A solid of revolution is formed by revolving this region about the \(y\)-axis.

Let’s try to apply the “Slice, Approximate, Integrate” procedure to find the volume of the solid.

Step 1: Slice The geometry of the base region suggests that it is advantageous to use horizontal slices, so we should integrate with respect to \(x\)\(y\).

We draw a slice of thickness \(\Delta y\) at a fixed, but unspecified \(y\)-value on the region of revolution.

Step 2: Approximate We approximate the slice in the region by a rectangle.

The result of rotating the slice appears on the solid just as before.

Notice that the outer radius and inner radius are finite, but the thickness \(\Delta y\) is thought of as quite small. To find the volume of the hollow cylinder, recall

\begin{align*} \left < \textrm {Volume of the hollow cylinder } \right > &= \\ \left < \textrm { Volume of the outer } \right . & \left . \textrm {cylinder } \right > - \left < \textrm { Volume of the inner cylinder } \right >. \end{align*}

The outer cylinder has radius \(R(y)\) and its volume is \(\Delta V_{outer} = \pi [R(y)]^2 \Delta y\), while the volume of the inner cylinder has radius \(r(y)\), so its volume is \(\Delta V_{inner} = \pi [r(y)]^2 \Delta \). Here, we have explicitly noted that these radii will certainly depend at which \(y\)-value they are drawn, but to make the notation cleaner the rest of the way, we will only write \(R\) and \(r\) instead of \(R(y)\) and \(r(y)\). The volume of our hollow cylinder is thus

Our goal is now to express both \(R\) and \(r\) in terms of the unspecified \(y\)-value of the slice. Using the picture is helpful to think about how to find \(R\) and \(r\).

• The outer radius \(R\) is the distance from the axis of rotation to the outer curveinner curve.
• The inner radius \(r\) is the distance from the axis of rotation to the outer curveinner curve.

Also, both of these distances are horizontalvertical distances.

We now label these on the image of the region of rotation.

We can find both \(R\) and \(r\) now the way we always find horizontal distances.

For the outer radius \(R\):

• The righthand curve is given by \(x_{right} = \sqrt {y+4}\)\(x_{right} = 1\)\(x_{right} = 0\)
• The lefthand curve is given by \(x_{left} = \sqrt {y+4}\)\(x_{left} = 1\)\(x_{left} = 0\)

Thus \(R = x_{right}-x_{left} = \answer [given]{\sqrt {y+4}-0}\).

Following similar logic for the inner radius \(r\) gives \(r = x_{right}-x_{left} = \answer [given]{1-0}\).

The volume of our single approximate slice is thus

and the approximate total volume using \(n\) slices is found by adding up the volumes of each slice, so

where \(y_k\) is the \(y\)-value chosen for the \(k\)-th slice and \(\Delta y_k\) is the thickness of the slice.

Step 3: Integrate In order to find the exact volume, we simultaneously must shrink the width of our slices while adding all of the volumes together. As usual, the definite integral allows us to do this, and we may write