Question

Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the line \(y=4\). $$ y=\frac{1}{2} x^{3}, \quad y=4, \quad x=0 $$

Short answer

By performing the definite integration described in the steps, we find the volume of the solid to be \(V = \frac{256\pi}{5}\) units^3.

Step-by-step solution

Identify the Bounds

We set the equation \(y = \frac{1}{2}x^3\) equal to \(y = 4\), and then solve for x to find the bounds of the integration. We get \(x = \sqrt[3]{8}\) and \(x = 0\) as the limits of our integral.

Express the Disk Radii

We will revolve our region about \(y = 4\), which means the radius of each disk will be \(4 - y\). We then substitute \(y = \frac{1}{2}x^3\) into the equation to get \(4 - \frac{1}{2}x^3\) as the radius.

Express the Disk Areas

Each disk will have the area \(\pi r^2\). Substitute our expression for the radius into this formula to get the area of each disk as \(\pi (4 - \frac{1}{2}x^3)^2 \).

Find the Volume

We add up all the disk areas from \(x = 0\) to \(x = \sqrt[3]{8}\) using the formula for the volume of a solid of revolution. So, Volume \(V = \int^{sqrt[3]{8}}_0 \pi (4 - \frac{1}{2}x^3)^2 dx\). Carry out this calculus integration to get the solid's volume.