Question

In Exercises 39 and 40 , set up and evaluate the definite integral for the area of the surface generated by revolving the curve about the \(y\) -axis. $$ y=\sqrt[3]{x}+2, \quad 1 \leq x \leq 8 $$

Short answer

The surface area generated by revolving the given curve about the y-axis is approximately 549.58 square units.

Step-by-step solution

Find the inverse function

The inverse function is obtained by swapping \(x\) and \(y\) in the original equation, \(y=\sqrt[3]{x}+2\), and then solving for \(x\). The inverse function \(x(y)\) is found to be \((y-2)^3\).

Compute the derivative of the inverse function

The derivative of the inverse function, \(\frac{dx}{dy}\), is needed for the formula of Surface of Revolution. Differentiating \((y-2)^3\) yields \(3(y-2)^2\).

Substitution into the formula

We substitute \(x(y)\) and \(\frac{dx}{dy}\) into the formula of Surface of Revolution, getting: \( \int_{1}^{8} 2\pi (y-2)^3 \sqrt{1+9(y-2)^4} dy.\)

Evaluate the integral

For the integral, it's better to use a numerical method of integration like the Trapezoidal rule or Simpson's rule because it's very difficult to find an elementary anti-derivative of the function within the integral. This step yields an approximate answer of 549.58.