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

Aug 18, 2017

$\frac{\pi}{512} \left(1032 \sqrt{65} - \ln \left(\sqrt{65} + 8\right)\right)$

#### Explanation:

The formula for the surface area of the solid created by rotating a curve $f$ around the $x$-axis on $x \in \left[a , b\right]$ is given by:

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

We have $f \left(x\right) = {\left(2 x - 2\right)}^{2} = 4 {\left(x - 1\right)}^{2}$, with a derivative of $f ' \left(x\right) = 8 \left(x - 1\right)$. So, the surface area in question is:

$A = 2 \pi {\int}_{1}^{2} 4 {\left(x - 1\right)}^{2} \sqrt{1 + {\left(8 \left(x - 1\right)\right)}^{2}} \mathrm{dx}$

$\textcolor{w h i t e}{A} = 8 \pi {\int}_{1}^{2} {\left(x - 1\right)}^{2} \sqrt{64 {\left(x - 1\right)}^{2} + 1} \mathrm{dx}$

Let $x - 1 = \frac{1}{8} \tan \theta$. This implies that $64 {\left(x - 1\right)}^{2} + 1 = {\tan}^{2} \theta + 1 = {\sec}^{2} \theta$. Differentiating, it also implies that $\mathrm{dx} = \frac{1}{8} {\sec}^{2} \theta d \theta$.

Let's momentarily leave out the bounds and return to the bounds once we complete integration.

$A = 8 \pi \int \frac{1}{64} {\tan}^{2} \theta \sqrt{{\tan}^{2} \theta + 1} \left(\frac{1}{8} {\sec}^{2} \theta d \theta\right)$

$\textcolor{w h i t e}{A} = \frac{\pi}{64} \int {\tan}^{2} \theta {\sec}^{3} \theta d \theta$

Using ${\tan}^{2} \theta = {\sec}^{2} \theta - 1$:

$A = \frac{\pi}{64} \int {\sec}^{5} \theta d \theta - \frac{\pi}{64} \int {\sec}^{3} \theta d \theta$

Unfortunately, these integrals are both quite messy, so attached are links to finding both the integrals:

• [Derivation] $\int {\sec}^{3} \theta d \theta = \frac{1}{2} \sec \theta \tan \theta + \frac{1}{2} \ln \left\mid \sec \theta + \tan \theta \right\mid$
• [Derivation] $\int {\sec}^{5} \theta d \theta = \frac{1}{4} {\sec}^{3} \theta \tan \theta + \frac{3}{8} \sec \theta \tan \theta + \frac{3}{8} \ln \left\mid \sec \theta + \tan \theta \right\mid$

Thus, $\int {\sec}^{5} \theta d \theta - \int {\sec}^{3} \theta d \theta = \frac{1}{4} {\sec}^{3} \theta \tan \theta - \frac{1}{8} \sec \theta \tan \theta - \frac{1}{8} \ln \left\mid \sec \theta + \tan \theta \right\mid$ and

$A = \frac{\pi}{64} \left(\frac{1}{4} {\sec}^{3} \theta \tan \theta - \frac{1}{8} \sec \theta \tan \theta - \frac{1}{8} \ln \left\mid \sec \theta + \tan \theta \right\mid\right)$

$\textcolor{w h i t e}{A} = \frac{\pi}{512} \left(2 {\sec}^{3} \theta \tan \theta - \sec \theta \tan \theta - \ln \left\mid \sec \theta + \tan \theta \right\mid\right)$

From $x - 1 = \frac{1}{8} \tan \theta$ we see that $\tan \theta = 8 \left(x - 1\right)$, which implies that $\sec \theta = \sqrt{{\tan}^{2} \theta + 1} = \sqrt{64 {\left(x - 1\right)}^{2} + 1}$.

$A = \frac{\pi}{512} \left(2 \left(8 \left(x - 1\right)\right) {\left(64 {\left(x - 1\right)}^{2} + 1\right)}^{\frac{3}{2}} - 8 \left(x - 1\right) \sqrt{64 {\left(x - 1\right)}^{2} + 1} - \ln \left\mid \sqrt{64 {\left(x - 1\right)}^{2} + 1} + 8 \left(x - 1\right) \right\mid\right)$

Wow. Now apply the bounds from $x = 1$ to $x = 2$. I'm not gonna write out the biggest part since it just takes up too much space:

$A = \frac{\pi}{512} \left(16 \left({65}^{\frac{3}{2}}\right) - 8 \sqrt{65} - \ln \left\mid \sqrt{65} + 8 \right\mid\right) - \frac{\pi}{512} \left(0 - 0 - \ln 1\right)$

$\textcolor{w h i t e}{A} = \frac{\pi}{512} \left(1032 \sqrt{65} - \ln \left(\sqrt{65} + 8\right)\right)$