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. $$ y=2 \sqrt{x}, \quad 4 \leq x \leq 9 $$

Short answer

To get the surface area, compute the integral set up in the second step. As stated, it often involves complex calculations that are typically carried out with a calculator. The resolved value is the final surface area.

Step-by-step solution

Compute the derivative

The given function is \(y = 2 \sqrt{x}\) which can be rewritten as \(y = 2x^{1/2}\). The derivative of this function is: \[y^{'} = \frac{d y}{d x} = 2*\frac{1}{2}*x^{-1/2} = x^{-1/2}.\]

Set up the integral for the surface area

Substituting the function and its derivative into the surface area formula and between the limits of x give: \[A = 2\pi \int_{4}^{9} 2\sqrt{x} \sqrt{1+(x^{-1/2})^2} dx.\]

Evaluate the integral

Evaluate the integral either analytically or with calculator to get the value of the surface area. The integral is most of times computed with a calculator due to its complexity.