Question

Use the shell method to find the volume of the solid generated by revolving the plane region about the given line. $$ y=x^{2}, \quad y=4 x-x^{2}, \text { about the line } x=2 $$

Short answer

The volume of the solid generated by revolving the given region about the line \(x = 2\) is \(\frac{16\pi}{3}\) units cubed.

Step-by-step solution

Define the Radius and the Height of the Shell

For the shell method, we have to define the radius \(r(x)\) which is the horizontal distance from the cylindrical shell to the line \(x=2\). The radius \(r(x)\) will be \(2 - x\). The heigh \(h(x)\) of the shell will be the difference between the two functions \(4x - x^2 - x^2 = 4x - 2x^2\).

Determine the Limits of Integration

The limits of integration will correspond to where the two curves intersect. They intersect when \(x^2 = 4x - x^2\). Solving for \(x\), we get \(x = 0\) and \(x = 2\). So, the limits of integration will be from 0 to 2.

Set Up and Compute the Integral

Now that we have \(r(x)\), \(h(x)\), and the limits of integration (a and b), we can substitute these into the volume formula. So, we have \(V = 2\pi \int_0^2 (2 - x)(4x - 2x^2) dx\). Calculating this integral and multiplying by 2π gives us the volume.

Compute the Integral

Computing this integral gives \(V = \frac{16\pi}{3}\).