Some notes on calculating volumes by integration
Let’s start with an (easy) warm-up example. Let’s say we’re looking at the solid obtained by revolving
the area under the graph of y = x2 for 1 ≤ x ≤ 3 about the x-axis:
1
3
We approximate the volume using cylinders of thickness ∆x centered along the x-axis (as pictured above).
A cylinder at position x would have a radius of x2 (since this is how high the curve goes) and a thickness of
∆x, and hence has a volume of π(x2 )2 ∆x. TheP
total volume of the solid of revolution can be approximated
by adding up the volumes of the cylinders, so
π(x2 )2 ∆x is an approximation for the volume of the solid
R3
of revolution. The approximation is a Riemann sum corresponding to the integral 1 π(x2 )2 dx, so the exact
R3
volume of the solid of revolution is the integral 1 π(x2 )2 dx.
This is the basic argument that we will keep making throughout this chapter, and this is the way you’re
more-or-less required to think in most places. After doing a few problems like this, however, we quickly
realize that we can always calculate the volume V of any solid by integrating the area of the cross-sections.
More specifically:
Z b
V =
A(h) dh
a
where A(h) is the area of a cross section of the solid perpendicular to the h-axis (if it’s not obvious to you
that this should be a function of h, draw a picture and figure it out!). The only catch is that we need to be
looking at our problem in just the right way for A(h) to have a nice, easy-to-find and (if we want a number
at the end) easy-to-integrate formula. In the example above, the area of the cross-sections perpendicular to
the x-axis are extremely easy to find: the cross-sections are disks, with radius (depending on x) equal to x2 ,
so they each have area π(x2 )2 .
Let’s see a slightly more complicated example.
Example 1. Find the volume of the surface obtained by revolving the region between the curves y = x2 and
y = 2x about the line x = −1.
Solution. In any problem of this sort, a well-drawn picture should be your weapon of choice. In this case,
the region that we are revolving is:
1
4
−1
0
2
It’s a little more difficult to draw the solid in this case, but luckily all we really need is to figure out what the
areas of the cross-sections perpendicular to the y-axis are. Some of these cross-sections have been shaded in
the image below.
4
−1
0
2
A cross-section perpendicular to the y-axis is an annulus (a disk with a disk removed from the center),
whose outer radius and inner radius depends on the position along the y-axis.The outer radius is the xcoordinate of the point on the outer curve (i.e. the parabola) at height y plus 1, while the inner radius is
the y-coordinate of the point on the inner curve (i.e. the line) plus 1 (because the center of these disks are
√
along the line x = −1). The outer radius at height y is therefore y + 1 and the inner radius is therefore
√
0.5y + 1. So the area of the cross section at height y is π( y + 1)2 − π(0.5y + 1)2 . And from our earlier
remark, this gives us the integral:
Z 4
√
π( y + 1)2 − π(0.5y + 1)2 dy
0
Here’s a selection of exercises that should help, if you feel you need some more practice with this.
Exercise 2. Write an integral that gives the volume of the solid obtained by revolving the region between the
curves y = x2 and y = x4 about the line y = 5. Do not evaluate your integral.
Exercise 3. Write an integral (or a sum of integrals)
that gives the volume of the solid obtained by revolving
√
the region bounded by the curves y = x2 , y = x and y = 21 about the line y = 5. Do not evaluate your
integral.
Exercise 4. Write an integral (or a sum of integrals)
that gives the volume of the solid obtained by revolving
√
the region bounded by the curves y = x2 , y = x and y = 12 about the line x = −1. Do not evaluate your
integral.
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When we calculate the volume of a solid by integrating the areas of the cross sections (perpendicular to
the axis of revolution), as we have done in all our examples above, we are using what is often called the disk
or washer method. However, while this is a simple, straightforward and extremely useful approach, there
are some problems that simply cannot be done (or can only be done with great pain)
Example 5. Find the volume of the surface obtained by revolving the region between y = ln x (for 1 ≤ x ≤ 3)
and the x-axis about the x-axis.
Solution. Our first step, as always, is to draw a picture so that we understand what’s going on.
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3
From the picture, it’s easy to see that a cross section perpendicular to the x-axis is a disk of radius ln x. So
the area of a cross section is π(ln x)2 . The integral in this case is therefore:
Z 3
V =
π(ln x)2 dx
1
It is possible to evaluate this integral, but it requires several integrations by parts. If we’re only slightly
more clever, however, we can easily come up with a better integral to evaluate. Let’s imagine that our solid
of revolution is made up of concentric cylinders, centered at the x-axis.
1
3
These cylinders are all hollow and have been drawn without any thickness for the sake of clarity, but you
should imagine them as having some (small) thickness ∆y (we use ‘y’ because the thickness is along the
y-axis). A cylindrical shell whose top is a distance of y above the x-axis has horizontal length 3 − ey (because
its length is the difference in the x-coordinates of its endpoints, and the point on the curve y = ln x at height
y has x-coordinate ey ). This cylinder also has thickness ∆y and radius y, so its volume is (approximately)
2πy(3 − ey )∆y (if you’re lost here, take a few minutes to think about these last few lines and possibly ask
me for help).
P
From this, we see that the Riemann sum approximating the volume of this solid is
2πy(3 − ey )∆y
(imagine drawing many, many cylinders like this in the picture above), so the exact volume of the cylinder
R3
is 1 2πy(3 − ey ) dy. This is significantly easier to integrate (you still need to integrate by parts, but just
once this time).
Notice that in the example above, we ended up with
Z b
V =
S(h) dh
a
where S(h) is the surface area of a cylinder centered on the axis of revolution at a distance h from the axis
of revolution. When the integrand is not the area of a cross section perpendicular to the axis of revolution,
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but instead a cylinder with center on the axis of revolution, we’re using what’s often called the shell method.
The names are not so important, but you should know that there is a difference between these two, and
some cases may call for one approach rather than the other. It’s best to get good enough with both that
you can write down both integrals in very little time and see which one looks nicer.
Once again, you should take the time to complete a few exercises.
Exercise 6. Write an integral (using shells) that gives the volume of the solid obtained by revolving the
region bounded by the curve y = ln x, the line x = 5 and the x-axis about the y-axis. Do not evaluate your
integral.
Exercise 7. Write an integral (using shells) that gives the volume of the solid obtained by revolving the
region between the curves y = x2 and y = x4 about the line y = 5. Do not evaluate your integral.
Exercise 8. Write an integral (using shells) that gives the volume of the solid obtained by revolving the
region bouned by the curve y = −x2 + 6x and the x-axis about the line x = 10. Do not evaluate your
integral. (Optional: Try repeating this problem using the ‘washer’ method. Which method is easier?)
Exercise 9. Write an integral (using shells) that gives the volume of the solid obtained by revolving the
region bounded by the curve y = x5 − 3x2 + 5, the vertical line x = 2, the x axis and the y axis about the y
axis. Do not evaluate your integral. (Optional: Try repeating this problem using the ‘washer’ method. What
goes wrong?)
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