Question

Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the given lines. \(y=6-2 x-x^{2}, \quad y=x+6\) (a) the \(x\) -axis (b) the line \(y=3\)

Short answer

The volume \( V_a \) and \( V_b \) obtained by integrating and solving in step 3 and step 5, respectively, represent the volume of the solid when the region is revolved around the x-axis and the line \(y=3\).

Step-by-step solution

Find the intersection points

Solve the equation \(6 - 2x - x^2 = x + 6\) for \(x\) which yields \(x = -2\) and \(x = 3\). The region of interest is thus between -2 and 3.

Set up the integral for part (a)

By revolving the region around the x-axis, we can apply the disk method. The radius of each disk is the distance from the x-axis to the curve, i.e., the value of the function itself. Since \(y = x + 6\) is always below \(y = 6 - 2x - x^2\) in the interval from -2 to 3, we take this as the radius. Thus, the volume is given by the integral \( V_a = \pi \int_{-2}^{3} [(6 - 2x - x^2) ^2 - (x+6)^2] dx \).

Solve the integral for part (a)

Compute the integral to solve for \( V_a \).

Set up the integral for part (b)

When revolving around \(y = 3\) we need to use the washer method as there is a hole along the axis of revolution. The outer radius R is the distance from \(y = 3\) to \(y = 6 - 2x - x^2\) and the inner radius r is the distance from \(y = 3\) to \(y = x + 6\). Thus, the volume is given by the integral \( V_b = \pi \int_{-2}^{3} [(3 - (6 - 2x - x^2)) ^2 - (3 - (x+6))^2] dx \).

Solve the integral for part (b)

Compute the integral to solve for \( V_b \).