Question

Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the given lines. \(y=2 x^{2}, \quad y=0, \quad x=2\) (a) the \(y\) -axis (b) the \(x\) -axis (c) the line \(y=8\) (d) the line \(x=2\)

Short answer

The volumes of the solids generated by revolving the region around the y-axis, the x-axis, the line \(y = 8\), and the line \(x = 2\), are respectively \(\frac{4\pi}{3}\), \(\frac{32\pi}{5}\), \(\frac{192\pi}{5}\), and \(8\pi\).

Step-by-step solution

Sketch the region

First, sketch the parabola \(y = 2x^2\) from \(x = 0\) to \(x = 2\) and the x-axis.

Volume around the y-axis

Next, use the disk method to calculate the volume of the solid generated by revolving the region around the y-axis. The volume is given by \(V = \pi \int_{a}^{b}[g(x)]^2 dx\). Here, \(g(x) = \sqrt{y/2}\), \(a = 0\) and \(b = 8\). So, \(V = \pi \int_{0}^{8}(\sqrt{y/2})^2 dy = \frac{4\pi}{3}\).

Volume around the x-axis

Now, use the disk method to calculate the volume of the solid when the region is revolved about the x-axis. In this case, \(g(x) = 2x^2\), \(a = 0\) and \(b = 2\). So, \(V = \pi \int_{0}^{2}(2x^2)^2 dx = \frac{32\pi}{5}\).

Volume around the line y=8

To find the volume of the solid when the region is revolved about the line \(y = 8\), use the washer method. The volume is \(V = \pi \int_{a}^{b}([R(x)]^2-[r(x)]^2) dx\), where \(R(x) = \sqrt{y/2}\) and \(r(x) = 2x^2\). Here, \(R(x) = 8\) and \(r(x) = 2x^2\), and \(a = 0\) and \(b = 2\). The volume is \(V = \pi \int_{0}^{2}[(8)^2 - (2x^2)^2] dx = \frac{192\pi}{5}\).

Volume around the line x=2

Finally, use the washer method for the solid obtained by rotating the region about the line \(x = 2\). Now, \(R(x) = 2 - 0 = 2\), and \(r(x) = 2 - x\), and \(a = 0\) and \(b = 8\), yielding \(V = \pi \int_{0}^{8}[(2)^2 - (2 - x)^2] dy = 8\pi\).