# What is the surface area of the solid created by revolving f(x) = (x-9)^2 , x in [2,3] around the x axis?

Aug 29, 2017

$\frac{- 867 \pi \sqrt{145}}{8} + \frac{2751 \pi \sqrt{197}}{16} + \frac{\pi}{32} \ln \left(\frac{\sqrt{197} - 14}{\sqrt{145} - 12}\right)$

#### Explanation:

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

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

Here, $f \left(x\right) = {\left(x - 9\right)}^{2}$, so $f ' \left(x\right) = 2 \left(x - 9\right)$. We're working on the interval $x \in \left[2 , 3\right]$. Thus, our surface area is given by:

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

Let $u = x - 9$. This will make what we need to do more obvious by making the integrand not so ugly. Note that $\mathrm{du} = \mathrm{dx}$, so this is an easy substitution. Don't forget to change the bounds by plugging $x = 2$ and $x = 3$ into $u = x - 9$.

$S = 2 \pi {\int}_{- 7}^{- 6} {u}^{2} \sqrt{1 + 4 {u}^{2}} \mathrm{du}$

Now we should use the trigonometric substitution $u = \frac{1}{2} \tan \theta$. This was chosen because $1 + 4 {u}^{2} = 1 + {\tan}^{2} \theta = {\sec}^{2} \theta$. This substitution also implies that $\mathrm{du} = \frac{1}{2} {\sec}^{2} \theta d \theta$. Let's ignore the bounds for now and come back later:

$I = 2 \pi \int {u}^{2} \sqrt{1 + 4 {u}^{2}} \mathrm{du}$

$I = 2 \pi \int \frac{1}{4} {\tan}^{2} \theta \sqrt{1 + {\tan}^{2} \theta} \left(\frac{1}{2} {\sec}^{2} \theta d \theta\right)$

$I = \frac{\pi}{4} \int {\tan}^{2} \theta {\sec}^{3} \theta d \theta$

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

$I = \frac{\pi}{4} \int \left({\sec}^{5} \theta - {\sec}^{3} \theta\right) d \theta$

These are two known integrals. They can be found using iterative integration by parts and are quite cumbersome. I recommend looking them up or committing them to memory, since they're fairly common in trigonometric substitution questions:

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

Then:

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

Let's revert to the variable $u$, which our definite integral is in. Recall that $\tan \theta = 2 u$ so $\sec \theta = \sqrt{1 + {\tan}^{2} \theta} = \sqrt{1 + 4 {u}^{2}}$.

$I = \frac{\pi}{16} {\left(1 + 4 {u}^{2}\right)}^{\frac{3}{2}} \left(2 u\right) - \frac{\pi}{32} \sqrt{1 + 4 {u}^{2}} \left(2 u\right) - \frac{\pi}{32} \ln \left\mid \sqrt{1 + 4 {u}^{2}} + 2 u \right\mid$

$I = \frac{\pi}{8} u {\left(1 + 4 {u}^{2}\right)}^{\frac{3}{2}} - \frac{\pi}{16} u \sqrt{1 + 4 {u}^{2}} - \frac{\pi}{32} \ln \left\mid \sqrt{1 + 4 {u}^{2}} + 2 u \right\mid$

Thus:

$S = {\left[\frac{\pi}{8} u {\left(1 + 4 {u}^{2}\right)}^{\frac{3}{2}} - \frac{\pi}{16} u \sqrt{1 + 4 {u}^{2}} - \frac{\pi}{32} \ln \left\mid \sqrt{1 + 4 {u}^{2}} + 2 u \right\mid\right]}_{- 7}^{- 6}$

Moving through this quickly:

$S = \frac{- 3 \pi}{4} {\left(145\right)}^{\frac{3}{2}} + \frac{3 \pi}{8} \sqrt{145} - \frac{\pi}{32} \ln \left\mid \sqrt{145} - 12 \right\mid - \left(\frac{- 7 \pi}{8} {\left(197\right)}^{\frac{3}{2}} + \frac{7 \pi}{16} \sqrt{197} - \frac{\pi}{32} \ln \left\mid \sqrt{197} - 14 \right\mid\right)$

$S = \pi \sqrt{145} \left(\frac{3}{8} - \frac{3}{4} \left(145\right)\right) + \pi \sqrt{197} \left(\frac{7}{8} \left(197\right) - \frac{7}{16}\right) + \frac{\pi}{32} \ln \left(\frac{\sqrt{197} - 14}{\sqrt{145} - 12}\right)$

$S = \frac{- 867 \pi \sqrt{145}}{8} + \frac{2751 \pi \sqrt{197}}{16} + \frac{\pi}{32} \ln \left(\frac{\sqrt{197} - 14}{\sqrt{145} - 12}\right)$

$S \approx 3481.65496968$