52 Math - Winter 2007 - Midterm Exam I
Problem 1. (8 pts.) Let R be a region of R2 symmetrical about the y -axis, and R+ be the half of the region R contained in the half plane x 0. Mark as TRUE/FALSE the following statements:
a)
R
dA = 2
R+
dA
TRUE
FALSE
b)
R
x dA = 2
R+
x dA
TRUE
FALSE
c)
R
sin(y ) dA = 2
R+
sin(y ) dA
TRUE
FALSE
d)
R
x2 dA = 2
R+
x2 dA
TRUE
FALSE
Problem 2. (10 pts.) Setup, but do not evaluate the triple iterated integral representing the volume of the bounded solid W bounded by the graphs of the paraboloid x2 + y 2 = z and the plane z = 2x + 3. Solution: It is easy to see that the solid must be inside the paraboloid (i.e. x2 + y 2 z ) and below the plane (i.e. z 2x + 3). Thus bounds for z are 2x + 3 z x2 + y 2 . The bounds for x and y are obtained by removing z from above inequalities, i.e. 2x + 3 x2 + y 2 , which after completion of the square becomes 4 (x 1)2 + y 2 , which is the circle with the center (1, 0) and radius 2. Thus the volume is: 2
3 4(x1) 2x+3 1
dz dy dx
x2 + y 2
4(x1)2
Problem 3. (12 pts.) Evaluate
1 0 1 3 3 3
y cos x3 dx dy +
1 y
y cos x3 dx dy
Solution: After the change of variables the above integral becomes:
3 1 0 x 3
y cos(x ) dy dx =
1
3
12 x cos(x3 ) dx 2
1
After substitution u = x3 this gives
1 6
(sin 27 sin 1)
Problem 4. (12 pts.) Change the order of integration in
2 1 0 x2
sin(x + 3y ) dy dx DO NOT EVALUATE THE INTEGRAL Solution: The region of integration is not horizontally simple, thus we need to use the sum of two integrals:
1 0 y 4 2
sin(x + 3y ) dx dy +
1 0
sin(x + 3y ) dx dy
y
Problem 5. 1. (4 pts.) Sketch the region in the xy plane described by the inequality |x| + |y | 1 Solution: The region is a square with vertices (1, 0), (0, 1), (1, 0), (0, 1). 2. (6 pts.) Show that
1
f (x + y ) dx dy =
|x|+|y |1 1
f (t) dt,
for any continuous real-valued function f . Hint: Perform some change of variables... Solution: Set u = x + y and v = x y . The rst variable is kind of obvious from the statement of the problem, the second is suggested by the boundaries of the region of integration. Then the square over which we integrate is 1 u 1 and 1 v 1. The Jacobian is: (u, ) v 11 = det = 2 1 1 (x, y ) Thus f (x + y ) dx dy =
|x|+|y |1 1 1 1 1 1
1 f (u) dv du = 2
1
1 2f (u) du 2
Problem 6. (12 pts.) Consider the region W on xyz space dened by inequalities: 1 x 2, 0 xy 2, 0 z 1. (x, y, z ) a) Find the Jacobian , where u = x, v = xy and w = z . (u, v, w) 2
Solution: 100 (u, v, w) = det y x 0 (x, y, z ) 001 Thus (x, y, z ) 1 = (u, v, w) u (x2 y + 3xyz ) dx dy dz
W
=x=u
b) Evaluate the integral
by applying the above change of coordinates. Solution: After the change of variables the integral becomes:
2 2 1 2 2 1 2 2 1
1 (uv + 3vw) dw dv du = u
1 0 0 1
du
0
v dv
0
dw +
1
1 du u
0
3v dv
0
w dw =
= 2 + ln 8 Problem 7. (12 pts.) Find the coordinates of the centroid of the solid in the rst octant that is bounded from above by the cone z = x2 + y 2 , below by the z = 0 and on the sides by the cylinder x2 + y 2 = 4. (It is part of a cylinder hollowed out by a cone) Hint 1: use cylindrical coordinates. Use: The volume of a cylinder with radius of the base r and the height h is r2 h. The 1 volume of a the cone with radius of the base r and the height h is 3 r2 h. Use these to nd the total mass of your solid. Solution: By symmetry x = y . The total mass can be computed as 1/4 of the volume of the cylinder ( 22 2) minus the 1 volume of the cone ( 3 22 2). Thus the mass is 4 . 3
/2 2 r
mx=
V
x dV =
0 0 0
r cos r dz dr d = 4
Thus x = y = . Similarly: 3
/2 2 r
mz =
V
x dV =
0 0 0
z r dz dr d =
Thus z = 3 . 4 Problem 8. (12 pts.) Let S be the part of the ball x2 + y 2 + z 2 r2 that lies above the xy plane. Find the moment of inertia of S about the z axis. 3
Hint: use spherical coordinates. /2 Use: 0 sin3 x dx = 2 . 3 Solution:
2 /2 r
Iz =
V
x + y dV =
0 0 0
2
2
( sin )2 2 sin d d d =
4r5 15
Problem 9. (12 pts.) Assume that a region R with density 1 in the xy coordinates is symmetrical about the x axis. Show that I(0,b) = I(0,0) + m b2 where I(a,b) is the moment of inertia about an axis perpendicular to xy plane at the point (a, b) and m is the total mass of the region. Solution: By the symmetry y = 0. Then we have: I(0,b) =
R
x2 + (y b)2 dA =
R
x2 + y 2 dA 2b
R I(0,0)
y dA +b2
R y =0 m
dA
4