We now study the integral of real-valued functions of more than one variable. Most of our time is devoted to functions of two variables. In this section, we introduce the integral informally by thinking about how to visualize it. This uses an idea likely to be familiar from first-year calculus.
It won’t be until the next section that we define what the integral actually is.
LetDbe a subset ofR2. At this point, we won’t be too careful about putting any conditions on D, but it’s best to think of it as some sort of two-dimensional blob, as opposed to a set of discrete points or a curve. Let f:A → R be a real-valued function whose domain contains D. For the moment, assume that f(x)≥0 for allx inD.
Let W be the region in R3 that lies below the graph z = f(x, y) and above D as shown in Figure 5.1. That is:
W ={(x, y, z)∈R3 : (x, y)∈Dand 0≤z≤f(x, y)}.
Figure 5.1: The regionW under the graphz=f(x, y) and aboveD Consider the volume of W, which we denote variously by RR
Df(x, y)dA,RR
Df(x, y)dx dy, or justRR
Df dA. To calculate it, we apply the following principle from first-year calculus:
Volume = Integral of cross-sectional area.
For instance, in first-year calculus, the cross-sections may turn out to be disks, washers, or, in a slight variant, cylinders.
Example 5.1. Find the volume under the planez = 4−x+ 2y and above the square 1≤x≤3, 2≤y≤4, in the xy-plane. See Figure5.2.
121
Figure 5.2: The region under z= 4−x+ 2y and above the square 1≤x≤3, 2≤y≤4 Here, f(x, y) = 4−x+ 2y, and, if Ddenotes the specified square, we are looking for RR
D(4− x+ 2y)dA. As above, letW denote the region below the plane and above the square. We consider first cross-sections perpendicular to the x-axis, that is, intersections of W with planes of the form x= constant. There is such a cross-section for each value ofx from x= 1 to x= 3, so:
Volume = Z 3
1
A(x)dx, (5.1)
whereA(x) is the area of the cross-section atx. A typical cross-section is shown in Figure5.3. The
Figure 5.3: A cross-section perpendicular to the x-axis
base is a line segment fromy= 2 toy= 4, and the top is a curve lying in the graphz= 4−x+ 2y.
(In fact, the cross-sections in this example are trapezoids, but let’s not worry about that.) In other words, A(x) is an area under a curve, so it too is an integral: A(x) = R4
2(4−x+ 2y)dy, where, for the given cross-section, x is fixed. Integrating the cross-sectional area as in equation (5.1) and evaluating gives:
Volume = Z 3
1
Z 4 2
4−x+ 2y dy
dx
= Z 3
1
4y−xy+y2
y=4 y=2
dx
= Z 3
1
(16−4x+ 16)−(8−2x+ 4) dx
= Z 3
1
(20−2x)dx
= 20x−x2
3 1
= (60−9)−(20−1)
= 32.
5.1. VOLUME AND ITERATED INTEGRALS 123 Alternatively, we could have used cross-sections perpendicular to the y-axis, in which case the cross-sections occur fromy= 2 to y= 4 (Figure 5.4):
Volume = Z 4
2
A(y)dy. (5.2)
Figure 5.4: A cross-section perpendicular to they-axis
Each cross-section lies below the graph and above a line segment that goes from x= 1 to x = 3, soA(y) =R3
1(4−x+ 2y)dx. This time the integral of the cross-sectional area (5.2) gives:
Volume = Z 4
2
Z 3 1
4−x+ 2y dx
dy
= Z 4
2
4x− 1
2x2+ 2xy
x=3 x=1
dy
= Z 4
2
(12−9
2+ 6y)−(4−1
2 + 2y) dx
= Z 4
2
(4 + 4y)dy
= 4y+ 2y2
4 2
= 48−16
= 32.
According to this approach, the volume is the integral of an integral, or aniterated integral.
Evaluating the integral consists of a sequence of “partial antidifferentiations” with respect to one variable at a time, all other variables being held constant. Once you integrate with respect to a particular variable, that variable disappears from the remainder of the calculation.
Example 5.2. Find the volume of the solid that lies inside the cylindersx2+z2 = 1 andy2+z2 = 1 and above thexy-plane.
The cylinders have axes along the yand x-axis, respectively, and intersect one another at right angles, as in Figure 5.5.
Perhaps it’s simplest to focus on the quarter of the solid that lies in the first “octant,” i.e., the region of R3 in which x,y, andz are all nonnegative. The base of this portion is the unit square 0≤x≤1, 0≤y≤1 (Figure5.6, left). Half of the solid lies underx2+z2 = 1 (Figure5.6, right), running in they-direction above the triangle in thexy-plane bounded by thex-axis, the linex= 1, and the liney=x. Call this triangleD1. The other half runs in thex-direction, undery2+z2 = 1
Figure 5.5: The cylindersx2+z2 = 1 (left),y2+z2= 1 (middle), and the region contained in both (right)
Figure 5.6: The portion in the first octant (left) and the portion under onlyx2+z2 = 1 (right) and above a complementary triangle D2 within the unit square. The volume of the original solid is 8 times the volume of the portion over D1. This lies below x2+z2 = 1, i.e., below the graph z=√
1−x2. Hence:
Volume = 8 Z Z
D1
p1−x2dA.
To calculate the volume over D1, let’s try cross-sections perpendicular to the x-axis. These exist fromx= 0 to x= 1:
Volume = 8 Z 1
0
A(x)dx.
The base of each cross-section is a line segment perpendicular to the x-axis within D1. The lower endpoint is always aty= 0, but the upper endpoint depends onx. Indeed, the upper endpoint lies on the line y=x. See Figure5.7.
Figure 5.7: The base of the cross-section atx
5.1. VOLUME AND ITERATED INTEGRALS 125 Hence A(x) =Rx
0
√
1−x2dy, so:
Volume = 8 Z 1
0
Z x 0
p1−x2dy
dx
= 8 Z 1
0
p1−x2ãy
y=x y=0dx
= 8 Z 1
0
xp
1−x2dx (letu= 1−x2, du=−2x dx)
= 8
−1 2ã2
3(1−x2)3/2
1 0
=−8
3(03/2−13/2)
= 8 3.
Alternatively, suppose we had used cross-sections perpendicular to they-axis. The cross-sections exist from y = 0 to y = 1, and, for each such y, the base of the cross-section is a line segment in thex-direction that goes fromx=y tox= 1, as illustrated in Figure 5.8.
Figure 5.8: The base of the cross-section aty Consequently:
Volume = 8 Z 1
0
A(y)dy= 8 Z 1
0
Z 1 y
p1−x2dx
dy.
While not impossible, it is a little less obvious how to do the first partial antidifferentiation with respect tox by hand. In other words, our original order of antidifferentiation might be preferable.
Example 5.3. Consider the iterated integral Z 2
0
Z 4 x2
x3ey3dy
dx.
(a) Sketch the domain of integrationD in thexy-plane.
(b) Evaluate the integral.
(a) The domain of integration can be reconstructed from the endpoints of the integrals. The outermost integral says that x goes from x= 0 to x= 2. Geometrically, we are integrating areas of cross-sections that are perpendicular to the x-axis. Then the inner integral says that, for each x, y goes from y =x2 toy = 4. Figure 5.9 exhibits the relevant input. Hence D is described by the conditions:
D={(x, y)∈R2 : 0≤x≤2, x2 ≤y≤4}.
This is the region in the first quadrant bounded by the parabola y = x2, the line y = 4, and the y-axis.
Figure 5.9: The domain of integration: cross-sections perpendicular to the x-axis
(b) To evaluate the integral as presented, we would antidifferentiate first with respect toy, treating x as constant:
Z 2 0
Z 4 x2
x3ey3dy
dx= Z 2
0
x3 Z 4
x2
ey3dy
dx.
The innermost antiderivative looks hard. So, having nothing better to do, we try switching the order of antidifferentiation. Using cross-sections perpendicular to they-axis, we see from the description of Dthaty goes fromy = 0 toy= 4, and, for each y,x goes fromx= 0 to x=√
y. The thinking behind the switched order is illustrated in Figure5.10.
Figure 5.10: Reversing the order of antidifferentiation Therefore:
Z 2
0
Z 4 x2
x3ey3dy
dx= Z 4
0
Z √y 0
x3ey3dx
dy
= Z 4
0
1 4x4ey3
x=√ y x=0
dy
= Z 4
0
1
4y2ey3−0
dy (letu=y3, du= 3y2dy)
= 1 4ã1
3ey3
4 0
= 1
12(e64−1).