Let \(R\) be the region in the \(xy\)-plane bounded between \(y=x^2-1\) and \(y=1\). This exercise gives practice using the Shell Method to set up an integral that gives the volume of the solid of revolution obtained when \(R\) is rotated about several different axes.

When using the Shell Method, the slices must be:

parallel to the axis of rotation. perpendicular to the axis of rotation.
Suppose that \(R\) is revolved about the \(x\)-axis. To use the Shell Method:
the slices must be vertical. We should integrate with respect to \(x\). the slices must be vertical. We should integrate with respect to \(y\). the slices must be horizontal. We should integrate with respect to \(x\). the slices must be horizontal. We should integrate with respect to \(y\).

We thus have a helpful version of the picture of the region \(R\) below:

We see from the picture that \(\rho \) is a:

vertical distance horizontal distance
Since \(\rho \) is the distance from the axis of rotation to the slice, and this is a vertical distance, we find \(\rho = y_{top}-y_{bot}\).
\(y_{top} = 2\) \(y_{top} = y\) \(y_{top} = 1\) \(y_{top} = x^2-1\)
\(y_{bot} = 2\) \(y_{bot} = y\) \(y_{bot} = 1\) \(y_{bot} = x^2-1\)
Remember that the process of find the volume requires us too express the volume of each shell in terms of the \(y\)-value where the shell is located. An arbitrary shell is located at \(y\)!

So, \(\rho = \answer {2-y}\).

We see from the picture that \(h\) is a:

vertical distance horizontal distance
Since \(h\) is the height of the slice, and this is a vertical distance, we find \(h = x_{right}-x_{left}\).
\(x_{right} = \sqrt {y+1}\) \(x_{right} = -\sqrt {y+1}\)
\(x_{left} = \sqrt {y+1}\) \(x_{left} = -\sqrt {y+1}\)

So, \(h= \answer {2\sqrt {y+1}}\).

Using the formula \(V = \int _{y=c}^{y=d} 2\pi \rho h \d y\), an integral that gives the volume of this solid of revolution is:
\[ V = \int _{y=\answer {-1}}^{y=\answer {1}} \answer {4 \pi (2-y)\sqrt {y+1}} \d y \]

Evaluating this, we find the volume is \(\answer { \frac {48 \sqrt {2}}{5} \pi }\) cubic units.

Make the substitution \(u=y+1\) and do some algebra!
Suppose that \(R\) is revolved about the line \(x=3\).
the slices must be vertical. We should integrate with respect to \(x\). the slices must be vertical. We should integrate with respect to \(y\). the slices must be horizontal. We should integrate with respect to \(x\). the slices must be horizontal. We should integrate with respect to \(y\).

We thus have a helpful version of the picture of the region \(R\) below:

We see from the picture that \(h\) is a:

vertical distance horizontal distance
Since \(h\) is the distance from the axis of rotation to the outer curve, and this is a vertical distance, we find \(h = y_{top}-y_{bot}\).
\(y_{top} = 1\) \(y_{top} = x^2-1\) \(y_{top} = y\) \(y_{top} = x\)
\(y_{bot} = 1\) \(y_{bot} = x^2-1\) \(y_{bot} = y\) \(y_{bot} = x\)

So, \(h= \answer {2-x^2}\).

We see from the picture that \(\rho \) is a:

vertical distance horizontal distance
Since \(\rho \) is the distance from the axis of rotation to the outer curve, and this is a horizontal distance, we find \(\rho = x_{right}-x_{left}\).
\(x_{right} = x\) \(x_{right} = 3\) \(x_{right} = x^2-1\)
\(x_{left} = x\) \(x_{left} = 3\) \(x_{left} = x^2-1\)

So, \(\rho = \answer {3-x}\).

Using the formula \(V = \int _{x=a}^{x=b} 2 \pi \rho h \d x\), an integral that gives the volume of this solid of revolution is:
\[ V = \int _{x=\answer {-\sqrt {2}}}^{x=\answer {\sqrt {2}}} 2\pi \answer {(3-x)(2-x^2)} \d x \]