# What is the surface area of the solid created by revolving f(x) = x^3 , x in [1,2] around the x axis?

Feb 25, 2017

$S = 2 \pi {\int}_{1}^{2} {x}^{3} \sqrt{1 + 9 {x}^{4}} \mathrm{dx} = \frac{\pi}{27} \left({145}^{\frac{3}{2}} - {10}^{\frac{3}{2}}\right) \approx 199.4805$

#### Explanation:

The surface area of a solid created by a function revolved the $x$-axis on $x \in \left[a , b\right]$ is:

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

Here we see that $f \left(x\right) = {x}^{3}$ and $f ' \left(x\right) = 3 {x}^{2}$ so:

$S = 2 \pi {\int}_{1}^{2} {x}^{3} \sqrt{1 + 9 {x}^{4}} \mathrm{dx}$

We can integrate this. Let $u = 1 + 9 {x}^{4}$ so $\mathrm{du} = 36 {x}^{3} \mathrm{dx}$. Due to this, we need to multiply the interior of the integral by $36$ and the exterior by $1 / 36$.

$S = \frac{2 \pi}{36} {\int}_{1}^{2} 36 {x}^{3} \sqrt{1 + 9 {x}^{4}} \mathrm{dx}$

Before going from $x$ to $u$, we will have to transform the bounds. $x = 1$ becomes $u = 1 + 9 \left({1}^{4}\right) = 10$. $x = 2$ becomes $u = 1 + 9 \left({2}^{4}\right) = 145$.

$S = \frac{\pi}{18} {\int}_{10}^{145} \sqrt{u} \mathrm{du}$

$S = \frac{\pi}{18} {\int}_{10}^{145} {u}^{\frac{1}{2}} \mathrm{du}$

$S = \frac{\pi}{18} {\left[\frac{2}{3} {u}^{\frac{3}{2}}\right]}_{10}^{145}$

$S = \frac{\pi}{18} \left(\frac{2}{3} {\left(145\right)}^{\frac{3}{2}} - \frac{2}{3} {\left(10\right)}^{\frac{3}{2}}\right)$

$S = \frac{\pi}{27} \left({145}^{\frac{3}{2}} - {10}^{\frac{3}{2}}\right) \approx 199.4805$