Let \(R\) be the region in the \(xy\)-plane bounded by \(y=0\), \(y=\ln x\), \(y=2\), and \(x=1\). This exercise will walk you through setting up an integral using the Shell Method that will give the volume of the solid generated when \(R\) is revolved about the line \(x=-1\).

Since we are using the Shell Method, the slices must be:

parallel perpendicular
to the axis of rotation.

Slices that are parallel to the axis of rotation \(x=-1\) are:

vertical horizontal

Since the slices are vertical, we must:

integrate with respect to \(x\). integrate with respect to \(y\).

Since we must integrate with respect to \(x\), we will use the result:

\[V = \int _{x=a}^{x=b}2\pi \rho h \d x \]

to set up the volume. We must now find the limits of integration as express the radius \(\rho \) and the height \(h\) in terms of the variable of integration \(x\).

The limits of integration are: \(a= \answer {1}\) and \(b = \answer {e^2}\).

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 horizontal distance, we find \(\rho = x_{right}-x_{left}\).
\(x_{right} = x\) \(x_{right} = \ln (x)\) \(x_{right} = -1\)
\(x_{left} = \ln (x)\) \(x_{left} = 1\) \(x_{left} = -1\)
Remember that the process of find the volume requires us to express the volume of each shell in terms of the \(x\)-value where the shell is located. An arbitrary shell is located at \(x\)!

So, \(\rho = \answer {x-(-1)}\).

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 \(h = y_{top}-y_{bot}\).
\(y_{top} = 2\) \(y_{top} = \ln (x)\)
\(y_{bot} =2\) \(y_{bot} = \ln (x)\)

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

Using

\[V = \int _{x=a}^{x=b} 2\pi \rho h \d x, \]
we find that an integral that gives the volume of the solid of revolution is:
\[ V= \int _{x=\answer {1}}^{x=\answer {e^2}} \answer {2\pi (x+1)(2-\ln (x))}\d x \]