Question

Use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving the plane region about the \(x\) -axis. $$ y=x^{2}, \quad x=0, \quad y=9 $$

Short answer

The volume of the solid generated by revolving the plane region about the x-axis is \(81\pi\).

Step-by-step solution

Identifying the shell's width, height and radius

In the shell method, the shells are parallel to the axis of rotation (the x-axis in this case). The width of the shells is an infinitesimally small change in x, \(dx\). The height of each shell is given by the function, \(y = x^{2}\). The radius is the x-coordinate of the function, \(x\). Hence the volume of an elemental shell is \(2\pi \cdot x \cdot x^{2} \cdot dx\).

Setting up the integral

To find the total volume, we integrate the volume of the elemental shell with respect to \(x\) (since \(dx\) is the width of the shell) from \(x=0\) to \(x=3\) (since \(y=9\) when \(x=3\)). Hence the integral set up is \(\int_{0}^{3} 2\pi \cdot x \cdot x^{2} \cdot dx\).

Evaluating the integral

This integral is evaluated by first simplifying to get \(\int_{0}^{3} 2\pi \cdot x^{3} \cdot dx\), which leads to \(\left[ \frac{1}{2} \cdot 2\pi \cdot x^{4} \right]_{0}^{3} = 81\pi\).