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James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

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996 Chapter 15 Multiple Integrals

Example 7 Find the volume of the solid S that is bounded by the elliptic paraboloid

x 2 1 2y 2 1 z − 16, the planes x − 2 and y − 2, and the three coordinate planes.

16

12

z 8

4

0

0

1

y

2 2

1

x

0

SOLUTIon We first observe that S is the solid that lies under the surface

z − 16 2 x 2 2 2y 2 and above the square R − f0, 2g 3 f0, 2g. (See Figure 15.) This

solid was considered in Example 1, but we are now in a position to evaluate the double

integral using Fubini’s Theorem. Therefore

V − yy s16 2 x 2 2 2y 2 d dA − y 2

0 y2 s16 2 x 2 2 2y 2 d dx dy

0

R

− y 2

0 f16x 2 1 3 x 3 2 2y 2 xg x−0

x−2

dy

FIGURE 15

− y 2

0 (88

3 2 4y 2 ) dy − f 88

3 y 2 4 3 y3 g 0

2

− 48 ■

In the special case where f sx, yd can be factored as the product of a function of x only

and a function of y only, the double integral of f can be written in a particularly simple

form. To be specific, suppose that f sx, yd − tsxdhsyd and R − fa, bg 3 fc, dg. Then

Fubini’s Theorem gives

yy f sx, yd dA − y d

c

yb tsxdhsyd dx dy − y d

Fy b

tsxdhsyd dxG dy

a

c a

R

In the inner integral, y is a constant, so hsyd is a constant and we can write

y d

c

Fy b

tsxdhsyd dxG dy − y d

FhsydSy b

tsxd dxDG dy − y b

tsxd dx y d

hsyd dy

a

c

a

a

c

since y b tsxd dx is a constant. Therefore, in this case the double integral of f can be written

as the product of two single

a

integrals:

11 yy tsxd hsyd dA − y b

tsxd dx y d

hsyd dy where R − fa, bg 3 fc, dg

a c

R

Example 8 If R − f0, y2g 3 f0, y2g, then, by Equation 11,

yy sin x cos y dA − y y2

sin x dx y y2

cos y dy

0

0

R

− f2cos xg 0

y2

fsin yg 0

y2

− 1 ? 1 − 1 ■

The function f sx, yd − sin x cos y in

Example 8 is positive on R, so the

integral represents the volume of the

solid that lies above R and below the

graph of f shown in Figure 16.

z

0

y

FIGURE 16

x

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