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. $$ y=x^{3}, \quad x=0, \quad y=8 $$

Short answer

The volume of the solid is \(\frac{64 \pi}{5}\) cubic units.

Step-by-step solution

Identify the Regions

The plane region is bound by the curve \(y = x^{3}\), the line \(x = 0\) (the y-axis), and horizontally by \(y = 8\), which is the point where \(x = 2\) on the curve \(y = x^{3}\). So, the solid is rotated about the \(x\)-axis from \(x = 0\) to \(x = 2\).

Set Up the Integral for Volume

For the shell method, the height of the shell \(h(x)\) is given by 'y' from the curve \(y = x^{3}\), and the radius 'r(x)' from the \(x\)-axis is \(x\). Thus the volume can be given by \[V = 2 \pi \int_{0}^{2} x \cdot x^{3} dx\]

Evaluate the Integral

To evaluate the integral, first simplify inside the integral to \[V = 2 \pi \int_{0}^{2} x^{4} dx.\] Using the power rule for integration, the integral of \(x^{4}\) is \(\frac{1}{5}x^{5}\). Evaluate this from \(0\) to \(2\), which gives us: \[V = 2 \pi \left[ \frac{1}{5} \cdot (2^{5}) - \frac{1}{5} \cdot (0^{5}) \right].\]

Simplify

On simplifying, we get: \[V = 2 \pi \left[ \frac{32}{5} - 0 \right] = 2 \pi \cdot \frac{32}{5} = \frac{64 \pi}{5}.\]