2350H1 MATH SOLUTIONS TO SECOND IN-CLASS EXAM Problem 1: Find the centroid of the quadrilateral with corners at (0, 0), (0, 1), (3, 1) and (3, 3). Answer: Call the quadrilateral Q. Q is the domain between the lines 1 2 y = 3 x and y = 3 x + 1, for 0 x 3. Thus, it is of Type I, and we nd
3
2 x+1 3
dA =
Q 0 3
1 3x 2 x+1 3
dy dx =
15 , 2 27 , 2
x dA =
Q 0 3
1 3x 2 x+1 3
x dy dx =
dy A =
Q 0
1 3x
y dy dx = 6.
Thus, x= 9 27/2 = , 15/2 5 y= 6 4 = . 15/2 5
1
2
MATH 2350H1 SOLUTIONS TO SECOND IN-CLASS EXAM
Problem 2: Let S be the domain consisting of all those points in space that are inside the sphere with center (0, 0, 0) and radius 1, and above the cone given by z = 1 x2 + y 2 . (a) Set up triple integrals in cylindrical and spherical coordinates for the volume of S. [Hint: The spherical equation for the cone is = 1/(cos() + sin()).] (b) Select one of the triple integrals from (a), and use it to nd the volume of S.
Answer: (a) In cylindrical coordinates, the cone is z = 1 r and the sphere is r2 + z 2 = 1. S consists of those points above the unit disc that are above the cone and below the upper hemisphere. Thus, in cylindrical coordinates, we have 2
2 1 1r
dV =
S 0 0 1r
r dz dr d.
In spherical coordinates, the sphere is given by = 1 and the cone by = 1/(cos()+sin()). Since S sits in the upper half-space (z 0), we get
/2 2 0 1
dV =
S 0 1/(cos()+sin())
2 sin() d d d.
(b) Cylindrical coordinates look easiest: 2
2 1 1r
dV =
S 0 2 0 1 1r
r dz dr d r(
0 2 0
= =
0 2
1 r2 1 + r) dr d
1 [ 1 (1 r2 )3/2 2 r2 + 1 r3 ]1 d 0 3 3 1 6
=
0
d =
. 3
(Here,
r
1
r2
dr was found by substituting u = 1 r2 .)
MATH 2350H1 SOLUTIONS TO SECOND IN-CLASS EXAM
3
Problem 3: Let S be the solid enclosed by the cylinder + x2 z 2 = 1 between the planes y = 1 and y = 1. (a) Set up the triple integral using that S is x-simple. (b) Set up the triple integral using that S is y-simple. (c) Set up the triple integral using that S is z-simple.
S
f dV in rectangular coordinates, f dV in rectangular coordinates, f dV in rectangular coordinates,
S
S
Answer: (a) The two halves of the cylinder are given by x = 1 z 2 on the domain 1 y, z 1. Thus, we get 2
1 1 1z
f dV =
S 1 1
f dx dy dz.
1z 2
(b) S consists of everything between the two planes y = 1 over the unit disc in the xz-plane. Therefore, 2
1 1x 1
f dV =
S 1
f dy dz dx.
1x2 1
(c) The two halves of the cylinder are given by z = the domain 1 x, y 1. Thus, we get 2
1 1 1x
1 x2 on
f dV =
S 1 1
f dz dy dx.
1x2
4
MATH 2350H1 SOLUTIONS TO SECOND IN-CLASS EXAM
Problem 4: Let C1 be the unit circle, and let C2 be the circle with 1 center ( 2 , 1 ) and radius 1/ 2. Set up and evaluate a double integral 2 to nd the area of the crescent D consisting of those points that are inside C...