} ^ 6} \mathrm{dx}$

$S = 2 \pi {\int}_{1}^{2} \frac{1}{x} ^ 5 \sqrt{{x}^{6} + 4} \mathrm{dx}$

This integral has no elementary solution.

But according to Wolfram Alpha, the surface area is approximately $4.46$ square units.

Hopefully this helps!