Question

In Exercises 23-28, find the volume of the solid generated by revolving the region about the y-axis. the region enclosed by $$x=y^{3 / 2}, x=0, y=2$$

Short answer

The volume of the solid is \(4\pi\) cubic units.

Step-by-step solution

Set Up the Integral

The integral will be from 0 to 2 because these are the limits of y. We will square the equation \(x = y^{3 / 2}\) and multiply by \(\pi\) to find the volume. So, the integral is \(\int_{0}^{2} \pi * (y^{3 / 2})^2 dy\).

Simplify the Integral

The integral simplifies to \(\int_{0}^{2} \pi * y^3 dy\).

Evaluate the Integral

The antiderivative of \(y^3\) is \(\frac{1}{4}y^4\). So, the integral becomes \(\frac{1}{4} \pi * y^4 |_{0}^{2}\)

Calculate the Result

The final result of the integral is \(\frac{1}{4} \pi * 2^4 - \frac{1}{4} \pi * 0^4 = \frac{1}{4} \pi * 16 = 4\pi\).