# Solid of Revolution – Finding Volume by Rotation

Finding the volume of a solid revolution is a method of calculating the volume of
a 3D object formed by a rotated area of a 2D space. Finding the volume is much like

finding the area
, but with an added component of rotating the area around
a line of symmetry – usually the x or y axis.

(1) Recall finding the area under a curve. Find the area of the definite
integral

Integrate across [0,3]:

Now, let’s rotate this area 360 degrees around the x axis. We will have a 3D solid
that looks like this:

To find this volume, we could take vertical slices of the solid (each dx wide and
f(x) tall) and add them up.

This is quite tedious, but thankfully we have calculus! Since the integrated area
is being rotated around the axis under the curve, we can use disk integration
to find the volume. Since the area is rotated full circle, we can use the formula
for area of a cylinder to find our volume.

Volume of a cylinder

We can merge the formula for volume of a cylinder and our definite integral to find
the volume of our solid. The radius for our cylinder would be the function f(x)
and the height of our cylinder would be the distance of each disk: dx

The volume of each slice would be

Adding the volumes of the disks with infinitely small dx would give us the formula

Using our function, we would get this integrand for the volume

Evaluating the integral, we get

We obtained 4Π units3 as our volume. Since our function is linear
and the radius is changing at a constant rate, it is easy to check this by plugging
in values to the formula for volume of a cone.

The answers are the same. Since our function was linear and shaped like a cone when
rotated around the x axis, it was okay to use the volume formula for a cone. Many
of the volumes we will be working with are not shaped like cone, so we cannot simply
substitute values in the formula. While algebra can take care of the nice straight
lines, calculus takes care of the not-so-nice curves.

## Volume by Rotation Examples

(2) Now lets try rotating the same area around the y axis.

The first rotated solid was integrated in terms x to find the area and rotated around
the x axis. Similarly, this solid is also integrated in terms of x for the
area, but it is now rotated around the y axis. Notice that this solid can
be obtained by subtracting a cone with radius 3 at y = 2 from the cylinder formed
from radius 3 and a height of 2.

Volume
of the Cylinder – Volume of the Cone

= area revolved around the y axis.

There are three ways to find this volume. We can do this by (a) using volume
formulas for the cone and cylinder, (b) integrating two different solids
and taking the difference, or (c) using shell integration (rotating
an area around a different axis than the axis the area touches). Let’s try all three
methods.

(a) Using the volume formulas, we would have

The radius for the cylinder and the cone would be 3 and the height would be 2.

The volume is 12Π units3. Let’s check it with integration.

(b) When integrating, we find the area from the curve to an axis. Since we
are revolving around the y axis, we need to integrate with respect to y. For the
Cylinder, our area before it is rotated would look like this:

The function of y is f(y) = 3 from [0,2]. Now we can set up our integral.

Now on to the cone. Since it is rotated around the y axis, we need to integrate
the original function with respect to y. All we have to do is solve our original
function for x instead of y, making it a function of y. The function of y would
look like this:

The function of y is f(y) = (32)y from [0,2]. Let’s set up our integral.

Now we subtract the volume of the cone from the volume of the cylinder. We get the

Finally, let’s carry out shell integration.

(c) Notice in disk integration the area was rotated around the same axis
that the area was integrated on. In other words, the axis the area touched was the
axis of rotation. In shell integration, it is the opposite. Notice that the area
is touching the x axis and the solid is rotating around the y axis.

The formula for shell integration is defined as:

where x is the distance to the y axis, or the radius, and f(x) is now the height
of the shell.

Simply substituting f(x) will give us

It seems like simply using the volume formulas was the best method, but let’s do
some different examples where that isn’t the case.

(3) Find the volume of the following function rotated around the x axis from
[0,2Π]

The rotated area would look like this:

Unless you know the formula for finding the volume of a vase, we must use integration
to find this volume. We cannot use the formula for any simple three dimensional
geometric figures like the first two examples. Revolving this solid about the x
axis, we would do the same as example (1) and set up an integral using the formula
for the volume of a cylinder. The radius of the cylinder is the curve, so we would
plug f(x) in for the radius, and then the height would be dx, which is from 0 to
2Π.

Volume of a cylinder

The total volume of the solid is 9Π2 units3.

What if we wanted to find the volume of the area rotated around the x axis of the
same function, but with some open space in the middle? This type of figure is called
a washer, or a donut. They are like discs because they are circular, but there is
space in the middle.

Consider the same function with f(x) = 1.

When rotated, it will look similar to our previous rotation but with a cylinder
removed in the middle.

To find the volume, we simply take the difference of our original area and the area
of the space in the center.

(4) It could also be beneficial to talk about a solid that is not rotated
a full 360 degrees. Think about a portion of a circle that has been shaded.

This circle has been shaded 240 degrees out of 360. How do we find the area? We
simply take a fraction of the total area, in this case, 240360, or two thirds.

This is the case with volume. If we have a portion of the area rotated, we find
out the volume of the solid out of the total volume if it were rotated 360 degrees.

This solid is also rotated 240 degrees around the x axis. What would the volume
be?

This is the volume for the rotated portion of the graph on the interval [a,b].

In the previous examples, we rotated areas about the x or y axis. What if we rotated