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

Feb 5, 2017

$2 \pi \left[\frac{1}{8} {17}^{\frac{3}{2}} - \frac{1}{16} \sqrt{17} - \frac{1}{64} \ln | \sqrt{17} + 4 | - \frac{1}{16} {5}^{\frac{3}{2}} + \frac{1}{32} \sqrt{5} + \frac{1}{64} \ln | \sqrt{5} + 2 |\right]$

#### Explanation:

Now, integrating by parts $\int {\sec}^{3} \theta = \int \sec \theta {\sec}^{2} \theta d \theta$

=$\sec \theta \int {\sec}^{2} \theta d \theta - \int \sec \theta \tan \theta \int {\sec}^{2} \theta d \theta$

=$\sec \theta \tan \theta - \int \sec \theta {\tan}^{2} \theta d \theta$

=$\sec \theta \tan \theta - \int {\sec}^{3} \theta d \theta + \int \sec \theta d \theta$

Now transposing $\int {\sec}^{3} \theta d \theta$to the left side and integrating $\sec \theta d \theta$,

$2 \int {\sec}^{3} \theta d \theta = \sec \theta \tan \theta + \ln | \sec \theta + \tan \theta |$

Thus $\int {\sec}^{3} \theta d \theta = \frac{1}{2} \sec \theta \tan \theta + \frac{1}{2} \ln | \sec \theta + \tan \theta |$

Like wise, using technique of integration by parts $\int {\sec}^{5} \theta d \theta = \int {\sec}^{3} \theta {\sec}^{2} \theta d \theta = {\sec}^{3} \theta \int {\sec}^{2} \theta d \theta - \int \left(3 {\sec}^{2} \theta \sec \theta \tan \theta\right) \int \left({\sec}^{2} \theta\right) d \theta$

=${\sec}^{3} \theta \tan \theta - 3 \int {\sec}^{3} \theta {\tan}^{2} \theta d \theta$

=${\sec}^{3} \theta \tan \theta - 3 \int \left({\sec}^{5} \theta - {\sec}^{3} \theta d \theta\right)$

Now transpose $\int {\sec}^{5} \theta d \theta$ to the right side to get

$\int {\sec}^{5} \theta d \theta = \frac{1}{4} {\sec}^{3} \theta \tan \theta + \frac{3}{4} \int {\sec}^{3} \theta d \theta$

Now using the integral of ${\sec}^{3} \theta$ derived earlier,

$\int {x}^{2} \sqrt{4 {x}^{2} + 1} \mathrm{dx} = \frac{1}{8} \left[\int {\sec}^{5} \theta d \theta - \int {\sec}^{3} \theta d \theta\right] = \frac{1}{8} \left[\frac{1}{4} {\sec}^{3} \theta \tan \theta - \frac{1}{4} \int {\sec}^{3} \theta d \theta\right]$

=$\frac{1}{32} \left[{\sec}^{3} \theta \tan \theta - \frac{1}{2} \sec \theta \tan \theta - \frac{1}{2} \ln | \sec \theta + \tan \theta |\right]$

Now substituting theta by x it would be $\tan \theta = 2 x$ and $\sec \theta = \sqrt{4 {x}^{2} + 1}$

$\int {x}^{2} \sqrt{4 {x}^{2} + 1} \mathrm{dx} = \frac{1}{16} x {\left(4 {x}^{2} + 1\right)}^{\frac{3}{2}} - \frac{1}{32} x \sqrt{4 {x}^{2} + 1} - \frac{1}{64} \ln | \sqrt{4 {x}^{2} + 1} + 2 x |$

The required surface area would thus be

$2 \pi {\left[\frac{1}{16} x {\left(4 {x}^{2} + 1\right)}^{\frac{3}{2}} - \frac{1}{32} x \sqrt{4 {x}^{2} + 1} - \frac{1}{64} \ln | \sqrt{4 {x}^{2} + 1} + 2 x |\right]}_{1}^{2}$

=$2 \pi \left[\frac{1}{8} {17}^{\frac{3}{2}} - \frac{1}{16} \sqrt{17} - \frac{1}{64} \ln | \sqrt{17} + 4 | - \frac{1}{16} {5}^{\frac{3}{2}} + \frac{1}{32} \sqrt{5} + \frac{1}{64} \ln | \sqrt{5} + 2 |\right]$

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