DOUBLE AND TRIPLE INTEGRALS

2. DOUBLE AND TRIPLE INTEGRALS

b Recall from calculus that b

a y dx = a f (x) dx

gives the area “under the curve” in Figure 2.1. Recall also the definition of the integral as the limit of a sum: We approximate the area by

a sum of rectangles as in Figure 2.1; a rep- resentative rectangle (shaded) has width ∆x. The geometry indicates that if we increase the number of rectangles and let all the widths

Figure 2.1 ∆x → 0, the sum of the areas of the rectan- gles will tend to the area under the curve. We define b

a f (x) dx as the limit of the sum of the areas of the rectangles; then we evaluate the integral as an antiderivative, and use b

a f (x) dx to calculate the area under the curve. We are going to do something very similar

in order to find the volume of the cylinder in Figure 2.2 under the surface z = f (x, y). We cut the (x, y) plane into little rectangles of area ∆A = (∆x) (∆y) as shown in Figure 2.2; above each ∆x ∆y is a tall slender box reaching up to the surface. We can approximate the desired volume by a sum of these boxes just as we ap- proximated the area in Figure 2.1 by a sum of rectangles. As the number of boxes increases and all ∆x and ∆y → 0, the geometry indi- cates that the sum of the volumes of the boxes will tend to the desired volume. We define the

Figure 2.2

Section 2 Double and Triple Integrals 243

double integral of f (x, y) over the area A in the (x, y) plane (Figure 2.2) as the limit of this sum, and we write it as A f (x, y) dx dy. Before we can use the double integral to compute volumes, however, we need to see how double integrals are eval- uated. Even though we may use a computer to do the work, we need to understand the process in order to set up integrals correctly and find and correct errors. Doing some hand evaluation is a good way to learn this.

Iterated Integrals We now show by some examples the details of evaluating double integrals.

Figure 2.3 Figure 2.4

Example 1. Find the volume of the solid (Figure 2.3) below the plane z = 1 + y, bounded by the coordinate planes and the vertical plane 2x + y = 2. From our discussion above, this is A z dx dy = A (1 + y) dx dy, where A is the shaded triangle in the (x, y) plane [A is shown also in Figure 2.4 (a and b)]. We are going to consider two ways of evaluating this double integral. We think of the triangle A cut up into little rectangles ∆A = ∆x ∆y (Figure 2.4) and the whole solid cut into vertical columns of height z and base ∆A (Figure 2.3). We want the (limit of the) sum of the volumes of these columns. First add up the columns (Figure 2.4a) for a

fixed value of x producing the volume of a slab (Figure 2.3) of thickness ∆x. This corresponds to integrating with respect to y (holding x constant, Figure 2.4a) from y = 0 to y on the line 2x + y = 2, that is y = 2 − 2x; we find

y 2 2−2x (2.1)

(What we have found is the area of the slab in Figure 2.3; its volume is the area times ∆x.) Now we add up the volumes of the slabs; this corresponds to integrating (2.1) with respect to x from x = 0 to x = 1:

(4 − 6x + 2x 2 ) dx = .

x=0

244 Multiple Integrals; Applications of Integration Chapter 5

We could summarize (2.1) and (2.2) by writing

We call (2.3) an iterated (repeated) integral. Multiple integrals are usually evaluated by using iterated integrals. Note that the large parentheses in (2.3) are not really necessary if we are always careful to state the variable in giving the limits on an

1 integral; that is, always write 1

x=0 , not just 0 .

Now we could also add up the volume z(∆A) by first integrating with respect to x (for fixed y, Figure 2.4b) from x = 0 to x = 1 − y/2 giving the volume of a slab perpendicular to the y axis in Figure 2.3, and then add up the volumes of the slabs by integrating with respect to y from y = 0 to y = 2 (Figure 2.4b). We write

2 1−y/2 2 1−y/2

(1 + y)(1 − y/2) dy

y=0

2 = 5 (1 + y/2 − y /2) dy = .

3 As the geometry would indicate, the results in (2.2) and (2.4) are the same; we have

y=0

two methods of evaluating the double integral by using iterated integrals. Often one of these two methods is more convenient than the other; we choose

whichever method is easier. To see how to decide, study the following sketches of areas A over which we want to find

A f (x, y) dx dy. In each case we think of combining little rectangles dx dy to form strips (as shown) and then combining the

strips to cover the whole area. Areas shown in Figure 2.5: Integrate with respect to y first. Note that the top and bottom of area A are curves whose equations we know; the boundaries at x = a and x = b are either vertical straight lines or else points.

Figure 2.5

We find

b y 2 (x)

f (x, y) dx dy =

f (x, y) dy dx.

x=a

y=y 1 (x)

Section 2 Double and Triple Integrals 245

Areas shown in Figure 2.6: Integrate with respect to x first. Note that the sides of area A are curves whose equations we know; the boundaries at y = c and y = d are either horizontal straight lines or else points.

Figure 2.6

We find

Areas shown in Figure 2.7: Integrate in either order. Note that these areas all satisfy the requirements for both (2.5) and (2.6).

We find

An important special case is a double integral over a rectangle (both x and y limits are constants) when f (x, y) is a product, f (x, y) = g(x)h(y). Then

f (x, y) dx dy = g(x)h(y) dy dx

x=a

y=c

g(x) dx

h(y) dy .

When areas are more complicated than those shown, we may break them into two or more simpler areas (Problems 9 and 10). We have seen how to set up and evaluate double integrals to find areas and volumes. Recall, however, that we use single integrals for other purposes than finding areas. Similarly, now that we know how to evaluate a double integral, we can use it to find other quantities besides areas and volumes.

246 Multiple Integrals; Applications of Integration Chapter 5

Example 2. Find the mass of a rectangular plate bounded by x = 0, x = 2, y = 0, y = 1, if its density (mass per unit area) is f (x, y) = xy. The mass of a tiny rectangle ∆A = ∆x ∆y is approximately f (x, y) ∆x ∆y, where f (x, y) is evaluated at some point in ∆A. We want to add up the masses of all the ∆A’s; this is what we find by evaluating the double integral of dM = xy dx dy. We call dM an element of mass and think of adding up all the dM ’s to get M .

V f (x, y, z) dx dy dz, is also defined as the limit of a sum and is evaluated by an iterated integral. If the

A triple integral of f (x, y, z) over a volume V , written

integral is over a box, that is, all limits are constants, then we can do the x, y, z integrations in any order. If the volume is complicated, then we have to consider the geometry as we did for double integrals to decide on the best order and find the limits. This process can best be learned from examples (below and Section 3) and practice (see problems).

Example 3. Find the volume of the solid in Figure 2.3 by using a triple integral. Here we imagine the whole solid cut into tiny boxes of volume ∆x ∆y ∆z; an element of volume is dx dy dz. We first add up the volumes of the tiny boxes to get the volume of a column; this means integrating with respect to z from 0 to 1 + y with x and y constant. Then we add up the columns to get a slab and the slabs to get the whole volume just as we did in Example 1. Thus:

1 2−2x 1+y

as in (2.1) and (2.2). Or, we could have used (2.4). Example 4. Find the mass of the solid in Figure 2.3 if the density (mass per unit volume)

is x + z. An element of mass is dM = (x + z) dx dy dz. We add up elements of mass just as we add up elements of volume; that is, the limits are the same as in Example 3.

where we evaluate the integrals as we did (2.1) to (2.4). (Check the result by hand and by computer.)

Section 2 Double and Triple Integrals 247

PROBLEMS, SECTION 2

In the problems of this section, set up and evaluate the integrals by hand and check your results by computer.

y=0 x=2y

y=1 x=√y

In Problems 7 to 18 evaluate the double integrals over the areas described. To find the limits, sketch the area and compare Figures 2.5 to 2.7.

7. RR A (2x − 3y) dx dy, where A is the triangle with vertices (0, 0), (2, 1), (2, 0). 8. RR A 6y 2 cos x dx dy, where A is the area inclosed by the curves y = sin x, the x axis, and the line x = π/2. 9. RR A sin x dx dy where A is the area shown in Figure 2.8. 10. RR A y dx dy where A is the area in Figure 2.8.

Figure 2.8 11. RR

A x dx dy, where A is the area between the parabola y = x and the straight line 2x − y + 8 = 0.

12. RR y dx dy over the triangle with vertices (−1, 0), (0, 2), and (2, 0). 13. RR 2xy dx dy over the triangle with vertices (0, 0), (2, 1), (3, 0).

2 x 14. 2 RR x e y dx dy over the area bounded by y = x −1 ,y=x −2 , and x = ln 4. 15. RR dx dy over the area bounded by y = ln x, y = e + 1 − x, and the x axis. 16. RR (9 + 2y 2 ) −1 dx dy over the quadrilateral with vertices (1, 3), (3, 3), (2, 6), (6, 6). 17. RR (x/y) dx dy over the triangle with vertices (0, 0), (1, 1), (1, 2).

18. RR y −1/2 dx dy over the area bounded by y = x 2 , x + y = 2, and the y axis. In Problems 19 to 24, use double integrals to find the indicated volumes.

19. Above the square with vertices at (0, 0), (2, 0), (0, 2), and (2, 2), and under the plane z = 8 − x + y.

20. Above the rectangle with vertices (0, 0), (0, 1), (2, 0), and (2, 1), and below the

surface z 2 = 36x 2 (4 − x 2 ).

21. Above the triangle with vertices (0, 0), (2, 0), and (2, 1), and below the paraboloid 2 z = 24 − x 2 −y .

22. Above the triangle with vertices (0, 2), (1, 1), and (2, 2), and under the surface z = xy.

23. Under the surface z = y(x + 2), and over the area bounded by x + y = 0, y = 1, y= √ x. 24. Under the surface z = 1/(y +2), and over the area bounded by y = x and y 2 +x = 2.

248 Multiple Integrals; Applications of Integration Chapter 5

In Problems 25 to 28, sketch the area of integration, observe that it is like the areas in Figure 2.7, and so write an equivalent integral with the integration in the opposite order. Check your work by evaluating the double integral both ways. Also check that your computer gives the same answer for both orders of integration.

Z 1 Z 3−3x

x=y/2

In Problems 29 to 32, observe that the inside integral cannot be expressed in terms of elementary functions. As in Problems 25 to 28, change the order of integration and so evaluate the double integral. Also try using your computer to evaluate these for both orders of integration.

x 33. A lamina covering the quarter disk x 2 +y 2 ≤ 4, x > 0, y > 0, has (area) density

y=0

x=y

x + y. Find the mass of the lamina. 34. A dielectric lamina with charge density proportional to y covers the area between

the parabola y = 16 − x 2 and the x axis. Find the total charge. 35. A triangular lamina is bounded by the coordinate axes and the line x + y = 6. Find

its mass if its density at each point P is proportional to the square of the distance from the origin to P .

36. A partially silvered mirror covers the square area with vertices at (±1, ±1). The fraction of incident light which it reflects at (x, y) is (x − y) 2 /4. Assuming a uniform intensity of incident light, find the fraction reflected.

In Problems 37 to 40, evaluate the triple integrals. Z 2 Z 2x Z y−x

Z 3 Z 2 Z 2y+z Z 2 Z 2x Z 1/z

39. 6y dx dz dy

40. z dy dz dx.

y=−2 z=1 x=y+z

41. Find the volume between the planes z = 2x + 3y + 6 and z = 2x + 7y + 8, and over the triangle with vertices (0, 0), (3, 0), and (2, 1).

42. Find the volume between the planes z = 2x + 3y + 6 and z = 2x + 7y + 8, and over the square in the (x, y) plane with vertices (0, 0), (1, 0), (0, 1), (1, 1).

43. Find the volume between the surfaces z = 2x 2 +y 2 + 12 and z = x 2 +y 2 + 8, and over the triangle with vertices (0, 0), (1, 0), and (1, 2).

44. Find the mass of the solid in Problem 42 if the density is proportional to y. 45. Find the mass of the solid in Problem 43 if the density is proportional to x. 46. Find the mass of a cube of side 2 if the density is proportional to the square of the

distance from the center of the cube.

Section 3 Applications of Integration; Single and Multiple Integrals 249

47. Find the volume in the first octant bounded by the coordinate planes and the plane x + 2y + z = 4.

48. Find the volume in the first octant bounded by the cone z 2 =x 2 −y 2 and the plane x = 4.

49. 2 Find the volume in the first octant bounded by the paraboloid z = 1 − x 2 −y , the plane x + y = 1, and all three coordinate planes.

50. Find the mass of the solid in Problem 48 if the density is z.