Question:

Consider the region bounded by the graph of the function $f(x) = 2x^2$ and the x-axis on the interval $[-1, 1]$.

a) Find the volume of the solid generated when this region is revolved about the x-axis.

b) Find the volume of the solid generated when this region is revolved about the y-axis.

Answer:

a) To find the volume of the solid generated when the given region is revolved about the x-axis, we can use the disk method. The volume of a solid generated by revolving a region bounded by the graph of a function $f(x)$ and the x-axis on an interval $[a, b]$ is given by:

In this case, the function is $f(x) = 2x^2$ and the interval is $[-1, 1]$. Thus, the volume can be expressed as:

Simplifying the integral:

Integrating:

Evaluating the definite integral:

Simplifying:

Therefore, the volume of the solid generated when the given region is revolved about the x-axis is 0.

b) To find the volume of the solid generated when the given region is revolved about the y-axis, we need to modify the formula. The modified formula for finding volume using the disk method is:

Since we are revolving around the y-axis, we need to find the inverse function of $f(x) = 2x^2$ to express it in terms of $y$. Solving for $x$:

The inverse function is $f^{-1}(y) = \sqrt{\frac{y}{2}}$.

Now, let's express the volume using the modified formula:

Simplifying the integral:

Integrating:

Evaluating the definite integral:

Simplifying:

Therefore, the volume of the solid generated when the given region is revolved about the y-axis is $4\pi$.