MATH 2110 – EXAM 4 REVIEW SOLUTIONS
1.
2
3 9
2 2 3/ 2
0 0
( ) dy dx
x
x y
−
+
∫ ∫
=
243
10
π
2.
2 2
R
x y dA
+
∫∫
=
9
π
3. Mass = 6c
Center of Mass = (8/3, 1)
4.
4 1
0 0 0
sin dz dy dx
x
x y
π
−
∫∫ ∫
=
80
3
−
5. V =
(2 2 ) (3 3 3 /2 )
1
0 0 0
dz dy dx
x x y− − −
∫ ∫ ∫
= 1
6.
( 1, 2 , 4)
−= (
3
, 2.18, 4)
7.
( 3,1, 2 3 )
= (4,
/ 6
π
,
/ 6
π
)
8. a.) Sphere centered at (0,0,0) with radius = 2
b.) Paraboloid
c.) Cylinder centered along z-axis with radius = 2
9.
12
π
10.
256 81
( )
16
e e
π
−
p.1058 - #22:
(
)
1
2 2 1
3
−