Practice Test 3 Key

Practice Test Ch. 15 1. The volume under the surface, z  x 2  y , and above the rectangle on the x-y plane with vertices (0,0), (0,6), (6,0) and (6,6) is to be approximated with six approximating rectangles as illustrated below. Using the centers of the subrectangles as the approximation points, write the approximate volume as a double sum.

6 3 0

0

  6   2 j  1 3

2

j 1 k 1

2

2

4

6



 1.5  3  k  1 

2. What is the approximate volume? 528 3. Find the exact volume using a double integral. 6 6

  x

6

2

0 0

 y  dxdy    6 y  72  dy  540 0

1

4. Given

 /4

 f  x, y  dA    D

0 tan 1  x 

f  x, y  dydx , sketch the region, D. Rewrite the

integral with the appropriate limits so that the order of integration is reversed.  / 4 tan  x 

  f  x, y  dxdy 0

0

5. Set up, without evaluating, double integrals in polar coordinates to find the shaded area in the figure below. It consists of the region which is outside the circle r  1 and inside the rose, r  2 cos  2  and the region that is outside the rose, r  2 cos  2  and inside the circle r  1 .  / 6 2cos 2 

8 0

 1

 /4

rdrd  8 

1



rdrd

 / 6 2cos 2 

6. Given that X and Y are independent, and their probability density functions are:  2  3y 1 1 y  3  2 1 x  2  f  x   x , and g  y    10 . Find the joint  0 otherwise  0 otherwise distribution function, and show that it is a probability distribution function.  3 y 1  h  x, y    5 x 2  0

  x, y  |1  x  2 and 1  y  3 Otherwise

h  x, y   0  x, y  in the domain 3 2

3 y 1 dxdy  1 5x2 1 1



7. Use the joint distribution function above to find the probability that x  y . 2 3 y 1 3x 2  2 x  1 1 ln  2  dydx  dx    0.1114 1 1 5 x 2 1 10 x 2 4 5 2 x

3 4 5

8. Evaluate the iterated integral,

   x 0 1 2

3

 y 2  z  dx dy dz

3 4 5

3

4

3

3 9 6399 2 0 1 2  x  y  z  dx dy dz  0 4 1  4 y  4 z  203 dydz  4 0  4 z  231 dz  4 3

2

9. The shaded region in the illustration has boundaries of, 1 x 2 y S1  , S2  , S3  , S 4  2 x . Given the transformation, u  xy and v  , 2x 2 x x sketch the transformed region in the u-v plane. 1  1  S1T : u  , v   , 2  2  2  1  1  S 2T : u   , 2  , v  2  2   1  ,2  2 

S3T : u  2, v  

 1  ,2 ,v  2  2 

S 4T : u   (1/2,2)

(2,2)

(1/2,1/2)

(2,1/2)

10. Find the Jacobean of this transformation. u x v x

u y y  y v x2 y

x 1 x