Quiz 3 (Section 001): Solutions
Problem 1. Find the volume of the solid in the first octant enclosed by surfaces
z= 4 −y2, z = 0, x = 0, y =x, and y= 2.
Solution. V=R RR(4 −y2)dA, where Ris the region in the xy-plane bounded
by the lines x= 0, y=xand y= 2, i.e., Ris determined by the inequalities
0≤x≤2, x≤y≤2. Clearly, on this region z= 4 −y2≥0. Then
V=
2
Z
0
2
Z
x
(4 −y2)dy dx =
2
Z
0
µ4y−y3
3¶¯
¯
2
y=xdx
=
2
Z
0
µ8−8
3−4x+x3
3¶dx =µ16
3x−2x2+x4
12 ¶¯
¯
2
0
=32
3−8 + 4
3= 4.
Problem 2. Evaluate
2
R
0
√4−x2
R
0
(x2+y2)dy dx by converting to an integral in
polar coordinates.
Solution. The region in the xy-plane given by the inequalities 0 ≤x≤2,
0≤y≤√4−x2is the quarter of the disk x2+y2≤4 centered at the origin of
radius 2. In polar coordinate this region is determined by the inequalities 0 ≤θ≤π
2,
0≤r≤2. Since in polar coordinates x=rcos θ, y =rsin θ, and thus x2+y2=r2,
we have
2
Z
0
√4−x2
Z
0
(x2+y2)dy dx =
π
2
Z
0
2
Z
0
r2·r dr dθ =
π
2
Z
0
r4
4¯
¯
2
0dθ = 4 ·π
2= 2π.
1