APPLICATIONS OF INTEGRATION; SINGLE AND MULTIPLE INTEGRALS

3. APPLICATIONS OF INTEGRATION; SINGLE AND MULTIPLE INTEGRALS

Many different physical quantities are given by integrals; let us do some problems to illustrate setting up and evaluating these integrals. The basic idea which we use in setting up the integrals in these problems is that an integral is the “limit of a sum.” Thus we imagine the physical object (whose volume, moment of inertia, etc., we are trying to find) cut into a large number of small pieces called elements. We write an approximate formula for the volume, moment of inertia, etc., of an element and then sum over all elements of the object. The limit of this sum (as the number of elements tends to infinity and the size of each element tends to zero) is what we find by integration and is what we want in the physical problem.

Using a computer to evaluate the integrals saves time and we will concentrate mainly on setting up integrals. However, in order to do a skillful job of finding limits, deciding order of integration, detecting and correcting errors, making useful changes of variables, and understanding the meaning of the symbols used, it is important to learn to evaluate multiple integrals by hand. So a good study method is to do some integrals both by hand and by computer. A computer is also very useful to plot graphs of curves and surfaces to help you find the limits in a multiple integral.

Example 1. Given the curve y = x 2 from x = 0 to x = 1, find

(a) the area under the curve (that is, the area bounded by the curve, the x axis, and the line x = 1; see Figure 3.1);

(b) the mass of a plane sheet of material cut in the shape of this area if its density (mass per unit area) is xy;

(c) the arc length of the curve; (d) the centroid of the area;

(e) the centroid of the arc; (f) the moments of inertia about the x, y, and z axes of the lamina in (b).

(a) The area is

A=

y dx =

x 2 dx =

x=0

250 Multiple Integrals; Applications of Integration Chapter 5

We could also find the area as a double integral of dA = dy dx (see Figure 3.1). We have then

as before. Although the double integral is entirely unnecessary in finding the area in this problem, we shall need to use a double integral to find the mass in part (b).

Figure 3.1 Figure 3.2 (b) The element of area, as in the double integral method in (a), is dA = dy dx.

Since the density is ρ = xy, the element of mass is dM = xy dy dx, and the total mass is

0 2 0 0 2 12 Observe that we could not do this problem as a single integral because the density

x=0 y=0

depends on both x and y. (c) The element of arc length ds is defined as indicated in Figures 3.1 and 3.2.

Thus we have

1 + (dy/dx) 2 dx = (dx/dy) 2 + 1 dy.

If y = f (x) has a continuous first derivative dy/ dx (except possibly at a finite num- ber of points), we can find the arc length of the curve y = f (x) between a and b by calculating b

a ds. For our example, we have

2 2 5 + ln(2 + s= 5) 1 + 4x dx =

(see Problem 32).

Section 3 Applications of Integration; Single and Multiple Integrals 251

(d) Recall from elementary physics that:

The center of mass of a body has coordinates ¯ x, ¯ y, ¯ z given by the equations (3.3)

z dM = ¯ z dM, where dM is an element of mass and the integrals are over the whole body.

Although we have written single integrals in (3.3), they may be single, double, or triple integrals depending on the problem and the method of evaluation. Since ¯ x, ¯ y, and ¯ z are constants, we can take them outside the integrals in (3.3) and solve for them. However, you may find it easier to remember the definitions in the form (3.3). For the example we are doing, ¯ z = 0 since the body is a sheet of material in the (x, y) plane. The element of mass is dM = ρ dA = ρ dx dy, where ρ is the density (mass per unit area in this problem). For a variable density as in (b), we would substitute the value of ρ into (3.3) and integrate both sides of each equation to find the coordinates of the center of mass. However, let us suppose the density is a constant. Then the first integral in (3.3) is

(3.4) ¯ xρ dA =

x dA. Similarly, a constant density ρ can be canceled from all the equations in (3.3). The

xρ dA or

¯ x dA =

quantities ¯ x, ¯ y, ¯ z, are then called the coordinates of the centroid of the area (or volume or arc).

The centroid of a body is the center of mass when we assume constant density.

In our example, we have

10 0 10 (Double integrals are not really necessary for any of these but the last.) Using the

x=0 y=0

x=0 y=0

value of A from part(a), we find ¯ x= 3 4 ,¯ y= 3 10 .

(e) The center of mass (¯ x, ¯ y) of a wire bent in the shape of the curve y = f (x) is given by

(3.6) xρ ds = ¯

yρ ds, where ρ is the density (mass per unit length), and the integrals are single integrals

xρ ds,

¯ yρ ds =

with ds given by (3.1). If ρ is constant, (3.6) defines the coordinates of the centroid.

252 Multiple Integrals; Applications of Integration Chapter 5

In our example we have

Note carefully here that it is correct to put y = x 2 in the last integral of (3.7), but it would not have been correct to do this in the last integral of (3.5); the reason is that over the area, y could take values from zero to x 2 , but on the arc, y takes only the value x 2 . By calculating the integrals in (3.7) we can find ¯ x and ¯ y.

(f) We need the following definition: The moment of inertia I of a point mass m about an axis is by definition the

product ml 2 of m times the square of the distance l from m to the axis. For an extended object we must integrate l 2 dM over the whole object, where l is the distance from dM to the axis.

In our example with variable density ρ = xy, we have dM = xy dy dx. The distance from dM to the x axis is y (Figure 3.3); similarly, the distance from dM to the y axis is x. The distance from dM to the z axis (the z axis is perpendicular to the paper in Figure 3.3) is

x 2 +y 2 . Then the three moments of inertia about the three coordinate axes are:

The fact that I x +I y =I z for a plane lamina in the (x, y) plane is known as the perpendicular axis theorem.

It is customary to write moments of inertia as multiples of the mass; using M = 1 12 from (b), we write

40 10 16 4 80 20 Example 2. Rotate the area of Example 1 about the x axis to form a volume and surface

of revolution, and find (a) the volume;

Section 3 Applications of Integration; Single and Multiple Integrals 253

(b) the moment of inertia about the x axis of a solid of constant density occupying the given volume;

(c) the area of the curved surface; (d) the centroid of the curved surface.

(a) We want to find the given volume. The easiest way to find a volume of revolution is to take as volume element a

thin slab of the solid as shown in Figure 3.4. The slab has circular cross section of radius y and thickness dx; thus the volume element is πy 2 dx.

Then the volume in our example is

We have really avoided part of the integration here because we knew the formula for the area of a circle. In finding volumes of solids which are not solids of revolution, we may have to use double or triple integrals. Even for a solid of revolution we might need multiple integrals to find the mass if the density is variable.

To illustrate setting up such integrals, let us do the above problem using triple integrals. For this we need the equation of the surface which is (see Problem 16)

Figure 3.4 Figure 3.5 To set up a multiple integral for the volume of a solid, we cut the solid into slabs

as in Figure 3.4 (not necessarily circular slabs, although they are in our example) and then as in Figure 3.5 we cut each slab into strips and each strip into tiny boxes of volume dx dy dz. The volume is

V=

dx dy dz;

the only problem is to find the limits! To do this, we start by adding up tiny boxes to get a strip; as we have drawn Figure 3.5, this means to integrate with respect to

y from one side of the circle y 2 +z 2 =x 4 to the other, that is, from y=− x 4 −z 2 to y = + x 4 −z 2 .

254 Multiple Integrals; Applications of Integration Chapter 5

Next we add all the strips in a slab. This means that, in Figure 3.5, we integrate with respect to z from the bottom to the top of the circle y 2 +z 2 =x 4 ; thus the z limits are z = ± radius of circle = ±x 2 . And finally we add all the slabs to obtain the solid. This means to integrate in Figure 3.4 from x = 0 to x = 1; this is just what we did in our first simple method. The final integral is then

1 x 2 √ x 4 −z 2

(3.10) V=

dy dz dx. (See Problem 33).

x=0 z=−x 2 y=− √ x 4 −z 2

Although the triple integral is an unnecessarily complicated way of finding a volume of revolution, this simple problem illustrates the general method of setting up an integral for any kind of volume. Once we have the volume as a triple integral, it is easy to write the integrals for the mass with a given variable density, for the coordinates of the centroid, for the moments of inertia, and so on. The limits of integration are the same as for the volume; we need only insert the proper expres- sions (density, etc.) in the integrand to get the mass, centroid, and so on.

(b) To find the moment of inertia of the solid about the x axis, we must integrate the quantity l 2 dM , where l is the distance from dM to the x axis; from Figure 3.5, since the x axis is perpendicular to the paper, l 2 =y 2 +z 2 . The limits on the integrals are the same as in (3.10). We are assuming constant density, so the factor ρ can be written outside the integrals. Then we have

) dy dz dx = z=−x 2 y=− x 4 −z 2 18

x=0

Since from (3.8) the mass of the solid is

π M = ρV = ρ

we can write I x (as is customary) as a multiple of M :

Figure 3.6 Figure 3.7 (c) We find the area of the surface of revolution by using as element the curved

surface of a thin slab as in Figure 3.6. This is a strip of circumference 2πy and width ds. To see this clearly and to understand why we use ds here but dx in the

Section 3 Applications of Integration; Single and Multiple Integrals 255

volume element in (3.8), think of the slab as a thin section of a cone (Figure 3.7) between planes perpendicular to the axis of the cone. If you wanted to find the

total volume V = 1 3 πr 2 h of the cone, you would use the height h perpendicular to the base, but in finding the total curved surface area S = 1 2 · 2πr · s, you would use the slant height s. The same ideas hold in finding the volume and surface elements. The approximate volume of the thin slab is the area of a face of the slab times its thickness (dh in Figure 3.7, dx in Figure 3.4). But if you think of a narrow strip of paper just covering the curved surface of the thin slab, the width of the strip of paper is ds, and its length is the circumference of the thin slab.

The element of surface area (in Figure 3.6) is then (3.11)

dA = 2πy ds.

The total area is [using ds from (3.2)]

(For more general surfaces, there is a way to calculate areas by double integration; we shall take this up in Section 5.)

(d) The y and z coordinates of the centroid of the surface area are zero by symmetry. For the x coordinate, we have by (3.4)

¯ x dA =

x dA,

or, using dA = 2πy ds and the total area A from (c), we have

PROBLEMS, SECTION 3

The following notation is used in the problems: M = mass, ¯ x, ¯ y, ¯ z = coordinates of center of mass (or centroid if the density is constant),

I = moment of inertia (about axis stated), I x ,I y ,I z = moments of inertia about x, y, z axes, I m = moment of inertia (about axis stated) through the center of mass.

Note: It is customary to give answers for I, I m ,I x , etc., as multiples of M (for

example, I = 1 Ml 3 2 ).

1. Prove the “parallel axis theorem”: The moment of inertia I of a body about a given axis is I = I m +Md 2 , where M is the mass of the body, I m is the moment of inertia of the body about an axis through the center of mass and parallel to the given axis, and d is the distance between the two axes.

256 Multiple Integrals; Applications of Integration Chapter 5

2. For a thin rod of length l and uniform density ρ find (a)

M, (b)

I m about an axis perpendicular to the rod, (c)

I about an axis perpendicular to the rod and passing through one end (see Problem 1).

3. A thin rod 10 ft long has a density which varies uniformly from 4 to 24 lb/ft. Find (a)

M, (b)

¯ x, (c)

I m about an axis perpendicular to the rod, (d)

I about an axis perpendicular to the rod passing through the heavy end. 4. Repeat Problem 3 for a rod of length l with density varying uniformly from 2 to 1. 5. For a square lamina of uniform density, find I about

(a) a side, (b)

a diagonal, (c)

an axis through a corner and perpendicular to the plane of the lamina. Hint: See the perpendicular axis theorem, Example 1f.

6. A triangular lamina has vertices (0, 0), (0, 6) and (6, 0), and uniform density. Find: (a)

¯ x, ¯ y, (b)

I x , (c)

I m about an axis parallel to the x axis. Hint: Use Problem 1 carefully. 7. A rectangular lamina has vertices (0, 0), (0, 2), (3, 0), (3, 2) and density xy. Find (a)

I m about an axis parallel to the z axis. Hint: Use the parallel axis theorem and the perpendicular axis theorem.

8. For a uniform cube, find I about one edge. 9. For the pyramid inclosed by the coordinate planes and the plane x + y + z = 1:

(a) Find its volume. (b)

Find the coordinates of its centroid. (c)

If the density is z, find M and ¯ z. 10. A uniform chain hangs in the shape of the catenary y = cosh x between x = −1 and

x = 1. Find (a) its length,

(b) ¯ y. 11. A chain in the shape y = x 2 between x = −1 and x = 1 has density |x|. Find

(a) M , (b) ¯ x, ¯ y. Prove the following two theorems of Pappus:

12. The area A inside a closed curve in the (x, y) plane, y ≥ 0, is revolved about the x axis. The volume of the solid generated is equal to A times the circumference of the circle traced by the centroid of A. Hint: Write the integrals for the volume and for the centroid.

Section 3 Applications of Integration; Single and Multiple Integrals 257

13. An arc in the (x, y) plane, y ≥ 0, is revolved about the x axis. The surface area generated is equal to the length of the arc times the circumference of the circle traced by the centroid of the arc.

14. Use Problems 12 and 13 to find the volume and surface area of a torus (doughnut). 15. Use Problems 12 and 13 to find the centroids of a semicircular area and of a semi-

circular arc. Hint: Assume the formulas A = 4πr 2 ,V= 4 3 πr 3 for a sphere. 16. Let a curve y = f (x) be revolved about the x axis, thus forming a surface of

revolution. Show that the cross sections of this surface in any plane x = const. [that is, parallel to the (y, z) plane] are circles of radius f (x). Thus write the

general equation of a surface of revolution and verify the special case f (x) = x 2 in (3.9).

In Problems 17 to 30, for the curve y = x, between x = 0 and x = 2, find: 17. The area under the curve. 18. The arc length. 19. The volume of the solid generated when the area is revolved about the x axis. 20. The curved area of this solid.

21, 22, 23. The centroids of the arc, the volume, and the surface area. 24, 25, 26, 27. The moments of inertia about the x axis of a lamina in the shape of

the plane area under the curve; of a wire bent along the arc of the curve; of the solid of revolution; and of a thin shell whose shape is the curved surface of the solid (assuming constant density for all these problems).

28. The mass of a wire bent in the shape of the arc if its density (mass per unit length) √ is x.

29. The mass of the solid of revolution if the density (mass per unit volume) is |xyz|. 30. The moment of inertia about the y axis of the solid of revolution if the density is

|xyz|. 31. (a)

Revolve the curve y = x −1 , from x = 1 to x = ∞, about the x axis to create a surface and a volume. Write integrals for the surface area and the volume. Find the volume, and show that the surface area is infinite. Hint: The surface area integral is not easy to evaluate, but you can easily show that it is greater R

than ∞ 1 x −1 dx which you can evaluate. (b)

The following question is a challenge to your ability to fit together your math- ematical calculations and physical facts: In (a) you found a finite volume and an infinite area. Suppose you fill the finite volume with a finite amount of paint and then pour off the excess leaving what sticks to the surface. Apparently you have painted an infinite area with a finite amount of paint! What is wrong? (Compare Problem 15.31c of Chapter 1.)

32. Use a computer or tables to evaluate the integral in (3.2) and verify that the answer is equivalent to the text answer. Hint: See Problem 1.4 and also Chapter 2, Sections

15 and 17. 33. Verify that (3.10) gives the same result as (3.8).

258 Multiple Integrals; Applications of Integration Chapter 5