CHAPTER 44
Multiple Integrals and their Applications
44.1 Evaluate the iterated integral (x + 2y) dx dy.
Therefore,
44.2 Evaluate the iterated integral (x2 + y2) dy dx.
Therefore,
44.3 Evaluate the iterated integral sin e dr d6.
Therefore, (Problem 20.48).
44.4 Evaluate the iterated integral
Hence,
44.5 Evaluate the iterated integral dx dy dz.
Hence,
Therefore,
44.6 Evaluate
In this case the double integral may be replaced by a product:
6. (See Problem 44.71.)
44.7 Evaluate
Therefore,
44.8 Evaluate
dx cannot be evaluated in terms of standard functions. Therefore, we change the order of integration.
using Fig. 44-1.
Fig. 44-1
44.9 Evaluate sin y dx dy.
s(sin y)ecos
Therefore,
405
,406 CHAPTER 44
44.10 Evaluate
Therefore,
44.11 Evaluate where 9? is the region bounded by y —x and y = x2.
The curves y = x and y = x2 intersect at (0,0) and (1,1), and, for 0<*<1, y =x is above
y = x (see Fig. 44-2).
Fig. 44-2 Fig. 44-3
44.12 Evaluate where 91 is the region bounded by y = 2x, y = 5x, and x = 2.
The lines y = 2x and y = 5x intersect at the origin. For 0 < j c < l , the region runs from y = 2x
up to y = 5x (Fig. 44-3). Hence,
44.13 Evaluate where &i is the region above the x-axis bounded by y 2 = 3x and y2 = 4 — x
(see Fig. 44-4).
It is convenient to evaluate / by means of strips parallel to the *-axis.
Fig. 44-4
44.14 Evaluate where £% is the region in the first quadrant bounded by x2 = 4 - 2y.
, MULTIPLE INTEGRALS AND THEIR APPLICATIONS 407
The curve x2 = 4 - 2y is a parabola with vertex at (0, 2) and passing through the A:-axis at
x =2 (Fig. 44-5). Hence,
Note that, if we integrate using strips
parallel to the y-axis, the integration is difficult.
Fig. 44-5 Fig. 44-6
44.15 Let 91 be the region bounded by the curve y = Vic and the line y = x (Fig. 44-6). Let
if y^O and f ( x , 0) = 1. Compute
dy. Integration by parts yields J y sin y dy =
sin y - y cos y. Hence, / = (-cos y + y cos y - sin y) (-sin 1)-(-!) = !-sin 1.
44.16 Find the volume V under the plane z = 3x + 4y and over the rectangle 91: l<x£2, O s y < 3 .
44.17 Find the volume V in the first octant bounded by z = y2, x = 2, and y = 4.
44.18 Find the volume V of the solid in the first octant bounded by y = 0, z = 0, y = 3, z = x, and z + x = 4
(Fig. 44-7).
For given x and y, the z-value in the solid varies from z =x to z = — x + 4. So V=
Fig. 44-7
Multiple Integrals and their Applications
44.1 Evaluate the iterated integral (x + 2y) dx dy.
Therefore,
44.2 Evaluate the iterated integral (x2 + y2) dy dx.
Therefore,
44.3 Evaluate the iterated integral sin e dr d6.
Therefore, (Problem 20.48).
44.4 Evaluate the iterated integral
Hence,
44.5 Evaluate the iterated integral dx dy dz.
Hence,
Therefore,
44.6 Evaluate
In this case the double integral may be replaced by a product:
6. (See Problem 44.71.)
44.7 Evaluate
Therefore,
44.8 Evaluate
dx cannot be evaluated in terms of standard functions. Therefore, we change the order of integration.
using Fig. 44-1.
Fig. 44-1
44.9 Evaluate sin y dx dy.
s(sin y)ecos
Therefore,
405
,406 CHAPTER 44
44.10 Evaluate
Therefore,
44.11 Evaluate where 9? is the region bounded by y —x and y = x2.
The curves y = x and y = x2 intersect at (0,0) and (1,1), and, for 0<*<1, y =x is above
y = x (see Fig. 44-2).
Fig. 44-2 Fig. 44-3
44.12 Evaluate where 91 is the region bounded by y = 2x, y = 5x, and x = 2.
The lines y = 2x and y = 5x intersect at the origin. For 0 < j c < l , the region runs from y = 2x
up to y = 5x (Fig. 44-3). Hence,
44.13 Evaluate where &i is the region above the x-axis bounded by y 2 = 3x and y2 = 4 — x
(see Fig. 44-4).
It is convenient to evaluate / by means of strips parallel to the *-axis.
Fig. 44-4
44.14 Evaluate where £% is the region in the first quadrant bounded by x2 = 4 - 2y.
, MULTIPLE INTEGRALS AND THEIR APPLICATIONS 407
The curve x2 = 4 - 2y is a parabola with vertex at (0, 2) and passing through the A:-axis at
x =2 (Fig. 44-5). Hence,
Note that, if we integrate using strips
parallel to the y-axis, the integration is difficult.
Fig. 44-5 Fig. 44-6
44.15 Let 91 be the region bounded by the curve y = Vic and the line y = x (Fig. 44-6). Let
if y^O and f ( x , 0) = 1. Compute
dy. Integration by parts yields J y sin y dy =
sin y - y cos y. Hence, / = (-cos y + y cos y - sin y) (-sin 1)-(-!) = !-sin 1.
44.16 Find the volume V under the plane z = 3x + 4y and over the rectangle 91: l<x£2, O s y < 3 .
44.17 Find the volume V in the first octant bounded by z = y2, x = 2, and y = 4.
44.18 Find the volume V of the solid in the first octant bounded by y = 0, z = 0, y = 3, z = x, and z + x = 4
(Fig. 44-7).
For given x and y, the z-value in the solid varies from z =x to z = — x + 4. So V=
Fig. 44-7
