Questions: Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the x-axis y = x^2 + 2 y = -x^2 + 2x + 6 x = 0 x = 3

Solution Steps

To find the volume of the solid generated by revolving the region bounded by the given equations about the x-axis, we can use the method of disks or washers. The volume \( V \) is given by the integral of the area of the cross-sectional disks or washers along the axis of revolution. Here, we will use the washer method since we have two functions defining the region.

1. Identify the outer radius \( R(x) \) and inner radius \( r(x) \) of the washers.
2. Set up the integral for the volume using the formula: \[ V = \pi \int_{a}^{b} \left[ R(x)^2 - r(x)^2 \right] \, dx \]
3. Evaluate the integral from \( x = 0 \) to \( x = 3 \).

We are given the functions: \[ y = x^2 + 2 \] \[ y = -x^2 + 2x + 6 \] and the limits of integration: \[ x = 0 \] \[ x = 3 \]

To find the volume of the solid generated by revolving the region bounded by these functions about the \( x \)-axis, we use the washer method. The volume \( V \) is given by: \[ V = \pi \int_{a}^{b} \left[ R(x)^2 - r(x)^2 \right] \, dx \] where \( R(x) \) is the outer radius and \( r(x) \) is the inner radius.

For our functions: \[ R(x) = -x^2 + 2x + 6 \] \[ r(x) = x^2 + 2 \]

The integrand for the volume is: \[ \pi \left[ (-x^2 + 2x + 6)^2 - (x^2 + 2)^2 \right] \]

We evaluate the definite integral from \( x = 0 \) to \( x = 3 \): \[ V = \pi \int_{0}^{3} \left[ (-x^2 + 2x + 6)^2 - (x^2 + 2)^2 \right] \, dx \]

After evaluating the integral, we find: \[ V = 15\pi \]

Final Answer