Section 15.5 Surface Area 2. Find the area of the indicated part of the surface (above the region

Section 15.5 Surface Area

2. Find the area of the indicated part of the surface (above the region D).

SECTION 15.5 Surface Area 1081

EXAMPLE 1 Find the surface area of the part of the surface z − x

²

1 2y 1 2 that lies

above the triangular region T in the xy-plane with vertices s0, 0d, s1, 0d, and s1, 1d.

SOLUTION

The region T is shown in Figure 3 and is described by

T −

hsx, yd | 0 < x < 1, 0 < y < xj Using Formula 2 with f sx, yd − x

²

1 2y 1 2 , we get

A − y

T

y ^ss2xd

²

^{1 s2d}

²

^{1 1 dA −} y

0¹

y

0^x

s4x

²

1 5 dy dx − y

0¹

x s4x

²

1 5 dx −

¹₈



²₃

s4x

²

1 5 d

^3y2

g

¹⁰

⁻

¹²¹

^{( 27 2 5 s5 )}

Figure 4 shows the portion of the surface whose area we have just computed.

■

EXAMPLE 2 Find the area of the part of the paraboloid z − x

²

1

y²

that lies under the plane z − 9.

SOLUTION

The plane intersects the paraboloid in the circle x

²

1

y²

− 9, z − 9.

Therefore the given surface lies above the disk D with center the origin and radius 3.

(See Figure 5.) Using Formula 3, we have A − y

D

y Î ^{1 1} S ^{−x −z} D

²

¹ S ^{−z −y} D

²

^{dA − y}

_D

^{y s1 1 s2xd}

²

^{1 s2yd}

²

^dA

− y

D

y ^{s1 1 4sx}

²

¹

^y2

^{d dA}

Converting to polar coordinates, we obtain

A − y

0²

y

0³

s1 1 4r

²

r dr d − y

0²

d y

0³

18

s1 1 4r

²

s8rd dr

− 2 (

¹₈

)

²₃

s1 1 4r

²

d

^3y2

g

⁰³

⁻



₆ ^{( 37 s37 2 1 )}

^■

x y=x

T (1, 0)

(1, 1)

(0, 0) y

FIGURE 3

x 0

z

T

FIGURE 4

9

x

z

3 y D

FIGURE 5

15.5 Exercises

1–2 Find the area of the indicated part of the surface (above the region D).

1. 2.

y (2, _2, 0)

x z

z=10+x+¥

D 2 _2

D

z=3+xy

≈+¥¯1, z=0 y x

z

3–14 Find the area of the surface.

3. The part of the plane 5x 1 3y 2 z 1 6 − 0 that lies above the rectangle f1, 4g 3 f2, 6g

4. The part of the plane 6x 1 4y 1 2z − 1 that lies inside the cylinder x²1y²− 25

5. The part of the plane 3x 1 2y 1 z − 6 that lies in the first octant

6. The part of the surface 2y 1 4z 2 x²− 5 that lies above the triangle with vertices s0, 0d, s2, 0d, and s2, 4d

7. The part of the paraboloid z − 1 2 x²2y² that lies above the plane z − 22

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Solution:

1536 ¤ CHAPTER 15 MULTIPLE INTEGRALS

35. (a) If ( ) is the probability that an individual at  will be infected by an individual at  , and   is the number of infected individuals in an element of area , then ( )  is the number of infections that should result from exposure of the individual at  to infected people in the element of area . Integration over  gives the number of infections of the person at  due to all the infected people in . In rectangular coordinates (with the origin at the city’s center), the exposure of a person at  is

 =





 ( )  = 



 1

20[20 − ( )]  = 





1 −20¹

( − ⁰)²+ ( − ⁰)²



(b) If  = (0 0), then

 = 





 1 −20¹

²+ ²



= 

 2

0

 10 0

1 −20¹

   = 21

2²−60¹³¹⁰

0

= 2 50 −⁵⁰3

= ²⁰⁰₃  ≈ 209

For  at the edge of the city, it is convenient to use a polar coordinate system centered at . Then the polar equation for the circular boundary of the city becomes  = 20 cos  instead of  = 10, and the distance from  to a point  in the city is again  (see the figure). So

 = 

 2

−2

 20 cos  0

1 −20¹

   = 

 2

−2

₁

2²−60¹³^{=20 cos }

=0 

= 2

−2

200 cos² −⁴⁰⁰3 cos³

 = 2002

−2

1

2+¹₂cos 2 −²3

1 − sin² cos 



= 2001

2 +¹₄sin 2 −²3sin  + ²₃· ¹3sin³^2

−2 = 200

4 + 0 −²3 +²₉+^₄ + 0 − ²3+²₉

= 200 2 −⁸9

≈ 136

Therefore the risk of infection is much lower at the edge of the city than in the middle, so it is better to live at the edge.

15.5 Surface Area

1. Here  = ( ) = 10 +  + ²and  is the triangle with vertices (0 0), (0 −2), and (2 −2). By Formula 2, the area of the surface is

() =



[( )]²+ [( )]²+ 1  =0

−2

_−

0

1²+ (2)²+ 1  

=0

−2

_−

0

2 + 4²  =0

−2

2 + 4²[]^=_=0^− = −0

−2

2 + 4²

= −1 8·

2

3(2 + 4²)^32

0

−2

= 18^32− 2^32

12 = 54√

2 − 2√ 2

12 = 13√

2 3 2. Here  = ( ) = 3 +  and  is the circle ²+ ²≤ 1. By Formula 2, the area of the surface is

() =





[( )]²+ [( )]²+ 1  =



²+²≤1

²+ ²+ 1  [Switch to polar coordinates]

=2

0

1 0

√²+ 1    =2

0 1 0

√²+ 1   = 2 ·¹2

2

3(²+ 1)^32=1

=0= ^2₃ (2^32− 1)

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6. Find the area of the surface. The part of the surface 2y − 4z − x² = 5 that lies above the triangle with vertices (0, 0), (2, 0), and (2, 4).

Solution:

SECTION 15.5 SURFACE AREA ¤ 563 is again  (see the figure). So

 = 

 2

−2

 20 cos  0

 1 − 

20



   = 

 2

−2

² 2 − ³

60

^{=20 cos }

=0



= 2

−2

200 cos² −⁴⁰⁰3 cos³

 = 2002

−2

1

2+ ¹₂cos 2 − ²3

1 − sin² cos 



= 2001

2 +¹₄sin 2 −²3sin  +²₃· ¹3sin³2

−2 = 200

4 + 0 − ²3+²₉ +^₄ + 0 −²3+²₉

= 200_

2 −⁸9

≈ 136

Therefore the risk of infection is much lower at the edge of the city than in the middle, so it is better to live at the edge.

15.5 Surface Area

1. Here  = ( ) = 5 + 3 + 6 and  is the rectangle [1 4] × [2 6], so by Formula 2 the area of the surface is

() =



[( )]²+ [( )]²+ 1  =



√5²+ 3²+ 1  =√ 35



=√

35 () =√

35 (3)(4) = 12√ 35

2.  = ( ) = ¹₂− 3 − 2 and  is the disk ²+ ²≤ 25, so by Formula 2

() =



(−3)²+ (−2)²+ 1  =√ 14

 =√

14 () =√

14 ( · 5²) = 25√ 14  3. The surface  is given by  = ( ) = 6 − 3 − 2 which intersects the -plane in the line 3 + 2 = 6, so  is the

triangular region given by ( ) 

 0 ≤  ≤ 2, 0 ≤  ≤ 3 −³2. By Formula 2, the surface area of  is

() =



(−3)²+ (−2)²+ 1  =√ 14

 =√

14 () =√ 141

2· 2 · 3

= 3√ 14

4.  = ( ) = ¹₄²−¹2 +⁵₄, and  is the triangular region given by {( ) | 0 ≤  ≤ 2 0 ≤  ≤ 2}. By Formula 2,

() =



₁

22

+

−¹2

2

+ 1  =2 0

2

0

1

4²+⁵₄  =2 0

1 2

√²+ 5

=2

=0 

= ¹₂2 0 2√

²+ 5  = ¹₂ ·²3(²+ 5)^322

0= ¹₃(9^32− 5^32) = 9 − ⁵3

√5

5. The paraboloid intersects the plane  = −2 when 1 − ²− ²= −2 ⇔ ²+ ²= 3, so  =

( ) | ²+ ²≤ 3. Here  = ( ) = 1 − ²− ² ⇒ = −2, ^ = −2 and

() =



(−2)²+ (−2)²+ 1  =



4(²+ ²) + 1  =2

0

^√3 0

√4²+ 1   

=2

0 ^√3 0 √

4²+ 1  =

2

0

1

12(4²+ 1)^32^√3

0 = 2 · 12¹

13^32− 1

= ^₆ 13√

13 − 1

6. ²+ ²= 4 ⇒  =√

4 − ²(since  ≥ 0), so ^= −(4 − ²)^−12,  = 0 and

() =

 1 0

 1 0



[−(4 − ²)^−12]²+ 0²+ 1   =

 1 0

 1 0



²

4 − ² + 1  

=

 1 0

√ 2

4 − ²

 1 0

 =

2 sin⁻¹ 2

1 0

1 0=

2 ·^6 − 0

(1) =^₃

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13. Find the area of the surface. The part of the sphere x²+ y²+ z² = a² that lies within the cylinder x²+ y²= ax and above the xy-plane.

Solution:

564 ¤ CHAPTER 15 MULTIPLE INTEGRALS 7. ²+ ²= 9 ⇒  =

9 − ². = 0,  = −(9 − ²)^−12 ⇒

() =

 4 0

 2 0



0²+ [−(9 − ²)^−12]²+ 1   =

 4 0

 2 0



²

9 − ² + 1  

=

 4 0

 2 0

 3

9 − ²  = 3

 4 0

 sin⁻¹

3

=2

=0  = 3

sin⁻¹2 3

⁴

0= 12 sin⁻¹2 3



8.  = ( ) = 1 + 3 + 2²with 0 ≤  ≤ 2, 0 ≤  ≤ 1. Thus by Formula 2,

() =



1 + (3)²+ (4)² =1 0

2

0

10 + 16²  =1 0

10 + 16²

=2

=0 

=1 0 2

10 + 16² = 2 ·32¹ ·²3(10 + 16²)^321

0= ₂₄¹(26^32− 10^32) 9.  = ( ) = with ²+ ²≤ 1, so ^= , =  ⇒

() =



²+ ²+ 1  =2

0

1 0

√²+ 1    =2

0

₁

3(²+ 1)^32=1

=0 

=2

0 1 3

2√ 2 − 1

 =^2₃  2√

2 − 1

10.  = ( ) = 4 − ²− ²and  is the projection of the paraboloid  = 4 − ²− ²onto the -plane, that is,

 =

( ) | ²+ ²≤ 4. So = −2, ^ = −2 ⇒

() =



(−2)²+ (−2)²+ 1  =



4(²+ ²) + 1  =2

0

2 0

√4²+ 1   

=2

0

1

12(4²+ 1)^32=2

=0  =2

0 1 12

17√ 17 − 1

 = ^₆ 17√

17 − 1

11.  =

²− ²− ², = −(²− ²− ²)^−12,  = −(²− ²− ²)^−12,

 () =







²+ ²

²− ²− ² + 1 

=

 2

−2

  cos  0

 ²

²− ² + 1   

=

 2

−2

  cos  0

√ 

²− ² 

=

 2

−2

−

²− ²= cos 

=0 

=

 2

−2−

²− ²cos² − 

 = 2²

 2 0

 1 −

1 − cos²



= 2²

 2 0

 − 2²

 2 0

sin²  = ² − 2²

 2 0

sin   = ²( − 2)

12. To find the region :  = ²+ ²implies  + ²= 4or ²− 3 = 0. Thus  = 0 or  = 3 are the planes where the surfaces intersect. But ²+ ²+ ²= 4implies ²+ ²+ ( − 2)²= 4, so  = 3 intersects the upper hemisphere. Thus

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1

25. Find the area of the finite part of the paraboloid y = x²+ z²cut off by the plane y = 25. [Hint: Project the surface onto the xz-plane.]

Solution:

SECTION 15.6 TRIPLE INTEGRALS ¤ 567

21. Here  = ( ) =  +  + , ( ) = , ( ) = , so

() =



√²+ ²+ 1  =√

²+ ²+ 1

 =√

²+ ²+ 1 ().

22. Let  be the upper hemisphere. Then  = ( ) =

²− ²− ², so

() =



[−(²− ²− ²)^−12]²+ [−(²− ²− ²)^−12]²+ 1 

=







²+ ²

²− ²− ² + 1  = lim

→⁻

 2

0

  0

 ²

²− ² + 1   

= lim

→⁻

 2

0

  0

√ 

²− ²  = 2 lim

→⁻

−

²− ²

0= 2 lim

→⁻−

²− ²− 

= 2(−)(−) = 2². Thus the surface area of the entire sphere is 4².

23. If we project the surface onto the -plane, then the surface lies “above” the disk ²+ ²≤ 25 in the -plane.

We have  = ( ) = ²+ ²and, adapting Formula 2, the area of the surface is

() = 

²+²≤25

[( )]²+ [( )]²+ 1  = 

²+²≤25

√4²+ 4²+ 1 

Converting to polar coordinates  =  cos ,  =  sin  we have

() =2

0

5 0

√4²+ 1    =2

0  5

0 (4²+ 1)^12 =

2

0

1

12(4²+ 1)^325 0= ^₆

101√

101 − 1

24. First we find the area of the face of the surface that intersects the positive -axis. As in Exercise 23, we can project the face onto the -plane, so the surface lies “above” the disk ²+ ²≤ 1. Then  = ( ) =√

1 − ²and the area is

 () =



²+²≤1

[( )]²+ [( )]²+ 1  =



²+²≤1

 0 +

√−

1 − ²

2

+ 1 

=



²+²≤1



²

1 − ² + 1  =

 1

−1

 √1−²

−√

1−²

√ 1

1 − ²  

= 4

 1 0

 √1−² 0

√ 1

1 − ²  [by the symmetry of the surface]

This integral is improper (when  = 1), so

 () = lim

→1⁻

4

  0

 √1−² 0

√ 1

1 − ²   = lim

→1⁻

4

  0

√1 − ²

√1 − ² = lim

→1⁻

4

  0

 = lim

→1⁻4 = 4.

Since the complete surface consists of four congruent faces, the total surface area is 4(4) = 16.

15.6 Triple Integrals

1. 

² =1 0

3 0

2

−1²   =1 0

3 0

1

2²²=2

=−1  =1 0

3 0

3

2² 

=1 0

1

2³=3

=0 =1 0

27

2   = ²⁷₄²1 0= ²⁷₄

° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.c

26. The figure shows the surface created when the cylinder y²+ z² = 1 intersects the cylinder x²+ z² = 1. Find the area of this surface.

SeCtION15.6 Triple Integrals 1029

21. Show that the area of the part of the plane z − ax 1 by 1 c that projects onto a region D in the xy-plane with area AsDd is sa²1b²11 AsDd.

22. If you attempt to use Formula 2 to find the area of the top half of the sphere x²1y²1z²− a², you have a slight problem because the double integral is improper. In fact, the integrand has an infinite discontinuity at every point of the boundary circle x²1y²− a². However, the integral can be computed as the limit of the integral over the disk x²1y²< t² as t l a². Use this method to show that the area of a sphere of radius a is 4a².

23. Find the area of the finite part of the paraboloid y − x²1z² cut off by the plane y − 25. [Hint: Project the surface onto the xz-plane.]

24. The figure shows the surface created when the cylinder y²1z²− 1 intersects the cylinder x²1z²− 1. Find the area of this surface.

z

x y 15. (a) Use the Midpoint Rule for double integrals (see Sec-

tion 15.1) with four squares to estimate the surface area of the portion of the paraboloid z − x²1y² that lies above the square f0, 1g 3 f0, 1g.

(b) Use a computer algebra system to approximate the sur- face area in part (a) to four decimal places. Compare with the answer to part (a).

16. (a) Use the Midpoint Rule for double integrals with m − n − 2 to estimate the area of the surface z − xy 1 x²1y², 0 < x < 2, 0 < y < 2.

(b) Use a computer algebra system to approximate the sur- face area in part (a) to four decimal places. Compare with the answer to part (a).

17. Find the exact area of the surface z − 1 1 2x 1 3y 1 4y², 1 < x < 4, 0 < y < 1.

18. Find the exact area of the surface

z − 1 1 x 1 y 1 x² 22 < x < 1 21 < y < 1 Illustrate by graphing the surface.

19. Find, to four decimal places, the area of the part of the surface z − 1 1 x²y² that lies above the disk x²1y²<1.

20. Find, to four decimal places, the area of the part of the surface z −s1 1 x²dys1 1 y²d that lies above the square

|

^x

|

¹

|

^y

|

^<1. Illustrate by graphing this part of the surface.

CAS

CAS

CAS

CAS

CAS

CAS

Just as we defined single integrals for functions of one variable and double integrals for functions of two variables, so we can define triple integrals for functions of three variables. Let’s first deal with the simplest case where f is defined on a rectangular box:

1 B −

h sx, y, zd | a < x < b, c < y < d, r < z < s j

The first step is to divide B into sub-boxes. We do this by dividing the interval fa, bg into

l subintervals

fx

i21

, x

i

g of equal width Dx, dividing fc, dg into m subintervals of width Dy, and dividing fr, sg into n subintervals of width Dz. The planes through the endpoints of these subintervals parallel to the coordinate planes divide the box B into lmn sub-boxes

Bi jk

− fx

i21

, x

i

g 3 fy

j21

, y

j

g 3 fz

k21

, z

k

g

which are shown in Figure 1. Each sub-box has volume DV − Dx Dy Dz.

Then we form the triple Riemann sum

2

o

^l

i−1

o

^m

j−1

o

ⁿ

k−1

f sx

ij k

* , y

ij k

* , z

ij k

* d DV

where the sample point sx

i jk

* , y

i jk

* , z

i jk

* d is in B

i jk

. By analogy with the definition of a double integral (15.1.5), we define the triple integral as the limit of the triple Riemann sums in (2).

B

Bijk

Îy Îx Îz z

x y

z

x y

FIGURE 1

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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Solution:

SECTION 15.6 TRIPLE INTEGRALS ¤ 567

21. Here  = ( ) =  +  + , ( ) = , ( ) = , so

() =



√²+ ²+ 1  =√

²+ ²+ 1

 =√

²+ ²+ 1 ().

22. Let  be the upper hemisphere. Then  = ( ) =

²− ²− ², so

() =



[−(²− ²− ²)^−12]²+ [−(²− ²− ²)^−12]²+ 1 

=







²+ ²

²− ²− ² + 1  = lim

→⁻

 2

0

  0

 ²

²− ² + 1   

= lim

→⁻

 2

0

  0

√ 

²− ²  = 2 lim

→⁻

−

²− ²

0= 2 lim

→⁻−

²− ²− 

= 2(−)(−) = 2². Thus the surface area of the entire sphere is 4².

23. If we project the surface onto the -plane, then the surface lies “above” the disk ²+ ²≤ 25 in the -plane.

We have  = ( ) = ²+ ²and, adapting Formula 2, the area of the surface is

() = 

²+²≤25

[( )]²+ [( )]²+ 1  = 

²+²≤25

√4²+ 4²+ 1 

Converting to polar coordinates  =  cos ,  =  sin  we have

() =2

0

5 0

√4²+ 1    =2

0 5

0 (4²+ 1)^12 =

2

0

1

12(4²+ 1)^325 0= ^₆

101√

101 − 1

24. First we find the area of the face of the surface that intersects the positive -axis. As in Exercise 23, we can project the face onto the -plane, so the surface lies “above” the disk ²+ ²≤ 1. Then  = ( ) =√

1 − ²and the area is

 () =



²+²≤1

[( )]²+ [( )]²+ 1  =



²+²≤1

 0 +

√−

1 − ²

2

+ 1 

=



²+²≤1



²

1 − ² + 1  =

 1

−1

 √1−²

−√

1−²

√ 1

1 − ² 

= 4

 1 0

 √1−² 0

√ 1

1 − ²  [by the symmetry of the surface]

This integral is improper (when  = 1), so

 () = lim

→1⁻

4

  0

 √1−² 0

√ 1

1 − ²  = lim

→1⁻

4

  0

√1 − ²

√1 − ² = lim

→1⁻

4

  0

 = lim

→1⁻4 = 4.

Since the complete surface consists of four congruent faces, the total surface area is 4(4) = 16.

15.6 Triple Integrals

1. 

² =1 0

3 0

2

−1²   =1 0

3 0

1

2²²=2

=−1  =1 0

3 0

3

2² 

=1 0

1

2³=3

=0 =1 0

27

2   = ²⁷₄²1 0= ²⁷₄

° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.c

2