1. Evaluate the iterated integral

   ó õ p 0  ó õ x 0  sin( x) dydx

   Solution: First we evaluate inner integral:

   ó õ p 0  ó õ x 0  sin( x) dydx = ó õ p 0  sin( x) y| 0xdx = ó õ p 0  xsin(x) dx

   and then integration by parts with u = x and dv = sin( x) dx yields

   ó õ p 0  ó õ x 0  sin( x) dydx = -xcos(x)  p 0   +  ó õ p 0  cos( x) dx = p

2. Find the volume of the solid bound between z = 0 and z = x+2y over

   R: x = 0 y = 0 x = 2 y = x2

   Solution: As a triple integral, we have

   V = ó õ ó õ ó õ S  dV = ó õ 2 0  ó õ x2 0  ó õ x+2y 0  dzdydx = ó õ 2 0  ó õ x2 0  ( x+2y) dydx =  52 5

3. Evaluate the following iterated integral by changing it from a Type I to a type II or vice versa:

   ó õ p 0  ó õ p x   sin( y) y dydx

   Solution: First, we convert to double integral over region R.

   ó õ p 0  ó õ p x   sin( y) y dydx = ó õ ó õ R   sin( y) y dA

   The region R in type II is given by y = 0, y = p, x = 0, x = y, so that

   ó õ p 0  ó õ p x   sin( y) y dydx = ó õ ó õ R   sin( y) y dA = ó õ p 0  ó õ y 0   sin( y) y dxdy = ó õ p 0   sin( y) y y  dy = ó õ p 0  sin( y) dy = 2

4. Evaluate the following iterated integral by changing it from a Type I to a Type II or vice versa:

   ó õ 1 0  ó õ 1-x 0  sec2( 2y-y2) dydx

   Solution: Convert to double integral over region R, which in type II is given by y = 0, y = 1, x = 0, x = 1-y.

   ó õ 1 0  ó õ 1-x 0  sec2( 2y-y2) dydx = ó õ ó õ R  sec2( 2y-y2) dA = ó õ 1 0  ó õ 1-y 0  sec2( 2y-y2) dxdy = ó õ 1 0  sec2( 2y-y2)   ( 1-y) dy

   Now we let u = 2y-y² and du = 2-2y , so that u( 0) = 0 and u( 1) = 1 and

   ó õ 1 0  ó õ 1-x 0  sec2( 2y-y2) dydx = ó õ ó õ R  sec2( 2y-y2) dA = ó õ 1 0  sec2( 2y-y2)   ( 1-y) dy =  1 2 ó õ 1 0  sec2( u) du =  1 2 tan( 1)

5. Find the mass of the cylinder between z = 0 and z = 1 over the interior of the unit circle if its mass density is given by r(x,y,z) = | y| . Solution: The mass is given by the triple integral

   M = ó õ ó õ ó õ S  r( x,y,z) dV = ó õ 1 -1  ó õ Ö 1-x2 -Ö 1-x2 ó õ 1 0  |y| dzdydx = ó õ 1 -1  ó õ Ö 1-x2   -Ö 1-x2 | y| dydx = 2 ó õ 1 -1  ó õ Ö 1-x2 0  ydydx = 2 ó õ 1 -1   y2 2 ê ê Ö 1-x2 0  dx = ó õ 1 -1  ( 1-x2) dx =  4 3

6. What is the volume of the polyhedron with vertices (0,0,0) , ( 1,0,0) , ( 0,1,0) , (1,1,0) , ( 0,0,1) , and ( 1,0,1) ? Solution:The solid is a ''wedge'' bound between the xy-plane and the plane passing through the points ( 0,1,0) , (1,1,0) , ( 0,0,1) , and ( 1,0,1) . The equation of the latter plane is easily shown to be z = 1-y. The region in the xy-plane over which the solid is defined is simply the unit square, so that

   V = ó õ 1 0  ó õ 1 0  ó õ 1-y 0  dzdxdy =  1 2

7. Suppose that the probability density for the time required to complete the ``A'' component of an exam is given by

   pA( x) = ì ï í ï î 0 if x < 0  1 30 e-x/30 if x ³ 0

   (time in minutes). Suppose the event of completing the ``B'' component of the exam has the same density. If the completion of the A and B sections are independent events, then what is the probability that a student will complete the entire exam (i.e., both sections) in less than an hour? Solution: Independence of events ``completing  A'' and ``completing  B'' imply that the joint density of A and B is

   p( x,y) = pA( x) pB( y)

   Since the distributions are identical, the probability is

   Pr ( x+y £ 60) =  1 900 ó õ ó õ R  e-x/30e-y/30dA

   where R is the region defined by x = 0, x = 60, y = 0, and y = 60-x. Thus,

   Pr ( x+y £ 60) =  1 900 ó õ 60 0  ó õ 60-x 0  e-x/30e-y/30dydx = 1-3e-2

   and since 1-3e^-2 » 0.59399, there is about a 59.4% chance of completing the entire exam in less than an hour.
8. Evaluate by converting to polar coordinates:

   ó õ 1 0  ó õ Ö 2-x2 x  dydx

   Solution: The region x = 0, x = 1, y = x, y = Ö{2-x²} is given in polar coordinates by

   R: q =  p 4 r = 0 q =  p 2 r = Ö2

   so that changing to a double integral and using dA = rdrdq yields

   ó õ 1 0  ó õ Ö 2-x2 x  dydx = ó õ ó õ R  dA = ó õ p/2 p/4  ó õ Ö2 0  rdrdq =  p 4

9. Evaluate by converting to polar coordinates

   ó õ 1 0  ó õ Ö 1-x2 1-x   dydx ( x2+y2)3/2

   Solution: The curve y = 1-x is the same as x+y = 1, and thus has a pullback of

   rcos( q) +rsin( q) = 1,        r =  1 cos( q) +sin( q)

   The curve y = Ö{1-x²} is the same as x²+y² = 1, or r = 1 in polar. Thus, the region x = 0, x = 1, y = 1-x, y = Ö{1-x²} is given in polar coordinates by

   R: q = 0 r =  1 cos( q) +sin( q) q =  p 2 r = 1

   so that changing to a double integral and using dA = rdrdq yields

   ó õ 1 0  ó õ Ö 1-x2 1-x   dydx ( x2+y2)3/2 = ó õ ó õ R   1 ( x2+y2) 3/2 dA = ó õ p/2 0  ó õ 1 [ 1/(cos( q) +sin(q) )]   1 ( r2) 3/2   rdrdq = ó õ p/2 0  ó õ 1 [ 1/(cos( q) +sin(q) )]   1 r2 drdq = ó õ p/2 0   -1 r ê ê 1 [ 1/(cos( q) +sin( q) )]  dq = ó õ p/2 0  ( cos( q) +sin( q) -1) dq = 2-  p 2

10. Use the coordinate transformation T( u,v) = á u,Öv ñ to evaluate

    ó õ Öp 0  ó õ Öp 0  ysin( y2) dydx

    Solution: If x = 0, then u = 0. If x = Öp, then u = Öp. If y = 0, then v = 0. If y = Öp, then v = p. Thus, the pullback of the region of integration into uv-coordinates is

    S: u = 0   v = 0 u = Öp   v = p

    Since T_u = á 1,0 ñ and since T_v = á 0,[ 1/2]v^-1/2 ñ , the Jacobian determinant is

    ¶( x,y) ¶( u,v) = ê ê ê ê ê 1 0 0  1 2 v-1/2 ê ê ê ê ê =  1 2 v-1/2

    Thus, we have

    ó õ Öp 0  ó õ Öp 0  ysin( y2) dydx = ó õ ó õ R  ysin( y2) dA = ó õ Öp 0  ó õ p 0  v1/2sin( v)  1 2 v-1/2dvdu =  1 2 ó õ Öp 0  ó õ p 0  sin( v)  dvdu = Öp

11. Use the coordinate transformation T( u,v) = áu,ve^-u ñ to evaluate

    ó õ 1 0  ó õ 1 0  yexdydx

    Solution: The pullback of x = 0 is u = 0. The pullback of x = 1 is u = 1. The pullback of y = 0 is ve^-u = 0, or v = 0. The pullback of y = 1 is ve^-u = 1, or v = e^u. Thus, the pullback of the region of integration into uv-coordinates is

    S: u = 0 v = 0 u = 1 v = eu

    Since T_u = á 1,-ve^-u ñ and since T_v = á 0,e^-u ñ , the Jacobian determinant is

    ¶( x,y) ¶( u,v)  =   ¶x ¶u  ¶y ¶v  -   ¶y ¶v  ¶x ¶u  =  e-u

    Thus, dA = e^-udvdu and we have

    ó õ 1 0  ó õ 1 0  yexdydx = ó õ ó õ R  yexdA = ó õ 1 0  ó õ eu 0  ve-ueu  e-udvdu = ó õ 1 0  ó õ eu 0  ve-udvdu =  1 2 e-  1 2

12. Suppose r( x,y,z) = xz( 1-y) coulombs per cubic meter is the charge density of a ''charge cloud'' contained in the ''box'' given by [ 0,1] ×[ 0,1] ×[0,1] . What is the total charge inside the box? Solution: If W denotes the box [ 0,1] ×[ 0,1] ×[ 0,1] , then

    Q = ó õ ó õ ó õ W  xz( 1-y) dV = ó õ 1 0  ó õ 1 0  ó õ 1 0  xz( 1-y) dzdydx =  1 2 ó õ 1 0  ó õ 1 0  ( x-xy) dydx =  1 8   Coulombs

13. Evaluate by converting to spherical coordinates

    ó õ 1 -1  ó õ Ö 1-x2 -Ö 1-x2 ó õ Ö 2-x2-y2 Ö x2+y2  dzdydz z x2+y2+z2

    Solution: As a triple integral, we have

    ó õ ó õ ó õ S   dV z x2+y2+z2

    where S is the solid over the unit circle which is above the cone z = Ö{x²+y²} and below the sphere with radius Ö2. However, the cone in spherical coordinates is f = p/4, while the sphere is r = Ö2. Moreover, z = rcos( f) , x²+y²+z² = r², and dV = r²sin( f) dr df dq yields

    ó õ ó õ ó õ S   dV z x2+y2+z2 = ó õ 2p 0  ó õ p/4 0  ó õ Ö2 0   r2sinfdrdfdq r2cos( f) = Ö2 ó õ 2p 0  ó õ p/4 0   sin( f) cos( f) dfdq = pÖ2  ln( 2)

14. What is the mass of the cone x²+y² = z² between z = 0 and z = 1 if the mass density is constant at m = 4 kg per cubic meter? Solution: In spherical coordinates, the cone x²+y² = z² corresponds to the constant angle f = p/4. However, the plane z = 1 corresponds to

    rcos( f) = 1,        or        r = sec(f)

    Thus, the mass of the cone is

    M = ó õ ó õ ó õ Cone    4  dV = ó õ 2p 0  ó õ p/4 0  ó õ sec( f) 0  4r2sin( f)drdfdq

    Evaluating and simplifying the inner integral yields

    M = ó õ 2p 0  ó õ p/4 0   4r3 3 sin(f) ê ê sec( f) 0    dfdq =  4 3 ó õ 2p 0  ó õ p/4 0  sec3( f)sin( f)   dfdq =  4 3 ó õ 2p 0  ó õ p/4 0  sec( f)    sec( f) tan( f)   dfdq

    since sin( f) sec( f) = tan( f) . Letting u = sec( f) , du = sec( f) tan( f) , u( 0) = 1, and u( p/4) = Ö2 yields

    M =  4 3 ó õ 2p 0  ó õ Ö2 1  ududq =  4 3 ó õ 2p 0   1 2 dq =  4p 3