Question

In Exercises \(35-38\), set up and evaluate the definite integral for the area of the surface generated by revolving the curve about the \(x\) -axis. $$ y=\frac{1}{3} x^{3}, \quad 0 \leq x \leq 3 $$

Short answer

The surface area of the surface generated by revolving the curve \(y=\frac{1}{3}x^{3}\), when \(0 \leq x \leq 3\), about the x-axis is \(81\pi\).

Step-by-step solution

Set Up the Integral

To solve for the surface area of revolution, the formula is: \[ A = 2\pi \int_{a}^{b} y \sqrt{1+ \left(\frac{dy}{dx}\right)^{2}} dx \]Let's first find the derivative of \(y = \frac{1}{3} x^{3}\) which is:\(\frac{dy}{dx} = x^{2}\)Then substitute \(y\) and \(\frac{dy}{dx}\) into the formula, the lower limit \(a\) is 0 and the upper limit \(b\) is 3, which gives us the integral to find the surface area.

Compute the Integral

The set up integral becomes:\[ A = 2\pi \int_{0}^{3} \frac{1}{3} x^{3} \sqrt{1+ (x^{2})^{2}} dx \]By simplifying the integral, we get:\[ A = 2\pi \int_{0}^{3} \frac{1}{3} x^{5} dx \]Then, use the power rule for integration (which says that the integral of x^n dx is \(\frac{1}{n+1}x^{n+1}\)) to evaluate the integral from 0 to 3.

Evaluate the Integral

Evaluate the integral to find the surface area:\[ A = 2\pi [\frac{1}{18} x^{6}]_{0}^{3} \]This simplifies to:\[ A = 2\pi (\frac{1}{18}*3^{6} - 0) \]This simplifies to:\[ A = 2\pi*\frac{729}{18} \]Finally, this simplifies to:\[ A = 81\pi \]