El Jhdejhed De Jksajk

Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.1 Double Integrals over Rectangles

1. (a) The subrectangles are shown in the figure. The surface is the graph of f (x,y)=xy and A=4 , so we estimate

V

3 i =1

2 j=1

f x ,y
i

( )
j

A A+ f ( 6,4 ) A

= f ( 2,2 ) A+ f ( 2,4 ) A+ f ( 4,2 ) A+ f ( 4,4 ) A+ f ( 6,2 ) =4 ( 4 ) +8 ( 4 ) +8 ( 4 ) +16 ( 4 ) +12 ( 4 ) +24 (4 ) =288 (b) V
3 i =1 2 j=1

f x ,y
i

( )
j

A A+ f ( 5,1 ) A+ f ( 5,3) A

= f ( 1,1 ) A+ f ( 1,3) A+ f ( 3,1 ) A+ f ( 3,3) =1 ( 4 ) +3 ( 4 ) +3 ( 4 ) +9 ( 4 ) +5 ( 4 ) +15 ( 4 ) =144 2. The subrectangles are shown in the figure. Since A=1 , we estimate

( y2 2x2) dA
R

4 i =1

2 j=1

f x ,y

* *

ij ij

A

= f ( 1,1 ) A+ f ( 1,2 ) A+ f ( 0,1 ) A+ f ( 0,2 ) A + f (1,1 ) A+ f ( 1,2 ) A+ f ( 2,1 ) A+ f ( 2,2 ) A = 1 ( 1 ) +2 ( 1 ) +1 ( 1 ) +4 ( 1 ) 1 ( 1 ) +2 ( 1 ) 7 ( 1 ) 4 ( 1 ) = 4 3. (a) The subrectangles are shown in the figure. Since sin (x+y)dA
R 2 i=1 2 j=1

A=

2

/4 , we estimate

f x ,y 0,

* *

ij ij

A A+ f 2 ,0 A+ f , 2 2 A
1

= f (0,0)

A+ f

2

Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.1 DoubleIntegrals over Rectangles

2

2

2

2

2

=0

4

+1

4

+1

4

+0

4

=

2

4.935

sin (x+y)dA
R

2 i=1

2 j=1

f(x ,y ) A
i j

=f (b) +f =1

4 4 3 , 4 4
2

,

A+f

3 4 4 3 3 A+f , 4 4 ,
2 2

A A
2

4

+0

4

+0

4

+( 1)

4

=0

4. (a) The subrectangles are shown in the figure. The surface is the graph of f (x,y)=x+2y and V =
R2

A=2 , so we estimate

(x+2y )dA

2

2 i=1

2 j=1

f x ,y

* *

ij ij

A A

= f (1,0) A+ f (1,2) A+ f (2,0) =1(2)+9(2)+2(2)+10(2)=44

A+ f (2,2)

2

Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.1 Double Integrals over Rectangles

(b) V =
R

(x+2y )dA

2

2 i=1

2 j=1

f (x ,y )
i j

A 3 ,1 2 A+ f 3 ,3 2 A

1 1 ,1 A+ f ,3 A+ f 2 2 5 377 39 = (2)+ (2)+ (2)+ (2)=88 2 2 2 2 =f

5. (a) Each subrectangle and its midpoint are shown in the figure. The area of each subrectangle is A=2 , so we evaluate f at each midpoint and estimate f (x,y)dA
R 2 i =1 2 j=1

f x ,y
i

( )
j

A

= f ( 1.5,1 ) A+ f ( 1.5,3) A + f ( 2.5,1 ) A+ f ( 2.5,3) A =1 ( 2 ) + ( 8 ) ( 2 ) +5 ( 2 ) + ( 1 ) ( 2 ) = 6

3

Stewart Calculus ET 5e0534393217;15. Multiple Integrals; 15.1 Double Integrals over Rectangles

(b) The subrectangles are shown in the figure. In each subrectangle, the sample point farthest from 1 the origin is the upper right corner, and the area of each subrectangle is A= . Thus we estimate 2 f ( x,y ) dA
R 4 i =1 4 j=1

f x ,y
i

( )
j

A A A

=f +f +f +f

( 1.5,1 ) ( 2,1 ) ( 2.5,1 ) ( 3,1 )

A+ f (1.5,2 ) A+ f ( 1.5,3) A+ f ( 1.5,4 ) A+ f ( 2,2 ) A+ f ( 2,3) A+ f ( 2,4 ) A A+ f ( 2.5,2 ) A+ f ( 2.5,3) A+ f ( 2.5,4 ) A+ f ( 3,2 ) A+ f ( 3,3) A+ f ( 3,4 ) A

=1 +5 = 3.5

1 2 1 2

+ ( 4) +3 1 2

1 2

+ ( 8) 1 2

1 2

+ ( 6) 1 2

1 2 +8

+3 1 2

1 2 +6

+0 1 2

1 2 +3

+ ( 5) 1 2

1 2 +0

+ ( 8) 1 2

1 2

+ ( 1)

+ ( 4)

6. To approximate the volume, let R bethe planar region corresponding to the surface of the water in the pool, and place R on coordinate axes so that x and y correspond to the dimensions given. Then we define f (x,y) to be the depth of the water at ( x,y ) , so the volume of water in the pool is the volume of the solid that lies above the rectangle R= 0,20 0,30 and below the graph of f (x,y) . We can estimate this volume using theMidpoint Rule with m=2 and n=3 , so A=100 . Each subrectangle with its midpoint is shown in the figure. Then

4

Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.1 Double Integrals over Rectangles

V

2 i =1

3 j=1

f x ,y
i

( )
j

A

= A f ( 5,5) + f ( 5,15) + f ( 5,25) + f ( 15,5) + f ( 15,15) + f ( 15,25) = 100 ( 3+7+10+3+5+8 ) =3600 Thus, we estimate that the...