Mobile Notice
You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best viewed in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (you should be able to scroll/swipe to see them) and some of the menu items will be cut off due to the narrow screen width.
Section 6.4 : Volume With Cylinders
For each of the following problems use the method of cylinders to determine the volume of the solid obtained by rotating the region bounded by the given curves about the given axis.
- Rotate the region bounded by \(x = {\left( {y - 2} \right)^2}\), the \(x\)-axis and the \(y\)-axis about the \(x\)-axis. Solution
- Rotate the region bounded by \(\displaystyle y = \frac{1}{x}\), \(\displaystyle x = \frac{1}{2}\), \(x = 4\)and the \(x\)-axis about the \(y\)-axis. Solution
- Rotate the region bounded by \(y = 4x\) and \(y = {x^3}\) about the \(y\)-axis. For this problem assume that \(x \ge 0\). Solution
- Rotate the region bounded by \(y = 4x\) and \(y = {x^3}\) about the \(x\)-axis. For this problem assume that \(x \ge 0\). Solution
- Rotate the region bounded by \(y = 2x + 1\), \(y = 3\) and \(x = 4\) about the line \(y = 10\). Solution
- Rotate the region bounded by \(x = {y^2} - 4\) and \(x = 6 - 3y\) about the line \(y = - 8\). Solution
- Rotate the region bounded by \(y = {x^2} - 6x + 9\) and \(y = - {x^2} + 6x - 1\) about the line \(x = 8\). Solution
- Rotate the region bounded by \(\displaystyle y = \frac{{{{\bf{e}}^{\frac{1}{2}x}}}}{{x + 2}}\), \(\displaystyle y = 5 - \frac{1}{4}x\), \(x = - 1\) and \(x = 6\) about the line \(x = - 2\). Solution