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CLP-3 Multivariable Calculus

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.

Definition 3.3.1.

The average (mean) of a set of n numbers f1, f2, , fn is
f¯=f=f1+f2++fnn
The notations f¯ and f are both commonly used to represent the average.
Now suppose that we want to take the average of a function f(x,y) with (x,y) running continuously over some region R in the xy-plane. A natural approach to defining what we mean by the average value of f over R is to
  • First fix any natural number n.
  • Subdivide the region R into tiny (approximate) squares each of width Δx=1n and height Δy=1n. 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) R1, R2, , RN, where N is the total number of squares.
  • Select, for each 1iN, one point in square number i and call it (xi,yi). So (xi,yi)Ri.
  • The average value of f at the selected points is
    1Ni=1Nf(xi,yi)=i=1Nf(xi,yi)i=1N1=i=1Nf(xi,yi)ΔxΔyi=1NΔxΔy
    We have transformed the average into a ratio of Riemann sums.
Once we have the Riemann sums it is clear what to do next. Taking the limit n, we get exactly Rf(x,y)dxdyRdxdy. That’s why we define

Definition 3.3.2.

Let f(x,y) be an integrable function defined on region R in the xy-plane. The average value of f on R is
f¯=f=Rf(x,y)dxdyRdxdy

Example 3.3.3. Average.

Let a>0. A mountain, call it Half Dome
 1 
There is a real Half-Dome mountain in Yosemite National Park. It has a=1445m.
, has height z(x,y)=a2x2y2 above each point (x,y) in the base region R={ (x,y) | x2+y2a2,x0 }. Find its average height.
Solution.
By Definition 3.3.2 the average height is
z¯=Rz(x,y)dxdyRdxdy=Ra2x2y2dxdyRdxdy
The integrals in both the numerator and denominator are easily evaluated by interpreting them geometrically.
  • The numerator Rz(x,y)dxdy=Ra2x2y2dxdy can be interpreted as the volume of
    { (x,y,z) | x2+y2a2, x0, 0za2x2y2 }={ (x,y,z) | x2+y2+z2a2, x0, z0 }
    which is one quarter of the interior of a sphere of radius a. So the numerator is 13πa3.
  • The denominator Rdxdy is the area of one half of a circular disk of radius a. So the denominator is 12πa2.
All together, the average height is
z¯=13πa312πa2=23a
Notice this this number is bigger than zero and less than the maximum height, which is a. 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 Ra2x2y2dxdy of Example 3.3.3 as an iterated integral.
Here is a sketch of the top view of the base region R.
Using the slicing in the figure
Ra2x2y2dxdy=aadya2y20dx a2x2y2
Note that, in the inside integral a2y20dx a2x2y2, the variable y is treated as a constant, so that the integrand a2y2x2=C2x2 with C being the constant a2y2. The standard protocol for evaluating this integral uses the trigonometric substitution
x=Csinθwith π2θπ2dx=Ccosθdθ
Trigonometric substitution was discussed in detail in Section 1.9 in the CLP-2 text. Since
x=0Csinθ=0θ=0x=a2y2=CCsinθ=Cθ=π2
and
a2x2y2=C2C2sin2θ=Ccosθ
the inner integral
a2y20dx a2x2y2=π/20C2cos2θ dθ=C2π/201+cos(2θ)2 dθ=C2[θ+sin(2θ)22]π/20=πC24=π4(a2y2)
and the full integral
Ra2x2y2dxdy=π4aa(a2y2) dy=π20a(a2y2) dy=π2[a3a33]=13πa3
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 π/20cos2θ dθ. Looking at the figures
we see that
π/20cos2θ dθ=π/20sin2θ dθ
Thus
π/20cos2θ dθ=π/20sin2θ dθ=π/2012[sin2θ+cos2θ]dθ=12π/20dθ=π4
It is not at all unusual to want to find the average value of some function f(x,y) with (x,y) running over some region R, but to also want some (x,y)’s to play a greater role in determining the average than other (x,y)’s. One common way to do so is to create a “weight function” w(x,y)>0 with w(x1,y1)w(x2,y2) giving the relative importance of (x1,y1) and (x2,y2). That is, (x1,y1) is w(x1,y1)w(x2,y2) times as important as (x2,y2). This leads to the definition

Definition 3.3.6.

Rf(x,y)w(x,y)dxdyRw(x,y)dxdy
is called the weighted average of f over R with weight w(x,y).
Note that if f(x,y)=F, a constant, then the weighted average of f is just F, just as you would want.

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 ρ(x)kg/m and with x running from a to b, is at
x¯=abx ρ(x)dxabρ(x)dx
That is, the centre of mass is at the average of the x-coordinate weighted by the mass density.
In two dimensions, the centre of mass of a plate that covers the region R in the xy-plane and that has mass density ρ(x,y) is the point (x¯,y¯) where
If the mass density is a constant, the centre of mass is also called the centroid, and is the geometric centre of R. In this case

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
R={ (x,y) | axb, B(x)yT(x) }
in the xy-plane is (x¯,y¯) where
x¯=abx[T(x)B(x)]dxAy¯=ab[T(x)2B(x)2]dx2A
and A=ab[T(x)B(x)]dx is the area of R. 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 see that the area of R is
A=Rdxdy=abdxB(x)T(x)dy=abdx [T(x)B(x)]
and that 3.3.8 gives
x¯=1ARx dxdy=1AabdxB(x)T(x)dy x=1Aabdx x[T(x)B(x)]y¯=1ARy dxdy=1AabdxB(x)T(x)dy y=1Aabdx [T(x)22B(x)22]
just as desired.
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
D={ (x,y) | x0, y0, x2+y2r2 }
by using the formulae of the last example. We’ll now find it again using 3.3.8.
Since the area of D is 14πr2, we have
x¯=Dx dxdy14πr2y¯=Dy dxdy14πr2
We’ll evaluate Dx dxdy by using horizontal slices, as in the figure on the left below.
Looking at that figure, we see that
  • y runs from 0 to r and
  • for each y in that range, x runs from 0 to r2y2.
So
Dx dxdy=0rdy0r2y2dx x=0rdy [x22]0r2y2=120rdy [r2y2]=12[r3r33]=r33
and
x¯=4πr2[r33]=4r3π
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 y¯=x¯. We’ll check that by evaluating Dy dxdy by using vertical slices slices, as in the figure on the right above. From that figure, we see that
  • x runs from 0 to r and
  • for each x in that range, y runs from 0 to r2x2.
So
Dy dxdy=0rdx0r2x2dy y=120rdx [r2x2]
This is exactly the integral 120rdy [r2y2] that we evaluated above, with y renamed to x. So Dy dxdy=r33 too and
y¯=4πr2[r33]=4r3π=x¯
as expected.

Example 3.3.11. Example 3.2.14, continued.

Find the centroid of the region that is inside the circle r=4cosθ and to the left of the line x=1.
Solution.
Recall that we saw in Example 3.2.14 that r=4cosθ was indeed a circle, and in fact is the circle (x2)2+y2=4. Here is a sketch of that circle and of the region of interest, R.
From the sketch, we see that R is symmetric about the x-axis. So we expect that its centroid, (x¯,y¯), has y¯=0. To see this from the integral definition, note that the integral Ry dxdy
  • has domain of integration, namely R, invariant under yy (i.e. under reflection in the x-axis), and
  • has integrand, namely y, that is odd under yy.
So Ry dxdy=0 and consequently y¯=0.
We now just have to find x¯:
x¯=Rx dxdyRdxdy
We have already found, in Example 3.2.14, that
Rdxdy=4π33
So we just have to compute Rx dxdy. Using R1 to denote the top half of R, and using polar coordinates, like we did in Example 3.2.14,
R1x dxdy=0π/3dθ01/cosθdr r(rcosθ)x+π/3π/2dθ04cosθdr r(rcosθ)x=0π/3dθ cosθ01/cosθdr r2+π/3π/2dθ cosθ04cosθdr r2=0π/3dθ sec2θ3+π/3π/2dθ 643cos4θ
The first integral is easy, provided we remember that tanθ is an antiderivative for sec2θ. For the second integral, we’ll need the double angle formula cos2θ=1+cos(2θ)2:
cos4θ=(cos2θ)2=[1+cos(2θ)2]2=14[1+2cos(2θ)+cos2(2θ)]=14[1+2cos(2θ)+1+cos(4θ)2]=38+cos(2θ)2+cos(4θ)8
so
R1x dxdy=13tanθ|0π/3+643[3θ8+sin(2θ)4+sin(4θ)32]π/3π/2=13×3+643[38×π634×2+332×2]=4π323
The integral we want, namely Rx dxdy,
  • has domain of integration, namely R, invariant under yy (i.e. under reflection in the x-axis), and
  • has integrand, namely x, that is even under yy.
So Rx dxdy=2R1x dxdy and, all together,
x¯=2(4π323)4π33=8π3434π330.59
As a check, note that 0x1 on R and more of R is closer to x=1 than to x=0. So it makes sense that x¯ is between 12 and 1.

Example 3.3.12. Reverse Centre of Mass.

Evaluate 022xx22xx2(2x+3y)dydx.
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,
Rx dxdy=x¯ Area(R)Ry dxdy=y¯ Area(R)
so that we can easily evaluate Rx dxdy and Ry dxdy provided R 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
022xx22xx2(2x+3y)dydx=202dx[2xx22xx2dy x]+302dx[2xx22xx2dy y]
On the domain of integration
  • x runs from 0 to 2 and
  • for each fixed 0x2, y runs from 2xx2 to +2xx2
Observe that y=±2xx2 is equivalent to
y2=2xx2=1(x1)2(x1)2+y2=1
Our domain of integration is exactly the disk
R={ (x,y) | (x1)2+y21 }
of radius 1 centred on (1,0).
So R has area π and centre of mass (x¯,y¯)=(1,0) and
022xx22xx2(2x+3y)dydx=2Rx dxdy+3Ry dxdy=2x¯Area(R)+3y¯Area(R)=2π

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 R1 and having mass density ρ1(x,y) and
  • a second end contained in a region R2 and having mass density ρ2(x,y) and
  • an infinitely strong, weightless (idealized) rod joining the two ends.
Suppose that both R1 and R2 are contained in the larger region R, that ρ1(x,y)=0 for all (x,y) not in R1 and that ρ2(x,y)=0 for all (x,y) not in R2.
Then the mass and centre of mass of the R1 end are
M1=R1ρ1(x,y)dxdy
and
(X¯1,Y¯1)withX¯1=R1xρ1(x,y)dxdyM1, Y¯1=R1yρ1(x,y)dxdyM1
and the mass and centre of mass of the R2 end are
M2=R2ρ2(x,y)dxdy
and
(X¯2,Y¯2)withX¯2=R2xρ2(x,y)dxdyM2, Y¯2=R2yρ2(x,y)dxdyM2
The mass and centre of mass of the entire dumbbell are
M=R{ρ1(x,y)+ρ2(x,y)}dxdy=Rρ1(x,y)dxdy+Rρ2(x,y)dxdy=R1ρ1(x,y)dxdy+R2ρ2(x,y)dxdy=M1+M2
and (X¯,Y¯) with
X¯=Rx{ρ1(x,y)+ρ2(x,y)}dxdyM=R1xρ1(x,y)dxdy+R2xρ2(x,y)dxdyM=M1X¯1+M2X¯2M1+M2Y¯=Ry{ρ1(x,y)+ρ2(x,y)}dxdyM=R1yρ1(x,y)dxdy+R2yρ2(x,y)dxdyM=M1Y¯1+M2Y¯2M1+M2
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 M1 located at the centre of mass (X¯1,Y¯1) of the R1 end, and one of mass M2 located at the centre of mass (X2,Y2) of the R2 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 1, of the region R1 in the sketch
We now again compute the centroid of R1, but this time we use the ideas of Example 3.3.13 above. We think of R1 as one end of the “dumbbell” whose other end, R2, is the quarter circular disk which fills out R1 to the 2×2 square in the sketch below. That is, the full dumbbell is
R=R1R2={ (x,y) | 0x2, 0y2 }
with
R1={ (x,y) | 0x2, 0y2, x2+y21 }R2={ (x,y) | x0, y0, x2+y21 }
  • The full dumbbell R is a 2×2 square and so has area, and hence mass, M=2×2=4. Just by symmetry, the centre of mass of R is (X¯,Y¯)=(1,1).
  • The end R2 is one quarter of a circular disk of radius 1 and so has area, and hence mass, M2=π4. In Example 3.3.10 above with r=1, we found that the centre of mass of R2 is (X¯2,Y¯2)=(43π,43π).
So M1=MM2=4π4 and just filling all of this data into the formulae at the end of Example 3.3.13 , above, gives
1=X¯=M1X¯1+M2X¯2M1+M2=(4π4)X¯1+(π4)(43π)4=(1π16)X¯1+1121=Y¯=M1Y¯1+M2Y¯2M1+M2=(4π4)Y¯1+(π4)(43π)4=(1π16)Y¯1+112
Solving gives
X¯1=Y¯1=11121π16=44483π
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 R in the xy-plane, that has mass density ρ(x,y)kg/m2, and that is rotating at ωrad/s about some axis. Let’s call the axis of rotation A. We are now going to determine the kinetic energy of that plate. Recall
 2 
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.
that, by definition, the kinetic energy of a point particle of mass m that is moving with speed v is 12mv2.
To get the kinetic energy of the entire plate, cut it up into tiny rectangles
 3 
The relatively small number of “rectangles” around the boundary of R 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.
, say of size dx×dy. Think of each rectangle as being (essentially) a point particle. If the point (x,y) on the plate is a distance D(x,y) from the axis of rotation A, then as the plate rotates, the point (x,y) sweeps out a circle of radius D(x,y). The figure on the right below shows that circle as seen from high up on the axis of rotation.
The circular arc that the point (x,y) sweeps out in one second subtends the angle ω radians, which is the fraction ω2π of a full circle and so has length ω2π(2πD(x,y))=ωD(x,y). Consequently the rectangle that contains the point (x,y)
  • has speed ωD(x,y), and
  • has area dxdy, and so
  • has mass ρ(x,y) dxdy, and
  • has kinetic energy
    12(ρ(x,y)dxdy)m(ωD(x,y))2v2=12ω2 D(x,y)2ρ(x,y) dxdy
So (via our usual Riemann sum limit procedure) the kinetic energy of R is
R12ω2 D(x,y)2ρ(x,y)dxdy=12ω2R D(x,y)2ρ(x,y)dxdy=12IAω2
where

Definition 3.3.15. Moment of Inertia.

IA=RD(x,y)2ρ(x,y) dxdy
is called the moment of inertial of R about the axis A. In particular the moment of inertia of R about the y-axis is
Iy=Rx2ρ(x,y) dxdy
and the moment of inertia of R about the x-axis is
Ix=Ry2ρ(x,y) dxdy
Notice that the expression 12IAω2 for the kinetic energy has a very similar form to 12mv2, just with the velocity v replaced by the angular velocity ω, and with the mass m replaced by IA, 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 xy-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, R, of the circle x2+y2=a2 about the x-axis. Assume that it has density one.
Solution.
The distance from any point (x,y) inside the disk to the axis of
rotation (i.e. the x-axis) is |y|. So the moment of inertia of the interior of the disk about the x-axis is
Ix=Ry2 dxdy
Switching to polar coordinates
 4 
See how handy they are!
,
Ix=02πdθ0adr r(rsinθ)2y2=02πdθ sin2θ0adr r3=a4402πdθ sin2θ=a4402πdθ 1cos(2θ)2=a48[θsin(2θ)2]02π=14πa4
For an efficient, sneaky, way to evaluate 02πsin2θ dθ, see Remark 3.3.5.

Example 3.3.17. Cardioid.

Find the moment of inertia of the interior, R, of the cardiod r=a(1+cosθ) about the z-axis. Assume that the cardiod lies in the xy-plane and has density one.
Solution.
We sketched the cardiod (with a=1) in Example 3.2.3.
As we said above, the formula for IA in Definition 3.3.15 is valid even when the axis of rotation is not contained in the xy-plane. We just have to be sure that our D(x,y) really is the distance from (x,y) to the axis of rotation. In this example the axis of rotation is the z-axis so that D(x,y)=x2+y2 and that the moment of inertia is
IA=R(x2+y2) dxdy
Switching to polar coordinates, using dxdy=rdrdθ and x2+y2=r2,
IA=02πdθ0a(1+cosθ)dr r×r2=02πdθ 0a(1+cosθ)dr r3=a4402πdθ (1+cosθ)4=a4402πdθ (1+4cosθ+6cos2θ+4cos3θ+cos4θ)
Now
02πdθ cosθ=sinθ|02π=002πdθ cos2θ=02πdθ 1+cos(2θ)2=12[θ+sin(2θ)2]02π=π02πdθ cos3θ=02πdθ cosθ[1sin2θ] =u=sinθ 00du (1u2)=0
To integrate cos4θ, we use the double angle formula
cos2θ=cos(2θ)+12cos4θ=(cos(2θ)+1)24=cos2(2θ)+2cos(2θ)+14=cos(4θ)+12+2cos(2θ)+14=38+12cos(2θ)+18cos(4θ)
to give
02πdθ cos4θ=02πdθ [38+12cos(2θ)+18cos(4θ)]=38×2π+12×0+18×0=34π
All together
IA=a44[2π+4×0+6×π+4×0+34π]=3516πa4

Exercises 3.3.4 Exercises

Exercise Group.

Exercises — Stage 1
1.
For each of the following, evaluate the given double integral without using iteration. Instead, interpret the integral in terms of, for example, areas or average values.
  1. D(x+3) dxdy, where D is the half disc 0y4x2
  2. R(x+y) dxdy where R is the rectangle 0xa, 0yb

Exercise Group.

Exercises — Stage 2
2. (✳).
Find the centre of mass of the region D in the xy--plane defined by the inequalities x2y1, assuming that the mass density function is given by ρ(x,y)=y.
3. (✳).
Let R be the region bounded on the left by x=1 and on the right by x2+y2=4. The density in R is
ρ(x,y)=1x2+y2
  1. Sketch the region R.
  2. Find the mass of R.
  3. Find the centre-of-mass of R.
Note: You may use the result sec(θ) dθ=ln|secθ+tanθ|+C.
4. (✳).
A thin plate of uniform density 1 is bounded by the positive x and y axes and the cardioid x2+y2=r=1+sinθ, which is given in polar coordinates. Find the x--coordinate of its centre of mass.
5. (✳).
A thin plate of uniform density k is bounded by the positive x and y axes and the circle x2+y2=1. Find its centre of mass.
6. (✳).
Let R be the triangle with vertices (0,2), (1,0), and (2,0). Let R have density ρ(x,y)=y2. Find y¯, the y--coordinate of the center of mass of R. You do not need to find x¯.
7. (✳).
The average distance of a point in a plane region D to a point (a,b) is defined by
1A(D)D(xa)2+(yb)2 dxdy
where A(D) is the area of the plane region D. Let D be the unit disk 1x2+y2. Find the average distance of a point in D to the center of D.
8. (✳).
A metal crescent is obtained by removing the interior of the circle defined by the equation x2+y2=x from the metal plate of constant density 1 occupying the unit disc x2+y21.
  1. Find the total mass of the crescent.
  2. Find the x-coordinate of its center of mass.
You may use the fact that π/2π/2cos4(θ) dθ=3π8.
9. (✳).
Let D be the region in the xy--plane which is inside the circle x2+(y1)2=1 but outside the circle x2+y2=2. Determine the mass of this region if the density is given by
ρ(x,y)=2x2+y2

Exercise Group.

Exercises — Stage 3
10. (✳).
Let a, b and c be positive numbers, and let T be the triangle whose vertices are (a,0), (b,0) and (0,c).
  1. Assuming that the density is constant on T, find the center of mass of T.
  2. The medians of T are the line segments which join a vertex of T 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 T. Show that the centroid of T is its centre of mass.