Section 3.3 Applications of Double Integrals
Double integrals are useful for more than just computing areas and volumes. Here are a few other applications that lead to double integrals.
Subsection 3.3.1 Averages
In Section 2.2 of the CLP-2 text, we defined the average value of a function of one variable. We’ll now extend that discussion to functions of two variables. First, we recall the definition of the average of a finite set of numbers.
Now suppose that we want to take the average of a function with running continuously over some region in the -plane. A natural approach to defining what we mean by the average value of over is to
- First fix any natural number
- Subdivide the region
into tiny (approximate) squares each of width and height This can be done by, for example, subdividing vertical strips into tiny squares, like in Example 3.1.11. - Name the squares (in any fixed order)
where is the total number of squares. - Select, for each
one point in square number and call it So - The average value of
at the selected points is
Once we have the Riemann sums it is clear what to do next. Taking the limit we get exactly That’s why we define
Definition 3.3.2.
Example 3.3.3. Average.
Let A mountain, call it Half Dome, has height above each point in the base region Find its average height.
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There is a real Half-Dome mountain in Yosemite National Park. It has
Solution.
By Definition 3.3.2 the average height is
The integrals in both the numerator and denominator are easily evaluated by interpreting them geometrically.
- The numerator
can be interpreted as the volume of So the numerator is - The denominator
is the area of one half of a circular disk of radius So the denominator is
All together, the average height is
Notice this this number is bigger than zero and less than the maximum height, which is That makes sense.
Example 3.3.4. Example 3.3.3, the hard way.
This last example was relatively easy because we could reinterpret the integrals as geometric quantities. For practice, let’s go back and evaluate the numerator of Example 3.3.3 as an iterated integral.
Here is a sketch of the top view of the base region
Using the slicing in the figure
Note that, in the inside integral the variable is treated as a constant, so that the integrand with being the constant The standard protocol for evaluating this integral uses the trigonometric substitution
Trigonometric substitution was discussed in detail in Section 1.9 in the CLP-2 text. Since
and
the inner integral
and the full integral
just as we saw in Example 3.3.3.
Remark 3.3.5.
We remark that there is an efficient, sneaky, way to evaluate definite integrals like Looking at the figures
It is not at all unusual to want to find the average value of some function with running over some region but to also want some ’s to play a greater role in determining the average than other ’s. One common way to do so is to create a “weight function” with giving the relative importance of and That is, is times as important as This leads to the definition
Definition 3.3.6.
Subsection 3.3.2 Centre of Mass
One important example of a weighted average is the centre of mass. If you support a body at its centre of mass (in a uniform gravitational field) it balances perfectly. That’s the definition of the centre of mass of the body. In Section 2.3 of the CLP-2 text, we found that the centre of mass of a body that consists of mass distributed continuously along a straight line, with mass density kg/m and with running from to is at
That is, the centre of mass is at the average of the -coordinate weighted by the mass density.
In two dimensions, the centre of mass of a plate that covers the region in the -plane and that has mass density is the point where
Equation 3.3.7. Centre of Mass.
If the mass density is a constant, the centre of mass is also called the centroid, and is the geometric centre of In this case
Equation 3.3.8. Centroid.
Example 3.3.9. Centre of Mass.
In Section 2.3 of the CLP-2 text, we did not have access to multivariable integrals, so we used some physical intuition to derive that the centroid of a body that fills the region
in the -plane is where
and is the area of Now that we do have access to multivariable integrals, we can derive these formulae directly from 3.3.8. Using vertical slices, as in this figure,
We’ll start with a simple mechanical example.
Example 3.3.10. Quarter Circle.
In Example 2.3.4 of the CLP-2 text, we found the centroid of the quarter circular disk
by using the formulae of the last example. We’ll now find it again using 3.3.8.
Since the area of is we have
We’ll evaluate by using horizontal slices, as in the figure on the left below.
Looking at that figure, we see that
runs from to and- for each
in that range, runs from to
So
and
This is the same answer as we got in Example 2.3.4 of the CLP-2 text. But because we were able to use horizontal slices, the integral in this example was a little easier to evaluate than the integral in CLP-2. Had we used vertical slices, we would have ended up with exactly the integral of CLP-2.
By symmetry, we should have We’ll check that by evaluating by using vertical slices slices, as in the figure on the right above. From that figure, we see that
runs from to and- for each
in that range, runs from to
So
This is exactly the integral that we evaluated above, with renamed to So too and
as expected.
Example 3.3.11. Example 3.2.14, continued.
Find the centroid of the region that is inside the circle and to the left of the line
Solution.
Recall that we saw in Example 3.2.14 that was indeed a circle, and in fact is the circle Here is a sketch of that circle and of the region of interest,
From the sketch, we see that is symmetric about the -axis. So we expect that its centroid, has To see this from the integral definition, note that the integral
- has domain of integration, namely
invariant under (i.e. under reflection in the -axis), and - has integrand, namely
that is odd under
So and consequently
We now just have to find
We have already found, in Example 3.2.14, that
So we just have to compute Using to denote the top half of and using polar coordinates, like we did in Example 3.2.14,
The first integral is easy, provided we remember that is an antiderivative for For the second integral, we’ll need the double angle formula
so
The integral we want, namely
- has domain of integration, namely
invariant under (i.e. under reflection in the -axis), and - has integrand, namely
that is even under
So and, all together,
As a check, note that on and more of is closer to than to So it makes sense that is between and
Example 3.3.12. Reverse Centre of Mass.
Evaluate
Solution.
This is another integral that can be evaluated without using any calculus at all. This time by relating it to a centre of mass. By 3.3.8,
so that we can easily evaluate and provided is sufficiently simple and symmetric that we can easily determine its area and its centroid.
That is the case for the integral in this example. Rewrite
On the domain of integration
runs from to and- for each fixed
runs from to
Observe that is equivalent to
Our domain of integration is exactly the disk
of radius centred on
So has area and centre of mass and
Example 3.3.13. Compound bodies.
In Section 2.3.1 of the CLP-2 text, we saw that it was possible to think of a body as being made up of a number of component parts, and then it was possible to compute the centre of mass of the body as a whole by using the centres of mass of the component parts. In that discussion, we could only consider bodies consisting of a finite number of point particles arrayed along a straight line. We will now see that the same procedure can be used with continuous mass distributions. Suppose that we have a dumbbell which consists of
- one end contained in a region
and having mass density and - a second end contained in a region
and having mass density and - an infinitely strong, weightless (idealized) rod joining the two ends.
Suppose that both and are contained in the larger region that for all not in and that for all not in
Then the mass and centre of mass of the end are
and
and the mass and centre of mass of the end are
and
The mass and centre of mass of the entire dumbbell are
and with
So we can compute the centre of mass of the entire dumbbell by treating it as being made up of two point particles, one of mass located at the centre of mass of the end, and one of mass located at the centre of mass of the end.
Example 3.3.14. Example 2.3.6 of CLP-2 revisited.
In Example 2.3.6 of the CLP-2 text, we found the centroid, i.e. the centre of mass with density of the region in the sketch
We now again compute the centroid of but this time we use the ideas of Example 3.3.13 above. We think of as one end of the “dumbbell” whose other end, is the quarter circular disk which fills out to the square in the sketch below. That is, the full dumbbell is
with
- The full dumbbell
is a square and so has area, and hence mass, Just by symmetry, the centre of mass of is - The end
is one quarter of a circular disk of radius and so has area, and hence mass, In Example 3.3.10 above with we found that the centre of mass of is
So and just filling all of this data into the formulae at the end of Example 3.3.13 , above, gives
Solving gives
which is exactly the answer that we found in Example 2.3.6 of the CLP-2 text.
Subsection 3.3.3 Moment of Inertia
Consider a plate that fills the region in the -plane, that has mass density and that is rotating at about some axis. Let’s call the axis of rotation We are now going to determine the kinetic energy of that plate. Recall that, by definition, the kinetic energy of a point particle of mass that is moving with speed is
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If you don’t recall, don’t worry. We wouldn’t lie to you. Or check it on Wikipedia. They wouldn’t lie to you either.
To get the kinetic energy of the entire plate, cut it up into tiny rectangles, say of size Think of each rectangle as being (essentially) a point particle. If the point on the plate is a distance from the axis of rotation then as the plate rotates, the point sweeps out a circle of radius The figure on the right below shows that circle as seen from high up on the axis of rotation.
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The relatively small number of “rectangles” around the boundary of won’t actually be rectangles. But, as we have seen in the optional §3.2.4, one can still make things rigorous despite the rectangles being a bit squishy around the edges.
The circular arc that the point sweeps out in one second subtends the angle radians, which is the fraction of a full circle and so has length Consequently the rectangle that contains the point
- has speed
and - has area
and so - has mass
and - has kinetic energy
So (via our usual Riemann sum limit procedure) the kinetic energy of is
where
Definition 3.3.15. Moment of Inertia.
Notice that the expression for the kinetic energy has a very similar form to just with the velocity replaced by the angular velocity and with the mass replaced by which can be thought of as being a bit like a mass.
So far, we have been assuming that the rotation was taking place in the -plane — a two dimensional world. Our analysis extends naturally to three dimensions, though the resulting integral formulae for the moment of inertia will then be triple integrals, which we have not yet dealt with. We shall soon do so, but let’s first do an example in two dimensions.
Example 3.3.16. Disk.
Find the moment of inertia of the interior, of the circle about the -axis. Assume that it has density one.
Solution.
The distance from any point inside the disk to the axis of
rotation (i.e. the -axis) is So the moment of inertia of the interior of the disk about the -axis is
Switching to polar coordinates,
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See how handy they are!
For an efficient, sneaky, way to evaluate see Remark 3.3.5.
Example 3.3.17. Cardioid.
Find the moment of inertia of the interior, of the cardiod about the -axis. Assume that the cardiod lies in the -plane and has density one.
Solution.
We sketched the cardiod (with ) in Example 3.2.3.
As we said above, the formula for in Definition 3.3.15 is valid even when the axis of rotation is not contained in the -plane. We just have to be sure that our really is the distance from to the axis of rotation. In this example the axis of rotation is the -axis so that and that the moment of inertia is
Switching to polar coordinates, using and
Now
To integrate we use the double angle formula
to give
All together
Exercises 3.3.4 Exercises
Exercise Group.
Exercises — Stage 1
Exercise Group.
Exercises — Stage 2
2. (✳).
Find the centre of mass of the region in the --plane defined by the inequalities assuming that the mass density function is given by
3. (✳).
- Sketch the region
- Find the mass of
- Find the centre-of-mass of
Note: You may use the result
4. (✳).
A thin plate of uniform density is bounded by the positive and axes and the cardioid which is given in polar coordinates. Find the --coordinate of its centre of mass.
5. (✳).
A thin plate of uniform density is bounded by the positive and axes and the circle Find its centre of mass.
6. (✳).
Let be the triangle with vertices and Let have density Find the --coordinate of the center of mass of You do not need to find
7. (✳).
8. (✳).
A metal crescent is obtained by removing the interior of the circle defined by the equation from the metal plate of constant density 1 occupying the unit disc
- Find the total mass of the crescent.
- Find the
-coordinate of its center of mass.
You may use the fact that
9. (✳).
Exercise Group.
Exercises — Stage 3
10. (✳).
- Assuming that the density is constant on
find the center of mass of - The medians of
are the line segments which join a vertex of to the midpoint of the opposite side. It is a well known fact that the three medians of any triangle meet at a point, which is known as the centroid of Show that the centroid of is its centre of mass.