Question

The volume of the solid generated by revolving the region bounded by the parabola \[ x = 2y^2 + 4 \quad \text{and the line} \quad x = 6 \quad \text{about the line} \quad x = 6 \] is

• A. \( \frac{78\pi}{15} \)
• B. \( \frac{91\pi}{15} \)
• C. \( \frac{64\pi}{15} \)
• D. \( \frac{117\pi}{15} \)

Hint:

For volumes of solids of revolution, use the disk method or washer method, depending on whether the region is revolved around the axis.

Answer 1

The Correct Option is C

Step 1: Set up the formula for the volume of revolution.
The volume of the solid generated by revolving a region about a vertical line is given by: \[ V = \pi \int_{y_1}^{y_2} \left[ R(y)^2 - r(y)^2 \right] \, dy \] where \( R(y) \) and \( r(y) \) are the outer and inner radii, respectively, at each point along the axis of revolution. Step 2: Define the radii.
For the given problem, the outer radius is \( R(y) = 6 - (2y^2 + 4) \), and the inner radius is \( r(y) = 6 - 6 = 0 \). Step 3: Set up the integral.
Thus, the volume is: \[ V = \pi \int_{y_1}^{y_2} \left[ (6 - 2y^2 - 4)^2 - 0^2 \right] \, dy \] where the limits \( y_1 \) and \( y_2 \) are determined by solving for the intersection points of the parabola and the line, i.e., \( 2y^2 + 4 = 6 \). Step 4: Solve the integral.
After solving the integral, we get: \[ V = \frac{64\pi}{15} \]