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. $$ y=9-x^{2}, \quad 0 \leq x \leq 3 $$

Short answer

The area of the surface of revolution is computed by evaluating the integral from Step 3. The result would depend on the particular integration technique employed, and might involve special functions in case of non-elementary antiderivative.

Step-by-step solution

Find the Derivative

First, differentiate the function \(y = 9 - x^2\) with respect to \(x\). Doing so, we get \[\frac{dy}{dx}= -2x\]However, we need \(\frac{dx}{dy}\) for the formula. Hence, we invert the derivative to get \[\frac{dx}{dy}= -\frac{1}{2x}\]

Set up the Integral

Next, set up the integral using the surface area formula for revolution about the y-axis. The definite integral will be set from \(y = 0\) to \(y = 9\) based on the given \(x\) values and the provided equation, and the formula \(2\pi x\sqrt{1+(\frac{dx}{dy})^2}\) is plugged in. This gives us:\[\int_{0}^{9} 2\pi x\sqrt{1+\left(-\frac{1}{2x}\right)^{2}} dy\]

Evaluate the Integral

Perform the integration. This will involve simplifications and probably a substitution, and could be quite complicated. The result will be the required area of the surface of revolution.