15.3 Double Integrals in Polar Coordinates

Section 15.3 Double Integrals in Polar Coordinates

8. Sketch the region whose area is given by the integral and evaluate the integral. Rπ π/2

R2 sin θ 0 r drdθ Solution:

SECTION 15.3 DOUBLE INTEGRALS IN POLAR COORDINATES ¤ 545

70. To find the equations of the boundary curves, we require that the

-values of the two surfaces be the same. In Maple, we use the command solve(4-xˆ2-yˆ2=1-x-y,y); and in Mathematica, we use Solve[4-xˆ2-yˆ2==1-x-y,y]. We find that the curves have equations  =1 ±√

13 + 4 − 4²

2 . To find the two points of intersection of these curves, we use the CAS to solve 13 + 4 − 4²= 0, finding that

 =^1±₂^√14. So, using the CAS to evaluate the integral, the volume of intersection is

 =

 (^{1 +√14})^2 (¹−√

14)^2

 ^1 +√

13 + 4− 4²

2

 1−√

13 + 4− 4²

2

[(4 − ²− ²) − (1 −  − )]   = 49

8

15.3 Double Integrals in Polar Coordinates

1. The region  is more easily described by polar coordinates:  =

( ) | 0 ≤  ≤ 4, 0 ≤  ≤^32

.

Thus

 ( )  =32 0

4

0  ( cos   sin )   .

2. The region  is more easily described by rectangular coordinates:  = {( ) | −1 ≤  ≤ 1, −  ≤  ≤ 1}.

Thus

 ( )  =1

−1

1

− ( )  .

3. The region  is more easily described by polar coordinates:  = {( ) | 0 ≤  ≤ 1,  ≤  ≤ 2}.

Thus

 ( )  =2



1

0  ( cos   sin )   .

4. The region  is more easily described by polar coordinates:  =

( ) | 0 ≤  ≤ 3, −^4 ≤  ≤ ^34

.

Thus

 ( )  =34

−4

3

0  ( cos   sin )   .

5. The integral34

4

2

1   represents the area of the region

 = {( ) | 1 ≤  ≤ 2, 4 ≤  ≤ 34}, the top quarter portion of a ring (annulus).

34

4

2

1    =34

4 2 1  

=

34

4

1 2²2

1=3

4 −^4

·¹2(4 − 1) =^2 ·³2 =^3₄

6. The integral

2

2 sin 

0   represents the area of the region  = {( ) | 0 ≤  ≤ 2 sin , 2 ≤  ≤ }. Since

 = 2 sin  ⇒ ²= 2 sin  ⇔ ²+ ²= 2 ⇔

²+ ( − 1)²= 1,  is the portion in the second quadrant of a disk of radius 1 with center (0 1).



2

2 sin 

0    =

2

₁

2²=2 sin 

=0  =

22 sin² 

=

2 2 ·¹2(1 − cos 2)  =

 −¹2sin 2

2

=  − 0 −^2 + 0 = ^₂

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13. Evaluate the given integral by changing to polar coordinates.

RR

De^−x2−y2dA, where D is the region bounded by the semicircle x =p

4 − y² and the y-axis.

Solution:

546 ¤ CHAPTER 15 MULTIPLE INTEGRALS

7.The half disk  can be described in polar coordinates as  = {( ) | 0 ≤  ≤ 5, 0 ≤  ≤ }. Then



 ²  = 0

5

0 ( cos )²( sin )    =

0 cos² sin  5 0 ⁴

=

−¹3cos³ 0

1 5⁵5

0= −¹3(−1 − 1) · 625 =¹²⁵⁰3

8.The region  is¹₈ of a disk, as shown in the figure, and can be described by  = {( ) | 0 ≤  ≤ 2, 4 ≤  ≤ 2}. Thus



(2 − )  =2

4

2

0(2 cos  −  sin )   

=2

4(2 cos  − sin ) 2 0 ²

=

2 sin  + cos 2

4

1 3³2

0

= (2 + 0 −√

2 −^√2²)8 3

=¹⁶₃ − 4√ 2

9.

sin(²+ ²)  =2 0

3

1 sin(²)    =2 0  3

1  sin(²)  =

2 0

−¹2cos(²)3 1

= 2

 −¹2(cos 9 − cos 1)

=^₄(cos 1 − cos 9)

10.





²

²+ ² =

 2

0

 



( sin )²

²    =

 2

0

sin² 

 



  =

 2

0

1

2(1 − cos 2) 

 



 

= ¹₂

 −¹2sin 22

0

₁

2²

= ¹₂(2 − 0 − 0) ·¹2

²− ²

=^₂(²− ²) 11.

^{−2−2} =2

−2

2

0 ^−2   =2

−22

0 ^−2

=

2

−2



−¹2^−22 0= 

−¹2

(⁻⁴− ⁰) = ^₂(1 − ⁻⁴)

12.

cos

²+ ² =2

0

2 0 cos√

²    =2

0 2

0  cos  . For the second integral, integrate by parts with

 = ,  = cos  . Then

cos

²+ ² =

2

0 [ sin  + cos ]²₀= 2(2 sin 2 + cos 2 − 1).

13.is the region shown in the figure, and can be described by  = {( ) | 0 ≤  ≤ 4 1 ≤  ≤ 2}. Thus



arctan()  =4 0

2

1 arctan(tan )   since  = tan .

Also, arctan(tan ) =  for 0 ≤  ≤ 4, so the integral becomes

4 0

2

1     =4 0  2

1   =₁

2²4 0

₁

2²2

1=^₃₂² ·³2 = ₆₄³². 14.





  =



²+ ²≤ 4

≥ 0,  ≥ 0

  −



(− 1)²+ ²≤ 1

≥ 0

 

=2 0

2

0 ²cos    −2 0

2 cos 

0 ²cos   

=2 0

1

3(8 cos )  −2 0

1

3(8 cos⁴) 

=⁸₃−12⁸

cos³ sin  +³₂( + sin  cos )2 0

=⁸₃−²3

0 +³₂ 2

=^{16− 3}₆

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20. Use a double integral to find the area of the region D.

SECTION 15.3 Double Integrals in Polar Coordinates 1067

15.3 Exercises

1–6 A region R is shown. Decide whether to use polar coordinates or rectangular coordinates and write yyR fsx, yd dA as an iterated integral, where f is an arbitrary continuous function on R.

1.

0 4

_4 y

x R

2.

_1 0 1

1 y

x R

3.

0 1 3

_3 _1 y

x R

4.

0 3

y

x R _3

5.

_2 0 2

1 y

x R

6.

0 8 10

10 8 y

x R

7–8 Sketch the region whose area is given by the integral and evaluate the integral.

7.

y

y4^3y4

y

1² r dr d 8.

y

y2

y

0^{2 sin} r dr d

9–16 Evaluate the given integral by changing to polar coordinates.

9. yy_D x²y dA, where D is the top half of the disk with center the origin and radius 5

10. yy_R s2x 2 yd dA, where R is the region in the first quadrant enclosed by the circle x²1y²− 4 and the lines x − 0 and y − x

11. yy_R sinsx²1y²d dA, where R is the region in the first quadrant between the circles with center the origin and radii 1 and 3

12.

y

R

y

_x^2y₁²_y² dA, where R is the region that lies between the circles x²1y²− a² and x²1y²− b² with 0 , a , b

13. yy_D e^2x22y2 dA, where D is the region bounded by the semicircle x −s4 2 y² and the y-axis

14. yy_D cos sx²1y² dA, where D is the disk with center the origin and radius 2

15. yy_R arctans yyxd dA,

where R −hsx, yd

|

^{1 < x21y2<}4, 0 < y < xj 16. yy_D x dA, where D is the region in the first quadrant that lies

between the circles x²1y²− 4 and x²1y²− 2x

17–22 Use a double integral to find the area of the region D.

17.

D r=1-cos ¨

r=1+cos ¨ 18.

D r=œ„¨„

19.

D

r=cos ¨ r=sin ¨ 20.

D

r@=cos 2¨

r=1/œ„2

21. D is the loop of the rose r − sin 3 in the first quadrant.

22. D is the region inside the circle sx 2 1d²1y²− 1 and outside the circle x²1y²− 1.

23–24

(a) Set up an iterated integral in polar coordinates for the volume of the solid under the surface and above the region D.

(b) Evaluate the iterated integral to find the volume of the solid.

23.

x z

y z=1+xy

D

≈+¥=4, z=0

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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 15.3 DOUBLE INTEGRALS IN POLAR COORDINATES ¤ 1519 18. The region  is described by  = {( ) | 0 ≤  ≤√

, 0 ≤  ≤ 2}, so the area is

() =

 2

0

 ^√ 0

   =

 2

0

² 2

^=√

=0

 =

2

0

 2 =

² 4

^2

0

= ².

19. By symmetry, the total area is twice the area defined by

 = {( ) | 0 ≤  ≤ sin , 4 ≤  ≤ } (see the figure).

The total area is 2() = 2

4

sin 

0    = 2 ·¹2



4

²=sin 

=0  =

4sin² 

=

4 1

2(1 − cos 2)  = ¹2

 −¹2sin 2

4

=¹₂( − 0) −¹2

_

4 −¹2

=^3₈ +¹₄

20. By symmetry, the area of the region is 4 times the area of the region  in the first quadrant between the circle  = 1√ 2and the curve ² = cos 2 ⇒  =√

cos 2. The curves intersect in the first quadrant when cos 2 =

√1 2

2

⇒ cos 2 = ¹₂ ⇒ 2 = ^3 ⇒  =^6. Thus,  = {( ) | 1√

2 ≤  ≤√

cos 2, 0 ≤  ≤ 6}, so the total area is

4() = 4

 6 0

 ^√cos 2

1√ 2

   = 4 · 1 2

6 0

²^{=√cos 2}

=1√

2  = 2

 6 0



cos 2 −1 2





= 21

2sin 2 −^2

6

0 =^√₂³−^6

21. One loop is given by the region

 = {( ) | 0 ≤  ≤ sin 3, 0 ≤  ≤ 3}, so the area is



 =3 0

sin 3

0    =¹₂3 0

²=sin 3

=0 

= ¹₂3

0 sin²3  =¹₂3 0

1

2(1 − cos 6) 

= ¹₄

 −¹6sin 63 0 =₁₂^

22. In polar coordinates the circle ( − 1)²+ ² = 1 ⇔ ²+ ²= 2 is ²= 2 cos  ⇒  = 2 cos , and the circle ²+ ² = 1 is  = 1. The curves intersect in the first quadrant when

2 cos  = 1 ⇒ cos  = ¹₂ ⇒  = 3, so the portion of the region in the first quadrant is given by

 = {( ) | 1 ≤  ≤ 2 cos  0 ≤  ≤ 3}. By symmetry, the total area is twice the area of :

2() = 2

 = 23 0

2 cos 

1    = 23 0

1

2²=2 cos 

=1 

=3 0

4 cos² − 1

 =3 0

4 ·¹2(1 + cos 2) − 1



=3

0 (1 + 2 cos 2)  = [ + sin 2]^3₀ =^₃ +^√₂³

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32. Use polar coordinates to find the volume of the given solid.

Inside the sphere x²+ y²+ z²= 16 and outside the cylinder x²+ y²= 4.

Solution:

1

SECTION 15.3 DOUBLE INTEGRALS IN POLAR COORDINATES ¤ 1521 30.  =

1≤ ²+ ²≤4

²+ ² =2

0

2 1

√²    =2

0 2

1 ² =

2

0

₁

3³2 1= 2₈

3−¹3

=¹⁴₃

31. 2 +  +  = 4 ⇔  = 4 − 2 − , so the volume of the solid is

 =

²+ ²≤1(4 − 2 − )  =2

0

1

0(4 − 2 cos  −  sin )   

=2

0

1 0

4 − ²(2 cos  + sin )

  =2

0

2²−¹3³(2 cos  + sin )=1

=0 

=2

0

2 −¹3(2 cos  + sin )

 =

2 −¹3(2 sin  − cos )2

0 = 4 +¹₃− 0 −¹3 = 4

32. The sphere ²+ ²+ ²= 16intersects the ­plane in the circle ²+ ²= 16, so

 = 2



4≤²+²≤16

16 − ²− ² [by symmetry] = 2

2

0

 4 2

16 − ²   

= 2

 2

0



 4 2

(16 − ²)^12 = 2

2

0

−¹3(16 − ²)^324 2

= −²3(2)(0 − 12^32) = ^4₃  12√

12

= 32√ 3  33. By symmetry,

 = 2



²+ ²≤ ²

²− ²− ² = 2

 2

0

  0

²− ²    = 2

 2

0



  0



²− ²

= 2

2

0



−¹3(²− ²)^32

0 = 2(2)

0 +¹₃³

=⁴₃³

34. The paraboloid  = 1 + 2²+ 2²intersects the plane  = 7 when 7 = 1 + 2²+ 2²or ²+ ²= 3and we are restricted to the first octant, so

 =



²+²≤ 3

≥0≥0

7 −

1 + 2²+ 2²

 =

 2 0

 ^√3 0

7 − (1 + 2²)

  

=2 0 ^√3

0

6 − 2³

 =

2 0

3²−¹2⁴^√3

0 = ^₂ ·⁹2 = ⁹₄ 35. The cone  =

²+ ²intersects the sphere ²+ ²+ ²= 1when ²+ ²+

²+ ²2

= 1or ²+ ²= ¹₂. So

 =



²+ ²≤ 12

1 − ²− ²−

²+ ²

 =

 2

0

 1√ 2

0

1 − ²− 

  

=2

0 1√ 2 0

√

1 − ²− ²

 =

2

0



−¹3(1 − ²)^32−¹3³1√ 2

0 = 2

−¹3

 ₁

√2− 1

=^₃ 2 −√

2

36. The two paraboloids intersect when 6 − ²− ²= 2²+ 2² or ²+ ² = 2. For ²+ ²≤ 2, the paraboloid

 = 6 − ²− ²is above  = 2²+ 2²so

 =



²+ ²≤ 2

[(6 − ²− ²) − (2²+ 2²)]  =



²+ ²≤ 2

[6 − 3(²+ ²)]  =

 2

0

 ^√2 0

(6 − 3²)   

=2

0 ^√2

0 (6 − 3³)  =

2

0

3²−³4⁴^√2

0 = 2 (6 − 3) = 6

° 2021 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

41. Evaluate the iterated integral by converting to polar coordinates. R¹₂

0

R

√

1−y²

√

3y xy²dxdy Solution:

550 ¤ CHAPTER 15 MULTIPLE INTEGRALS

31.The region  of integration is shown in the figure. In polar coordinates the line  =√

3 is  = 6, so

 12 0

 √1−²

√3 

²  =

 6 0

 1 0

( cos )( sin )²  

=6

0 sin² cos  1 0 ⁴

=1

3sin³6 0

1 5⁵1

0

=

1 3

₁

2

3

− 0 ₁

5 − 0

=₁₂₀¹

32. 

2



0 ( cos )²( sin )    =

2



0 ⁴cos² sin   

=

2cos² sin   0 ⁴

=

−¹3cos³

2

₁

5⁵ 0

= −¹3

cos³ − cos^{3 }2

 1 5⁵

=₁₅¹⁵

33. = {( ) | 0 ≤  ≤ 1, 0 ≤  ≤ 2}, so



 ^(2+2)2 =2

0

1

0 ^(2)2   =2

0 1

0 ^4 = 21

0 ^4. Using a calculator, we estimate 21

0 ^4 ≈ 45951.

34. = {( ) | 0 ≤  ≤ 1, 0 ≤  ≤ 2}, so





1 + ²+ ² =2 0

1

0 ( cos )( sin )√

1 + ²   

=2

0 sin  cos  1 0 ³√

1 + ² =1

2sin²2 0

1 0 ³√

1 + ²

=¹₂1 0 ³√

1 + ² ≈ 01609

35.The surface of the water in the pool is a circular disk  with radius 5 m. If we place  on coordinate axes with the origin at the center of  and define ( ) to be the depth of the water at ( ), then the volume of water in the pool is the volume of the solid that lies above  =

( ) | ²+ ²≤ 25and below the graph of ( ). We can associate north with the positive -direction, so we are given that the depth is constant in the -direction and the depth increases linearly in the

-direction from (0 −5) = 1 to (0 5) = 2. The trace in the -plane is a line segment from (0 −5 1) to (0 5 2). The slope of this line is_{5− (−5)}^{2− 1} =₁₀¹, so an equation of the line is  − 2 = 10¹( − 5) ⇒  = 10¹ +³₂. Since ( ) is independent of , ( ) = ₁₀¹ +³₂. Thus the volume is given by

 ( ) , which is most conveniently evaluated using polar coordinates. Then  = {( ) | 0 ≤  ≤ 5, 0 ≤  ≤ 2} and substituting  =  cos ,  =  sin  the integral becomes

2

0

5 0

1

10 sin  +³₂

   =2

0

1

30³sin  +³₄² = 5

 = 0  =2

0

25

6 sin  +⁷⁵₄



=

−²⁵6 cos  +⁷⁵₄2

0 = ⁷⁵₂ = 375

Thus the pool contains 375 ≈ 118 m³of water.

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