dvostrukistwartreš

Under the surface and above the triangle with vertices, , and

22. Enclosed by the paraboloid and the planes ,, ,

23. Bounded by the planes , , , and

24. Bounded by the planes , , , and

25. Enclosed by the cylinders , and the planes ,

26. Bounded by the cylinder and the planes , in the first octant

27. Bounded by the cylinder and the planes ,, in the first octant

28. Bounded by the cylinders and

; 29. Use a graphing calculator or computer to estimate the -coordinates of the points of intersection of the curves

and . If is the region bounded by these curves,estimate .

; 30. Find the approximate volume of the solid in the first octant that is bounded by the planes , , and and thecylinder . (Use a graphing device to estimate thepoints of intersection.)

31–32 |||| Find the volume of the solid by subtracting two volumes.

31. The solid enclosed by the parabolic cylinders ,and the planes ,

32. The solid enclosed by the parabolic cylinder and theplanes ,

33–36 |||| Use a computer algebra system to find the exact volumeof the solid.

33. Under the surface and above the regionbounded by the curves and for

34. Between the paraboloids and and inside the cylinder

35. Enclosed by

36. Enclosed by

37–42 |||| Sketch the region of integration and change the order ofintegration.

37. 38. y1

0 y

4

4x f �x, y� dy dxy

4

0 y

sx

0 f �x, y� dy dx

■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■

z � x 2 � y 2 and z � 2y

z � 1 � x 2 � y 2 and z � 0

x 2 � y 2 � 1z � 8 � x 2 � 2y 2z � 2x 2 � y 2

x � 0y � x 2 � xy � x 3 � xz � x 3y 4 � xy 2

CAS

■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■

z � 2 � yz � 3yy � x 2

2x � 2y � z � 10 � 0x � y � z � 2y � x 2 � 1

y � 1 � x 2

y � cos xz � xz � 0y � x

xxD x dADy � 3x � x 2

y � x 4x

■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■

y 2 � z2 � r 2x 2 � y 2 � r 2

z � 0x � 0y � zx 2 � y 2 � 1

z � 0x � 0x � 2y,y 2 � z2 � 4

y � 4z � 0y � x 2z � x 2

z � 0x � y � 2y � xz � x

x � y � z � 1z � 0y � 0x � 0

z � 0y � xy � 1x � 0z � x 2 � 3y 2

�1, 2��4, 1��1, 1�z � xy21.1–6 |||| Evaluate the iterated integral.

1. 2.

3. 4.

6.

7–18 |||| Evaluate the double integral.

7.

8.

9.

10.

11. ,

12. ,

, is bounded by , ,

14. , is bounded by

15. ,

is the triangular region with vertices (0, 2), (1, 1), and

16.

is bounded by the circle with center the origin and radius 2

18. is the triangular region with vertices ,

, and

19–28 |||| Find the volume of the given solid.

19. Under the plane and above the regionbounded by and

20. Under the surface and above the region boundedby and x � y 3x � y 2

z � 2x � y 2

y � x 4y � xx � 2y � z � 0

■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■

�0, 3��1, 2�

�0, 0�yyD

2xy dA, D

D

yyD

�2x � y� dA,17.

yyD

xy 2 dA, D is enclosed by x � 0 and x � s1 � y 2

�3, 2�D

yyD

y 3 dA

y � sx and y � x 2DyyD

�x � y� dA

x � 1y � x 2y � 0DyyD

x cos y dA13.

D � ��x, y� � 0 � y � 1, 0 � x � y�yyD

xsy 2 � x 2 dA

D � ��x, y� � 1 � y � 2, y � x � y 3 �yyD

e x y dA

yyD

e y2 dA, D � ��x, y� � 0 � y � 1, 0 � x � y�

yyD

2y

x 2 � 1 dA, D � {�x, y� � 0 � x � 1, 0 � y � sx}

yyD

4y

x 3 � 2 dA, D � ��x, y� � 1 � x � 2, 0 � y � 2x�

yyD

x 3y 2 dA, D � ��x, y� � 0 � x � 2, �x � y � x�

■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■

y1

0 y

v

0 s1 � v 2 du dvy

� 2

0 y

cos �

0 e sin � dr d�5.

y1

0 y

2�x

x �x 2 � y� dy dxy

1

0 y

e y

y sx dx dy

y2

1 y

2

y xy dx dyy

1

0 y

x 2

0 �x � 2y� dy dx

|||| 15.3 Exercises

1002 ❙ ❙ ❙ ❙ CHAPTER 15 MULTIPLE INTEGRALS

SECTION 15.4 DOUBLE INTEGRALS IN POLAR COORDINATES ❙ ❙ ❙ ❙ 1003

51–52 |||| Use Property 11 to estimate the value of the integral.

51. ,

52. , is the disk with center the origin and radius

53. Prove Property 11.

In evaluating a double integral over a region , a sum of iterated integrals was obtained as follows:

Sketch the region and express the double integral as an iterated integral with reversed order of integration.

55. Evaluate , where

[Hint: Exploit the fact that is symmetric with respect to bothaxes.]

56. Use symmetry to evaluate , where is the region bounded by the square with vertices and .

57. Compute , where is the disk, by first identifying the integral as the volume

of a solid.

58. Graph the solid bounded by the plane and the paraboloid and find its exact volume.(Use your CAS to do the graphing, to find the equations of theboundary curves of the region of integration, and to evaluatethe double integral.)

z � 4 � x 2 � y 2x � y � z � 1CAS

x 2 � y 2 � 1DxxD s1 � x 2 � y 2 dA

�0, �5��5, 0�

DxxD �2 � 3x � 4y� dA

D

D � ��x, y� � x 2 � y 2 � 2�.xxD �x 2 tan x � y 3 � 4� dA

D

yyD

f �x, y� dA � y1

0 y

2y

0 f �x, y� dx dy � y

3

1 y

3�y

0 f �x, y� dx dy

D54.

■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■

12Dyy

D

e x 2�y 2 dA

D � �0, 1� �0, 1�yyD

sx 3 � y 3 dA

39. 40.

42.

43–48 |||| Evaluate the integral by reversing the order of integration.

44.

45. 46.

47.

48.

49–50 |||| Express as a union of regions of type I or type II andevaluate the integral.

50.

■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■

x0

y

y=_1

x=_1

x=1

x=¥

y=1+≈

Dx0

y

1

_1

_1 1

D(1, 1)

yyD

xy dAyyD

x 2 dA49.

D

■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■

y8

0 y

2

sy3 e x4

dx dy

y1

0 y

� 2

arcsin y cos x s1 � cos2x dx dy

y1

0 y

1

x 2 x 3 sin�y 3 � dy dxy

3

0y

9

y 2 y cos�x 2 � dx dy

y1

0 y

1

sy sx 3 � 1 dx dyy

1

0 y

3

3y e x2

dx dy43.

■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■

y1

0 y

� 4

arctan x f �x, y� dy dxy

2

1 y

ln x

0 f �x, y� dy dx41.

y3

0 y

s9�y

0 f �x, y� dx dyy

3

0 y

s9�y 2 �s9�y 2

f �x, y� dx dy

|||| 15.4 D o u b l e I n t e g r a l s i n P o l a r C o o r d i n a t e s

Suppose that we want to evaluate a double integral , where is one of theregions shown in Figure 1. In either case the description of in terms of rectangular coor-dinates is rather complicated but is easily described using polar coordinates.

FIGURE 1

x0

y

R

≈+¥=1

(a) R=s(r, ¨) | 0¯r¯1, 0¯¨¯2πd

x0

y

R

≈+¥=4

≈+¥=1

(b) R=s(r, ¨) | 1¯r¯2, 0¯¨¯πd

RR

RxxR f �x, y� dA


Recall from Figure 2 that the polar coordinates of a point are related to the rect-angular coordinates by the equations

(See Section 10.3.)The regions in Figure 1 are special cases of a polar rectangle

which is shown in Figure 3. In order to compute the double integral , whereis a polar rectangle, we divide the interval into subintervals of equal

width and we divide the interval into subintervals ofequal width . Then the circles and the rays divide the polarrectangle R into the small polar rectangles shown in Figure 4.

The “center” of the polar subrectangle

has polar coordinates

We compute the area of using the fact that the area of a sector of a circle with radius and central angle is . Subtracting the areas of two such sectors, each of which hascentral angle , we find that the area of is

Although we have defined the double integral in terms of ordinary rect-angles, it can be shown that, for continuous functions , we always obtain the same answer using polar rectangles. The rectangular coordinates of the center of are

, so a typical Riemann sum is

�m

i�1 �

n

j�1 f �ri* cos � j*, ri* sin � j*� Ai � �

m

i�1 �

n

j�1 f �ri* cos � j*, ri* sin � j*� ri* r �1

�ri* cos � j*, ri* sin � j*�Rij

fxxR f �x, y� dA

� 12 �ri � ri�1 ��ri � ri�1 � � � r i* r �

Ai � 12 ri

2 � �12 ri�12 � � 1

2 �ri

2 � ri�12 � �

Rij� � � j � � j�1

12 r 2��

rRij

� j* � 12 ��j�1 � �j�ri* � 1

2 �ri�1 � ri �

Rij � ��r, �� ri�1 � r � ri, � j�1 � � � � j�

FIGURE 3 Polar rectangle FIGURE 4 Dividing R into polar subrectangles

O

∫å

r=a ¨=å

¨=∫r=b

R

Î¨

¨=¨j

¨=¨j_1

(ri*, ¨j*)

r=ri_1

r=ri

Rij

O

� � � jr � ri� � �� n��j�1, �j�n��, ��r � �b � a� m

�ri�1, ri�m�a, b�RxxR f �x, y� dA

R � ��r, �� a � r � b, � � � � ��

y � r sin �x � r cos �r 2 � x 2 � y 2

�x, y��r, ��

O

y

x

¨

x

yr

P(r, ̈ )=P(x, y)

FIGURE 2

If we write , then the Riemann sum in Equation 1 can be writ-ten as

which is a Riemann sum for the double integral

Therefore, we have

Change to Polar Coordinates in a Double Integral If is continuous on a polar rect-angle given by , , where , then

The formula in (2) says that we convert from rectangular to polar coordinates in a double integral by writing and , using the appropriate limits of

| integration for and , and replacing by . Be careful not to forget the additionalfactor r on the right side of Formula 2. A classical method for remembering this is shownin Figure 5, where the “infinitesimal” polar rectangle can be thought of as an ordinary rect-angle with dimensions and and therefore has “area”

EXAMPLE 1 Evaluate , where is the region in the upper half-planebounded by the circles and .

SOLUTION The region can be described as

It is the half-ring shown in Figure 1(b), and in polar coordinates it is given by , 1 � r � 2

R � ��x, y� � y � 0, 1 � x 2 � y 2 � 4�

R

x 2 � y 2 � 4x 2 � y 2 � 1RxxR �3x � 4y 2 � dA

O

d¨

r d¨

dr

dA

r

FIGURE 5

dA � r dr d�.drr d�

r dr d�dA�ry � r sin �x � r cos �

yyR

f �x, y� dA � y�

� y

b

a f �r cos �, r sin �� r dr d�

0 � � � � � 2�� 0 � a � r � bRf2

� y�

� y

b

a f �r cos �, r sin �� r dr d�

� lim m, n l

�m

i�1 �

n

j�1 t�ri*, � j*� r � � y

�

�y

b

a t�r, � � dr d�

yyR

f �x, y� dA � lim m, n l

�m

i�1 �

n

j�1 f �ri* cos � j*, ri* sin � j*� Ai

y�

� y

b

a t�r, �� dr d�

�m

i�1 �

n

j�1 t�ri*, � j*� r �

t�r, �� rf �r cos �, r sin ��


. Therefore, by Formula 2,

EXAMPLE 2 Find the volume of the solid bounded by the plane and the paraboloid.

SOLUTION If we put in the equation of the paraboloid, we get . Thismeans that the plane intersects the paraboloid in the circle , so the solid liesunder the paraboloid and above the circular disk given by [see Figures 6and 1(a)]. In polar coordinates is given by , . Since

, the volume is

If we had used rectangular coordinates instead of polar coordinates, then we would haveobtained

which is not easy to evaluate because it involves finding the following integrals:

What we have done so far can be extended to the more complicated type of regionshown in Figure 7. It’s similar to the type II rectangular regions considered in Section 15.3.In fact, by combining Formula 2 in this section with Formula 15.3.5, we obtain the fol-lowing formula.

If is continuous on a polar region of the form

then yyD

f �x, y� dA � y�

� y

h2��

h1�� f �r cos �, r sin �� r dr d�

D � ��r, �� , h1�� r � h2��

f3

y �1 � x 2 �3 2 dxy x 2s1 � x 2 dxy s1 � x 2 dx

V � yyD

�1 � x 2 � y 2 � dA � y1

�1 y

s1�x2

�s1�x2 �1 � x 2 � y 2 � dy dx

� y2�

0 d� y

1

0 �r � r 3 � dr � 2� r 2

2�

r 4

4 0

1

��

2

V � yyD

�1 � x 2 � y 2 � dA � y2�

0 y

1

0 �1 � r 2 � r dr d�

1 � x 2 � y 2 � 1 � r 20 � � � 2�0 � r � 1D

x 2 � y 2 � 1Dx 2 � y 2 � 1

x 2 � y 2 � 1z � 0

z � 1 � x 2 � y 2z � 0

� 7 sin � �15�

2�

15

4 sin 2�

0

�

�15�

2

� y�

0 [7 cos � �

152 �1 � cos 2��] d�

� y�

0 [r 3 cos � � r 4 sin2�]r�1

r�2 d� � y

�

0 �7 cos � � 15 sin2� � d�

� y�

0 y

2

1 �3r 2 cos � � 4r 3 sin2�� dr d�

yyR

�3x � 4y 2 � dA � y�

0 y

2

1 �3r cos � � 4r 2 sin2�� r dr d�

0 � � � �


FIGURE 6

y

0

(0, 0, 1)

Dx

z

O

∫å

r=h¡(¨)

¨=å

¨=∫ r=h™(¨)

D

FIGURE 7D=s(r, ¨) | å¯¨¯∫, h¡(̈ )¯r¯h™(̈ )d

|||| Here we use the trigonometric identity

See Section 7.2 for advice on integrating trigonometric functions.

sin2� � 12 �1 � cos 2��

In particular, taking , , and in this formula, we seethat the area of the region bounded by , , and is

and this agrees with Formula 10.4.3.

EXAMPLE 3 Use a double integral to find the area enclosed by one loop of the four-leavedrose .

SOLUTION From the sketch of the curve in Figure 8 we see that a loop is given by theregion

So the area is

EXAMPLE 4 Find the volume of the solid that lies under the paraboloid ,above the -plane, and inside the cylinder .

SOLUTION The solid lies above the disk whose boundary circle has equationor, after completing the square,

(see Figures 9 and 10). In polar coordinates we have and , sothe boundary circle becomes , or . Thus, the disk is given by

and, by Formula 3, we have

� 2[32 � � sin 2� �

18 sin 4�]0

� 2� 2�3

2��

2 � �3�

2

� 2 y� 2

0 �1 � 2 cos 2� �

12 �1 � cos 4�� d�

� 8 y� 2

0 cos4� d� � 8 y

� 2

0 �1 � cos 2�

2 �2

d�

� y� 2

�� 2 r 4

4 0

2 cos �

d� � 4 y� 2

�� 2 cos4� d�

V � yyD

�x 2 � y 2 � dA � y� 2

�� 2 y

2 cos �

0 r 2 r dr d�

D � {�r, � � � �� 2 � � � � 2, 0 � r � 2 cos �}Dr � 2 cos �r 2 � 2r cos �

x � r cos �x 2 � y 2 � r 2

�x � 1�2 � y 2 � 1

x 2 � y 2 � 2xD

x 2 � y 2 � 2xxyz � x 2 � y 2

� 14 y

� 4

�� 4 �1 � cos 4�� d� � 1

4 [� �14 sin 4�]�� 4

� 4�

�

8

� y� 4

�� 4 [ 1

2 r 2]0cos 2� d� � 1

2 y� 4

�� 4 cos2 2� d�

A�D� � yyD

dA � y� 4

�� 4 y

cos 2�

0 r dr d�

D � {�r, �� 4 � � � � 4, 0 � r � cos 2�}

r � cos 2�

� y�

�

r 2

2 0

h��

d� � y�

� 12 �h��2 d�

A�D� � yyD

1 dA � y�

� y

h��

0 r dr d�

r � h�� Dh2�� h��h1�� 0f �x, y� � 1


FIGURE 8

¨=π4

¨=_π4

FIGURE 9

0

y

x1 2

D

(x-1)@+¥=1 (or r=2 cos ¨)

FIGURE 10

y

x

z

12. ,where

, where D is the region bounded by thesemicircle and the y-axis

14. , where is the region in the first quadrant enclosedby the circle

15. ,where

16. , where is the region in the first quadrant that liesbetween the circles and

17–20 |||| Use a double integral to find the area of the region.

One loop of the rose

18. The region enclosed by the curve

19. The region within both of the circles and

20. The region inside the circle and outside the circle

21–27 |||| Use polar coordinates to find the volume of the givensolid.

Under the paraboloid and above the disk

22. Inside the sphere and outside the cylinder

23. A sphere of radius

24. Bounded by the paraboloid and the plane

Above the cone and below the sphere

26. Bounded by the paraboloids and

27. Inside both the cylinder and the ellipsoid

28. (a) A cylindrical drill with radius is used to bore a holethrough the center of a sphere of radius . Find the volumeof the ring-shaped solid that remains.

(b) Express the volume in part (a) in terms of the height ofthe ring. Notice that the volume depends only on , not on

or .

29–32 |||| Evaluate the iterated integral by converting to polar coordinates.

29. 30. ya

�a y

sa 2�y 2

0 �x 2 � y 2 �3�2 dx dyy

1

0 y

s1�x 2

0 e x 2�y 2 dy dx

r2r1

hh

r2

r1

■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■

4x 2 � 4y 2 � z2 � 64x 2 � y 2 � 4

z � 4 � x 2 � y 2z � 3x 2 � 3y 2

x 2 � y 2 � z2 � 1z � sx 2 � y 225.

z � 4z � 10 � 3x 2 � 3y 2

a

x 2 � y 2 � 4x 2 � y 2 � z 2 � 16

x 2 � y 2 � 9z � x 2 � y 221.

■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■

r � 2r � 4 sin �

r � sin �r � cos �

r � 4 � 3 cos �

r � cos 3�17.

■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■

x 2 � y 2 � 2xx 2 � y 2 � 4Dxx

D x dA

R � ��x, y� � 1 � x 2 � y 2 � 4, 0 � y � x�xx

R arctan� y�x� dA

x 2 � y 2 � 25Rxx

R yex dA

x � s4 � y 2

xxD e�x 2�y 2

dA13.

R � ��x, y� � x 2 � y 2 � 4, x � 0�xx

R s4 � x 2 � y 2 dA1–6 |||| A region is shown. Decide whether to use polar coordi-

nates or rectangular coordinates and write as an iter-ated integral, where is an arbitrary continuous function on .

1. 2.

3. 4.

6.

7–8 |||| Sketch the region whose area is given by the integral andevaluate the integral.

7. 8.

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

9. ,where is the disk with center the origin and radius 3

10. , where is the region that lies to the left ofthe -axis between the circles and

11. , where is the region that lies above the-axis within the circle x 2 � y 2 � 9x

RxxR cos�x 2 � y 2� dA

x 2 � y 2 � 4x 2 � y 2 � 1yRxx

R �x � y� dA

DxxD xy dA

■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■

y��2

0 y

4 cos �

0 r dr d�y

2�

� y

7

4 r dr d�

■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■

0

2

y

x

R

2052

5

2

y

x

R

5.

0 31

3

1

y

x

R

0 2

2

y

x

R

0 2

2

y

x

R

0 2

2

y

x

R

RfxxR f �x, y� dA

R

|||| 15.4 Exercises


SECTION 15.5 APPLICATIONS OF DOUBLE INTEGRALS ❙ ❙ ❙ ❙ 1009

32.

33. A swimming pool is circular with a 40-ft diameter. The depthis constant along east-west lines and increases linearly from2 ft at the south end to 7 ft at the north end. Find the volume ofwater in the pool.

34. An agricultural sprinkler distributes water in a circular patternof radius 100 ft. It supplies water to a depth of feet per hourat a distance of feet from the sprinkler.(a) What is the total amount of water supplied per hour to

the region inside the circle of radius centered at the sprinkler?

(b) Determine an expression for the average amount of waterper hour per square foot supplied to the region inside thecircle of radius .

Use polar coordinates to combine the sum

into one double integral. Then evaluate the double integral.

36. (a) We define the improper integral (over the entire plane

� lima l �

yyDa

e��x 2�y 2 � dA

I � yy� 2

e��x 2�y 2 � dA � y�

�� y

�

�� e��x 2�y 2 � dy dx

� 2�

y1

1�s2 y

x

s1�x 2 xy dy dx � y

s2

1 y

x

0 xy dy dx � y

2

s2 y

s4�x 2

0 xy dy dx

35.

R

R

re�r

■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■

y2

0 y

s2x�x 2

0 sx 2 � y 2 dy dxy

2

0 y

s4�y 2

�s4�y 2

x 2 y 2 dx dy31. where is the disk with radius and center the origin. Show that

(b) An equivalent definition of the improper integral in part (a)is

where is the square with vertices . Use this toshow that

(c) Deduce that

(d) By making the change of variable , show that

(This is a fundamental result for probability and statistics.)

37. Use the result of Exercise 36 part (c) to evaluate the followingintegrals.

(a) (b) y�

0 sx e�x dxy

�

0 x 2e�x 2 dx

y�

�� e�x 2�2 dx � s2�

t � s2 x

y�

�� e�x 2 dx � s�

y�

�� e�x 2 dx y

�

�� e�y 2 dy � �

�a, a�Sa

yy� 2

e��x 2�y 2 � dA � lima l �

yySa

e��x 2�y 2 � dA

y�

�� y

�

�� e��x 2�y 2 � dA � �

aDa

|||| 15.5 A p p l i c a t i o n s o f D o u b l e I n t e g r a l s

We have already seen one application of double integrals: computing volumes. Anothergeometric application is finding areas of surfaces and this will be done in the next section.In this section we explore physical applications such as computing mass, electric charge,center of mass, and moment of inertia. We will see that these physical ideas are also impor-tant when applied to probability density functions of two random variables.

D e n s i t y a n d M a s s

In Section 8.3 we were able to use single integrals to compute moments and the center ofmass of a thin plate or lamina with constant density. But now, equipped with the doubleintegral, we can consider a lamina with variable density. Suppose the lamina occupies aregion of the -plane and its density (in units of mass per unit area) at a point in

is given by , where is a continuous function on . This means that

where and are the mass and area of a small rectangle that contains and thelimit is taken as the dimensions of the rectangle approach 0. (See Figure 1.)

�x, y�Am

��x, y� � lim m

A

D��x, y�D�x, y�xyD

FIGURE 1

0 x

y

D

(x, y)

0

1

0

x2

(x+2y)dydx =0

1

xy+y2 y=x

2

y=0dx=

0

1

x(x2)+(x

2)2

0 0 dx

=0

1

(x3+x

4)dx=

14

x4+

15

x5 1

0=

920

1

2

y

2

xydxdy =1

212

x2y

x=2

x=ydy=

1

212

y(4 y2)dy=

12 1

2

(4y y3)dy

=12

2y2 1

4y

4 2

1=

12

8 4 2+14

=98

0

1

y

ey

x dxdy =0

123

x3/2 x=e

y

x=ydy=

23 0

1

(e3y/2

y3/2

)dy=23

23

e3y/2 2

5y

5/2 1

0

=23

23

e3/2 2

523

e0+0 =

49

e3/2 32

45

0

1

x

2 x

(x2

y)dydx =0

1

x2y

12

y2 y=2 x

y=xdx=

0

1

x2(2 x)

12

(2 x)2

x2(x)+

12

x2

dx

=0

1

( 2x3+2x

2+2x 2)dx=

12

x4+

23

x3+x

22x

1

0=

56

0

/2

0

cos

esin

dr d =0

/2

resin r=cos

r=0d =

0

/2

(cos )esin

d = esin /2

0

=esin ( /2)

e0=e 1

1.

2.

3.

4.

5.

6.

1

Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.3 Double Integrals over General Regions

0

1

0

v

1 v2

dudv =0

1

u 1 v2 u=v

u=0dv=

0

1

v 1 v2

dv=13

(1 v2)3/2 1

0

=13

(0 1)=13

Dx

3y

2dA =

0

2

x

x

x3y

2dydx=

0

213

x3y

3 y=x

y= xdx=

13 0

2

2x6dx

=23

17

x7 2

0=

221

27

0 =25621

D

4y

x3+2

dA =1

2

0

2x4y

x3+2

dydx=1

22y

2

x3+2

y=2x

y=0

dx=1

28x

2

x3+2

dx

=83

ln x3+2

2

1=

83

(ln 10 ln 3)=83

ln103

0

1

0

x2y

x2+1

dydx =0

1y

2

x2+1

y= x

y=0

dx=0

1x

x2+1

dx

=12

ln x2+1

1

0=

12

(ln 2 ln 1)=12

ln 2

0

1

0

y

ey

2

dxdy=0

1

xey

2 x=y

x=0dy=

0

1

yey

2

dy=12

ey

2 1

0=

12

(e 1)

1

2

y

y3

ex/y

dxdy=1

2

yex/y x= y

3

x= ydy=

1

2

yey

2

ey( )dy=12

ey

2 12

ey2

2

1=

12

(e4

4e)

0

1

0

y

x y2

x2

dxdy=0

113

(y2

x2)3/2 x=y

x=0dy=

13 0

1

y3dy=

13

14

y4 1

0=

112

7.

8.

9.

10.

11.

12.

2


0

1

0

x2

xcos ydydx=0

1

xsin yy=x

2

y=0dx=

0

1

xsin x2dx=

12

cos x2 1

0=

12

(1 cos 1)

0

1

x2

x

(x+y)dydx =0

1

xy+12

y2 y= x

y=x2

dx

=0

1

x3/2

+12

x x3 1

2x

4dx

=25

x5/2

+14

x2 1

4x

4 110

x5 1

0=

310

1

2

2 y

2y 1

y3dxdy =

1

2

xy3 x=2y 1

x=2 ydy

=1

2

(2y 1) (2 y) y3dy

=1

2

(3y4

3y3)dy=

35

y5 3

4y

4 2

1

=965

1235

+34

=14720

13.

14.

15.

16.

3


Dxy

2dA

=1

1

0

1 y2

xy2dxdy

=1

1

y2 1

2x

2 x= 1 y2

x=0dy=

12 1

1

y2(1 y

2)dy

=12 1

1

(y2

y4)dy=

12

13

y3 1

5y

5 1

1

=12

13

15

+13

15

=215

2

2

4 x2

4 x2

(2x y)dydx

=2

2

2xy12

y2 y= 4 x

2

y= 4 x2dx

=2

2

2x 4 x2 1

24 x

2( )+2x 4 x2

+12

4 x2( ) dx

=2

2

4x 4 x2

dx=43

4 x2( ) 3/2 2

2=0

4x 4 x2

17.

(Or, note that is an odd function, so

4


2

2

4x 4 x2

dx=0

DxydA =

0

1

2x

3 x

2xydydx=0

1

xy2 y=3 x

y=2xdx

=0

1

x[(3 x)2

(2x)2]dx

=0

1

( 3x3

6x2+9x)dx

=34

x4

2x3+

92

x2 1

0=

34

2+92

=74

V =0

1

x4

x

(x+2y)dydx

=0

1

xy+y2 y=x

y=x4dx=

0

1

(2x2

x5

x8)dx

=23

x3 1

6x

6 19

x9 1

0=

23

16

19

=7

18

.)

18.

19.

5


V =0

1

y3

y2

(2x+y2)dxdy

=0

1

x2+xy

2 x=y2

x=y3dy=

0

1

(2y4

y6

y5)dy

=25

y5 1

7y

7 16

y6 1

0=

19210

V =1

2

1

7 3y

xydxdy=1

212

x2y

x=7 3y

x=1dy

=12 1

2

(48y 42y2+9y

3)dy

=12

24y2

14y3+

94

y4 2

1=

318

20.

21.

22.

6


V =0

1

x

1

(x2+3y

2)dydx

=0

1

x2y+y

3 y=1

y=xdx=

0

1

(x2+1 2x

3)dx

=13

x3+x

12

x4 1

0=

56

V =0

1

0

1 x

1 x y( ) dydx

=0

1

y xyy

2

2

y=1 x

y=0dx

=0

1

(1 x)2 1

2(1 x)

2dx

=0

112

(1 x)2dx=

16

(1 x)3 1

0=

16

23.

24.

7


V =0

1

x

2 x

xdydx

=0

1

x yy=2 x

y=xdx=

0

1

(2x 2x2)dx

= x2 2

3x

3 1

0=

13

V =2

2

x2

4

x2dydx

=2

2

x2

yy=4

y=x2dx=

2

2

(4x2

x4)dx

=43

x3 1

5x

5 2

2=

323

325

+323

325

=12815

25.

26.

8


V =0

2

0

2y

4 y2

dxdy

=0

2

x 4 y2 x=2y

x=0dy=

0

2

2y 4 y2

dy

=23

4 y2( ) 3/2 2

0=0+

163

=163

V =0

1

0

1 x2

ydydx=0

1y

2

2

y= 1 x2

y=0dx

=0

11 x

2

2dx=

12

x13

x3 1

0=

13

V 8 V1

V1 =

0

r

0

r2

y2

r2

y2

dxdy

=0

r

x r2

y2 x= r

2y

2

x=0dy

27.

28.

By symmetry, the desired volume is times the volume in the first octant. Now

9


=0

r

(r2

y2)dy= r

2y

13

y3 r

0=

23

r3

V =163

r3

x=0 x 1.213

DxdA

0

1.213

x4

3x x2

xdydx=0

1.213

xyy=3x x

2

y=x4

dx

=0

1.213

(3x2

x3

x5)dx= x

3 14

x4 1

6x

6 1.213

0

0.713

y=cos xy=x x 0.7391

V0

0.7391

x

cos x

zdydx=0

0.7391

x

cos x

xdydx

Thus .

29.

From the graph, it appears that the two curves intersect at and at . Thus the desiredintegral is

30.

The desired solid is shown in the first graph. From the second graph, we estimate that intersects at . Therefore the volume of the solid is

10


=0

0.7391

xyy=cos x

y=xdx=

0

0.7391

(xcos x x2)dx

= cos x+xsin x13

x3 0.7391

00.1024

y=0

V0

0.7391

0

x

xdydx+0.7391

/2

0

cos x

xdydx 0.4684

y=1 x2

y=x2

1 1,0( ) 1 x2

x2

1 1,1z=2x+2y+10 z=2 x y

V =1

1

x2

1

1 x2

(2x+2y+10)dydx1

1

x2

1

1 x2

(2 x y)dydx

=1

1

x2

1

1 x2

(2x+2y+10 (2 x y))dydx=1

1

x2

1

1 x2

(3x+3y+8)dydx

=1

1

3xy+32

y2+8y

y=1 x2

y=x2

1dx

=1

1

3x(1 x2)+

32

(1 x2)2+8(1 x

2) 3x(x

21)

32

(x2

1)2

8(x2

1) dx

=1

1

( 6x3

16x2+6x+16)dx=

32

x4 16

3x

3+3x

2+16x

1

1

=32

163

+3+16+32

163

3+16=643

y=1 z=3

y=x2

y=1 2+y 3y 0 y 1z=2+y z=3y

Note: There is a different solid which can also be construed to satisfy the conditions stated in theexercise. This is the solid bounded by all of the given surfaces, as well as the plane . In case youcalculated the volume of this solid and want to check your work, its volume is

.

31. The two bounding curves and intersect at with on .Within this region, the plane is above the plane , so

32. The two planes intersect in the line , , so the region of integration is the plane region

enclosed by the parabola and the line . We have for , so the solid region isbounded above by and bounded below by .

11


V =1

1

x2

1

(2+y)dydx1

1

x2

1

(3y)dydx=1

1

x2

1

(2+y 3y)dydx=1

1

x2

1

(2 2y)dydx

=1

1

2y y2 y=1

y=x2dx=

1

1

(1 2x2+x

4)dx= x

23

x3+

15

x5 1

1=

1615

y=x3

x y=x2+x x=2 x

2+x>x

3x

0,2( )

V =0

2

x3

x

x2+x

zdydx=0

2

x3

x

x2+x

(x3y

4+xy

2)dydx=

13,984,735,61614,549,535

x 1 y 1 2x2+y

2<8 x

22y

2

1 x 1 1 x2

y 1 x2

V =1

1

1 x2

1 x2

(8 x2

2y2) (2x

2+y

2) dydx=

132

x2+y

2=1 z=0 D

x2+y

21

D(1 x

2y

2)dA=

1

1

1 x2

1 x2

(1 x2

y2)dydx=

2

xy x2+y

2=2y

x2+y

22y=0 x

2+(y 1)

2=1 1 x 1

33. The two bounding curves and intersect at the origin and at , with on . Using a CAS, we find that the volume is

34. For and , . Also, the cylinder is described by the inequalities

, . So the volume is given by

[using a CAS]

35. The two surfaces intersect in the circle , and the region of integration is the disk :

. Using a CAS, the volume is .

36. The projection onto the plane of the intersection of the two surfaces is the circle

, so the region of integration is given by ,

12


1 1 x2

y 1+ 1 x2

2y x2+y

2

V =1

1

1 1 x2

1+ 1 x2

[2y (x2+y

2)]dydx=

2

D = (x,y) |0 y x ,0 x 4{ }= (x,y) | y

2x 4,0 y 2{ }

0

4

0

x

f (x,y)dydx=D

f (x,y)dA=0

2

y2

4

f (x,y)dxdy

D = x,y( ) |4x y 4,0 x 1{ }

= x,y( ) |0 xy4

,0 y 4{ }

0

1

4x

4

f (x,y)dydx =D

f (x,y)dA

. In this region, so, using a CAS, the volume is

.

37.

Because the region of integration is

we have

.

38.


we have

13


=0

4

0

y/4

f (x,y)dxdy

D = (x,y) | 9 y2

x 9 y2

,0 y 3{ }= (x,y) |0 y 9 x

2, 3 x 3{ }

0

3

9 y2

9 y2

f(x,y)dxdy=D

f(x,y)dA=3

3

0

9 x2

f(x,y)dydx

D = (x,y) |0 x 9 y ,0 y 3{ }= (x,y) |0 y 3,0 x 6{ } x,y( ) |0 y 9 x

2, 6 x 3{ }

0

3

0

9 y

f (x,y)dxdy=

Df (x,y)dA

39.


we have

40.

To reverse the order, we must break the region into two separate type I regions. Because the region ofintegration is

we have

14


=0

6

0

3

f (x,y)dydx+6

3

0

9 x2

f (x,y)dydx

D = (x,y) |0 y ln x,1 x 2{ }

= (x,y) |ey

x 2,0 y ln 2{ }

1

2

0

ln x

f (x,y)dydx =D

f (x,y)dA

=0

ln 2

ey

2

f (x,y)dxdy

D = (x,y) |arctanx y4

,0 x 1{ }= (x,y) |0 x tan y,0 y

4{ }

0

1

arctanx

/4

f (x,y)dydx =D

f (x,y)dA

=0

/4

0

tan y

f (x,y)dxdy

41.


we have

42.


we have

15


0

1

3y

3

ex

2

dxdy =0

3

0

x/3

ex

2

dydx

=0

3

ex

2

yy=x/3

y=0dx=

0

3x3

ex

2

dx

=16

ex

2 3

0=

e9

16

0

1

y

1

x3+1 dxdy =

0

1

0

x2

x3+1 dydx

=0

1

x3+1 y

y=x2

y=0dx=

0

1

x2

x3+1 dx

=29

(x3+1)

3/2 1

0=

29

23/2

1( )

0

3

y2

9

ycos x2dxdy =

0

9

0

x

ycos x2dydx

43.

44.

45.

16


=0

9

cos x2 y

2

2

y= x

y=0dx=

0

912

xcos x2dx

=14

sin x2 9

0=

14

sin 81

0

1

x2

1

x3sin (y

3)dydx =

0

1

0

y

x3sin (y

3)dxdy

=0

1x

4

4sin (y

3)

x= y

x=0dy

=0

114

y2sin (y

3)dy

=112

cos (y3)

1

0=

112

(1 cos 1)

0

1

arcsiny

/2

cos x 1+cos2x dxdy

=0

/2

0

sin x

cos x 1+cos2x dydx

=0

/2

cos x 1+cos2x y

y=sin x

y=0dx

46.

47.

17


=0

/2

cos x 1+cos2x sin xdx

Let u=cos x,du= sin xdx,dx=du/( sin x)

=1

0

u 1+u2

du=13

1+u2( ) 3/2 0

1

=13

8 1( )=13

2 2 1( )

0

8

3y

2

ex

4

dxdy =0

2

0

x3

ex

4

dydx

=0

2

ex

4

yy=x

3

y=0dx=

0

2

x3e

x4

dx

=14

ex

4 2

0=

14

(e16

1)

D = (x,y) |0 x 1, x+1 y 1{ } (x,y) | 1 x 0,x+1 y 1{ }(x,y) |0 x 1, 1 y x 1{ } (x,y) | 1 x 0, 1 y x 1{ }

Dx

2dA =

0

1

1 x

1

x2dydx+

1

0

x+1

1

x2dydx+

0

1

1

x 1

x2dydx+

1

0

1

x 1

x2dydx

= 40

1

1 x

1

x2dydx[ by symmetry of the regions and because f (x,y)=x

20]

= 40

1

x3dx=4

14

x4 1

0=1

48.

49.

,

all type I.

18


D = (x,y) | 1 x 0, 1 y 1+x2{ } (x,y) |0 x 1, x y 1+x

2{ }(x,y) |0 x 1, 1 y x{ }

DxydA =

1

0

1

1+x2

xydydx+0

1

x

1+x2

xydydx+0

1

1

x

xydydx

=1

012

xy2 y=1+x

2

y= 1dx+

0

112

xy2 y=1+x

2

y= xdx+

0

112

xy2 y= x

y= 1dx

=1

0

x3+

12

x5

dx+0

112

(x5+2x

3x

2+x)dx+

0

112

(x2

x)dx

=14

x4+

112

x6 0

1+

12

16

x6+

12

x4 1

3x

3+

12

x2 1

0+

12

13

x3 1

2x

2 1

0

=13

+5

12112

=0

D= 0,1 0,1 0 x3+y

32 A(D)=1 0

Dx

3+y

3dA 2

D= (x,y) |x2+y

2 14{ } 1=e

0e

x2+y

2

e1/4

A(D)=4 4 D

ex

2+y

2

dA (e1/4

)4

m f (x,y) MD

mdAD

f (x,y)dAD

MdA

mD

1dAD

f (x,y)dA MD

1dA mA(D)D

f (x,y)dA MA(D)

50.

,

all type I.

51. For , and , so .

52. Since , and , so

.

53. Since , by (8)

by (7) by (10).

54.

19


Df (x,y)dA =

0

1

0

2y

f (x,y)dxdy+1

3

0

3 y

f (x,y)dxdy

=0

2

x/2

3 x

f (x,y)dydx

D(x

2tan x+y

3+4) dA=

Dx

2tan xdA+

Dy

3dA+

D4dA x

2tan x x D

yD

x2tan xdA=0 y

3y D

xD

y3dA=0

D(x

2tan x+y

3+4)dA=4

DdA=4(area of D)=4 2( ) 2

=8

D y

3x xD

3xdA=0 4y y D

xD

4ydA=0

D2 3x+4y( ) dA =

D2dA=2

DdA

= 2(area of D)=2(50)= 100

1 x2

y2

0D

1 x2

y2

dA

55. . But is an odd function of and is

symmetric with respect to the axis, so . Similarly, is an odd function of and

is symmetric with respect to the axis, so . Thus

56. First, 0in0.19in ] The region , shown in the figure, is symmetric with respect to the axis and

is an odd function of , so . Similarly, is an odd function of and is symmetric

with respect to the axis, so . Then

57. Since , we can interpret as the volume of the solid that lies below

20


z= 1 x2

y2

D xy z= 1 x2

y2

x2+y

2+z

2=1 z 0 xy x

2+y

2=1 D

112

43

(1)3

=23

z

y=1 13+4x 4x

2

2

13+4x 4x2=0

x=1 14

2

V =1 14( ) /2

1+ 14( ) /2

1 13+4x 4x2( ) /2

1+ 13+4x 4x2( ) /2

[(4 x2

y2) (1 x y)]dydx=

498

the graph of and above the region in the plane. is equivalent to

, which meets the plane in the circle , the boundary of . Thus, the

solid is an upper hemisphere of radius which has volume .

58. To find the equations of the boundary curves, we require that the values of the two surfaces bethe same. In Maple, we use the command solve(4 x^2 y^2=1 x y,y);and in Mathematica, we use Solve[4 x^2 y^ 2==1 x y,y] . We find that the curves have equations

.

To find the two points of intersection of these curves, we use the CAS to solve , finding

that . So, using the CAS to evaluate the integral, the volume of intersection is

.

21


R R= (r, ) |0 r 2,0 2{ }

Rf (x,y)dA=

0

2

0

2

f (rcos ,rsin )r dr d

R R= (x,y) |0 x 2,0 y 2 x{ }

Rf (x,y)dA=

0

2

0

2 x

f (x,y)dydx

R R= (x,y) | 2 x 2,x y 2{ }

Rf (x,y)dA=

2

2

x

2

f (x,y)dydx

R R= (r, ) |1 r 3,02{ }

Rf (x,y)dA=

0

/2

1

3


R R= (r, ) |2 r 5,0 2{ }

Rf (x,y)dA=

0

2

2

5


R

R= (r, ) |0 r 2 2,4

54{ }

Rf (x,y)dA=

/4

5 /4

0

2 2


2

4

7

r dr d R= (r, ) |4 r 7, 2{ }

2

4

7

r dr d =2

d4

7

r dr

=2 1

2r

2 7

4=

12

49 16( )=33

2

1. The region is more easily described by polar coordinates: . Thus

.

2. The region is more easily described by rectangular coordinates: .

Thus .

3. The region is more easily described by rectangular coordinates: .

Thus .

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

Thus .

5. The region is more easily described by polar coordinates: . Thus

.

6. The region is more easily described by polar coordinates:

. Thus .

7. The integral represents the area of the region , the lower

half of a ring.

1

Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.4 Double Integrals in Polar Coordinates

0

/2

0

4cos

r dr d R= (r, ) |0 r 4cos ,0 /2{ }

r=4cos r2=4rcos x

2+y

2=4x (x 2)

2+y

2=4 R

2,0( )

0

/2

0

4cos

r dr d =0

/212

r2 r=4cos

r=0d =

0

/2

8cos2

d

=0

/2

4(1+cos 2 )d =4 +12

sin 2/2

0=2

D D= (r, ) |0 r 3,0 2{ }

DxydA =

0

2

0

3

(rcos )(rsin )r dr d =0

2

sin cos d0

3

r3dr

=12

sin2 2

0

14

r4 3

0=0

R(x+y)dA =

/2

3 /2

1

2

(rcos +rsin )r dr d =/2

3 /2

1

2

r2(cos +sin dr d

=/2

3 /2

(cos +sin d1

2

r2dr = sin cos

3 /2

/2

13

r3 2

1

=( 1 0 1+0)83

13

=143

8. The integral represents the area of the region .

Since , is the portion in the first quadrant of acircle of radius 2 with center .

9. The disk can be described in polar coordinates as . Then

10.

2


Rcos (x

2+y

2)dA =

0 0

3

cos (r2)r dr d =

0d

0

3

rcos (r2)dr

=0

12

sin (r2)

3

0=

12

(sin 9 sin 0)=2

sin 9

R4 x

2y

2dA =

/2

/2

0

2

4 r2

r dr d =/2

/2

d0

2

r 4 r2

dr

=/2

/2

12

23

(4 r2)3/2 2

0=

2+

213

(0 43/2

) =83

De

x2

y2

dA =/2

/2

0

2

er

2

r dr d =/2

/2

d0

2

rer

2

dr

=/2

/2

12

er

2 2

0=

12

(e4

e0)=

2(1 e

4)

Rye

xdA=

0

/2

0

5

(rsin )ercos

r dr d =0

5

0

/2

r2sin e

rcosd dr

0

/2

r2sin e

rcosd u=rcos du= rsin d

0

/2

r2sin e

rcosd =

u=r

u=0

r eudu= r[e

0e

r]=re

rr

0

5

0

/2

r2sin e

rcosd dr=

0

5

(rer

r)dr= rer

er 1

2r

2 5

0=4e

5 232

R R= (r, ) |0 /4,1 r 2{ }

Rarctan(y/x)dA=

0

/4

1

2

arctan(tan )r dr d

y/x=tan (tan )= 0 /4

11.

12.

13.

14. . First we integrate

: Let , and

. Then

, where we integrated by parts in the

first term.

15. is the region shown in the figure, and can be described by .

Thus

since . Also, arctan for ,so the integral becomes

3


0

/4

1

2

r dr d =0

/4

d1

2

r dr=12

2 /4

0

12

r2 2

1=

2

3232

=3

642

DxdA

=

x2+ y

24

x 0,y 0

xdA

x 1( )2+ y

21

y 0

xdA

=0

/2

0

2

r2cos dr d

0

/2

0

2cos

r2cos dr d

=0

/213

(8cos )d0

/213

(8cos4

)d

=83

812

cos3

sin +32

( +sin cos )/2

0

=83

23

0+32 2

=16 3

6

D= (r, ) /6 /6,0 r 3{ }

DdA =

/6

/6

0

cos 3

r dr d =/6

/612

r2 r=cos 3

r=0d

=/6

/612

cos23 d =2

0

/612

1+cos 62

d

.

16.

17. One loop is given by the region , so the area is

4


=12

+16

sin 6/6

0=

12

D= (r, ) |0 2 ,0 r 4+3cos{ }

A(D) =D

dA=0

2

0

4+3cos

r dr d =0

212

r2 r=4+3cos

r=0d =

12 0

2

(4+3cos )2d

=12 0

2

(16+24cos +9cos2

)d =12 0

2

16+24cos +91+cos 2

2d

=12

16 +24sin +92

+94

sin 22

0=

412

A =20

/4

0

sin

r dr d =20

/412

r2 r=sin

r=0d

=0

/4

sin2

d =0

/412

(1 cos 2 )d

=12

12

sin 2/4

0

=12 4

12

sin2

0+12

sin 0 =18

2( )

18. , so

19. By symmetry,

5


2=4sin =6

56

A =/6

5 /6

2

4sin

r dr d =/6

5 /612

r2 r=4sin

r=2d =

/6

5 /6

(8sin2

2)d

=/6

5 /6

[4(1 cos 2 ) 2]d = 2 2sin 25 /6

/6=

43

+2 3

V =

x2+ y

29

(x2+y

2)dA=

0

2

0

3

(r2)r dr d =

0

2

d0

3

r3dr=

2

0

14

r4 3

0=2

814

=81

2

x2+y

2+z

2=16 xy x

2+y

2=16

V = 2

4 x2+y

216

16 x2

y2

dA [by symmetry]

= 20

2

2

4

16 r2

r dr d =20

2

d2

4

r(16 r2)1/2

dr

= 22

0

13

(16 r2)3/2 4

2=

23

(2 )(0 123/2

)=43

12 12( )=32 3

V =2

x2+ y

2a

2a

2x

2y

2dA=2

0

2

0

a

a2

r2

r dr d =20

2

d0

a

r a2

r2

dr

=22

0

13

(a2

r2)3/2 a

0=2(2 ) 0+

13

a3

=43

a3

z=10 3x2

3y2

z=4 4=10 3x2

3y2

x2+y

2=2

20. implies that or , so

.

21.

22. The sphere intersects the plane in the circle , so

23. By symmetry,

24. The paraboloid intersects the plane when or . So

6


V =x

2+ y

22

(10 3x2

3y2) 4 dA=

0

2

0

2

(6 3r2)r dr d

=0

2

d0

2

(6r 3r3)dr=

2

03r

2 34

r4 2

0=6

z= x2+y

2x

2+y

2+z

2=1 x

2+y

2+ x

2+y

2( ) 2

=1 x2+y

2=

12

V =x

2+ y

21/2

1 x2

y2

x2+y

2( )dA=0

2

0

1/ 2

1 r2

r( ) r dr d

=0

2

d0

1/ 2

r 1 r2

r2( )dr=

2

0

13

(1 r2)3/2 1

3r

3 1/ 2

0

= 213

1

21 =

32 2( )

3x2+3y

2=4 x

2y

2x

2+y

2=1

V =x

2+ y

21

[(4 x2

y2) 3(x

2+y

2)]dA=

0

2

0

1

4(1 r2)r dr d

=0

2

d0

1

(4r 4r3)dr=

2

02r

2r

4 1

0=2

x2+y

2=4 z= 64 4x

24y

2

z= 64 4x2

4y2

V =x

2+ y

24

64 4x2

4y2

64 4x2

4y2( ) dA

=x

2+y

24

2 64 4x2

4y2

dA=40

2

0

2

16 r2

r dr d

25. The cone intersects the sphere when or .

So

26. The two paraboloids intersect when or . So

27. The given solid is the region inside the cylinder between the surfaces

and . So

7


= 40

2

d0

2

r 16 r2

dr=42

0

13

(16 r2)3/2 2

0

= 813

(123/2

162/3

)=83

64 24 3( )

xy r2

1x

2+y

2r

2

2xy

V = 2

r2

1x

2+ y

2r

2

2

r2

2x

2y

2dA=2

0

2

r1

r2

r2

2r

2r dr d

= 20

2

dr

1

r2

r2

2r

2r dr=

43

(r2

2r

2)3/2 r

2

r1

=43

(r2

2r

2

1)3/2

r2

2=

12

h2+r

2

1

14

h2=r

2

2r

2

1

h V =43

14

h2 3/2

=6

h3

0

1

0

1 x2

ex

2+y

2

dydx =0

/2

0

1

er

2

r dr d

28. (a) Here the region in the plane is the annular region and the desired volume is

twice that above the plane. Hence

(b) A cross sectional cut is shown in the figure. So or .

Thus the volume in terms of is .

29.

8


=0

/2

d0

1

rer

2

dr

=/2

0

12

er

2 1

0=

14

(e 1)

/2

/2

0

a

(r2)3/2

r dr d =/2

/2

d0

a

r4dr

=/2

/2

15

r5 a

0

=15

a5

0 0

2

(rcos )2(rsin )

2r dr d =

0(sin cos )

2d

0

2

r5dr

=0

12

sin 22d

0

2

r5dr

=14

12

18

sin 40

16

r6 2

0

=14 2

646

=43

30.

31.

32.

9


0

/2

0

2cos

r2dr d =

0

/213

r3 r=2cos

r=0d

=0

/283

cos3

d

=83

sin13

sin3 /2

0=

169

D 20 DD f (x,y)

x,y( )

D= (x,y) |x2+y

2400{ } f (x,y) y

xy f (0, 20)=2 f (0,20)=7 yz

(0, 20,2) (0,20,7)7 2

20 20( ) =18

z 7=18

(y 20) z=18

y+92

f (x,y) x f (x,y)=18

y+92

Df (x,y)dA

D= (r, ) |0 r 20,0 2{ } x=rcos y=rsin

0

2

0

2018

rsin +92

r dr d =0

2124

r3sin +

94

r2 r =20

r =0d =

0

21000

3sin +900 d

=1000

3cos +900

2

0=1800

1800 56553

R

V =0

2

0

R

err dr d =

0

2

d0

R

rerdr=

2

0re

re

r R

0

= 2 [ ReR

eR+0+1]=2 (1 Re

Re

R)ft

3

33. The surface of the water in the pool is a circular disk with radius ft. If we place oncoordinate 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

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 linearlyin the direction from to . The trace in the plane is a line segment from

to . The slope of this line is , so an equation of the line is

. Since is independent of , . Thus the volume is

given by , which is most conveniently evaluated using polar coordinates. Then

and substituting , the integral becomes

Thus the pool contains ft of water.

34. (a) The total amount of water supplied each hour to the region within feet of the sprinkler is

10


R

Vareaofregion

=V

R2

=2 1 Re

Re

R( )R

2

3

1/ 2

1

1 x2

x

xydydx+1

2

0

x

xydydx+2

2

0

4 x2

xydydx

=0

/4

1

2

r3cos sin dr d =

0

/4r

4

4cos sin

r =2

r =1d

=154 0

/4

sin cos d =154

sin2

2

/4

0=

1516

Da

e(x

2+y

2)dA=

0

2

0

a

rer

2

dr d =212

er

2 a

0= 1 e

a2( ) a

lima

1 ea

2( )= ea

2

0 a e(x

2+y

2)dA=

Sa

e(x

2+y

2)dA=

a

a

a

a

ex

2

ey

2

dxdy=a

a

ex

2

dxa

a

ey

2

dy a

=

R2

(x2+y

2)dA

=lima S

a

e(x

2+y

2)dA=lim

a a

a

ex

2

dxa

a

ey

2

dy = ex

2

dx ey

2

dy .

lima a

a

ex

2

dxa

a

ey

2

dy

(b) The average amount of water per hour per square foot supplied to the region within feet of the

sprinkler is ft (per hour per square foot). See the definition

of the average value of a function on page 1022 [ET 986].

35.

36. (a) for each . Then

since as . Hence .

(b) for each .

Then, from (a), , so

To evaluate , we are using the fact that these integrals are bounded.

11


1,1 0<ex

2

1 , 1( ) 0<ex

2

ex

1,( ) 0<ex

2

<ex

0 ex

2

dx1

exdx+

1

1

dx+1

exdx=2(e

1+1)

ex

2

dx ey

2

dy = y x ex

2

dx

2

=

ex

2

dx= ex

2

0 x ex

2

dx=

t= 2 x ex

2

dx=1

2e

t2/2( )dt =

1

2e

t2/2

dt et2/2

dt= 2

u=x dv=xex

2

dx du=dx v=12

ex

2

0x

2e

x2

dx = limt 0

t

x2e

x2

dx=limt

12

xex

2 t

0+

0

t12

ex

2

dx

= limt

12

tet2

+12 0

ex

2

dx=0+12 0

ex

2

dx [by l’Hospital’s Rule]

=14

ex

2

dx [since ex

2

is an even function]

=14

[by Exercise 36(c)]

u= x u2=x dx=2udu

0x e

xdx = lim

t 0

t

x exdx=lim

t 0

t

ueu

2

2udu=20

u2e

u2

du

= 214

[by part(a)]=12

This is true since on , while on , and on , .

Hence .

(c) Since and can be replaced by , implies that

. But for all , so .

(d) Letting , , so that or .

37. (a) We integrate by parts with and . Then and , so

(b) Let . Then

12


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