Quiz 8
Name:
SOLUTIONS
Maths 114 - Calculus II
November 9, 2010
Note: In order to receive full credit, you must show work that justifies your answer.
1. Find the volume of the solid under the surface x2 + y − 2z = 0 and above the
region in the xy-plane bounded by y = x2 and y = 2 − x2 .
First, we need to know how to describe the region R in the xy-plane bounded
by y = x2 and y = 2 − x2 . These two curves intersect when x2 = 2 − x2 , i.e.
x = ±1, which gives (−1, 1) and (1, 1) as the points of intersection. Therefore R
can be described as all points (x, y) where −1 ≤ x ≤ 1 and x2 ≤ y ≤ 2 − x2 .
Next, we can rewrite the surface as z = 12 [x2 + y], so the volume we want is
given by
ZZ
R
1 2
[x + y]dA =
2
=
=
=
=
=
Z
1
Z
2−x2
1 2
[x + y]dy dx
2
−1 x2
y=2−x2
Z 1 1 2
1 2 dx
x y+ y 2 −1
2
y=x2
Z
1 1 2
1
x (2 − x2 ) − x2 (x2 ) +
(2 − x2 )2 − (x2 )2 dx
2 −1
2
Z 1
1
1
2x2 − 2x4 +
4 − 4x2 dx
2 −1
2
Z 1
1
2 − 2x4 dx
2 −1
x=1
x5 x−
5 x=−1
8
=
.
5
2. Use a double integral to find the area of the region in the first and fourth
quadrants bounded by r = 1 − cos(θ) and r = 1 + cos(θ).
The region we want is outside the region enclosed by r = 1 − cos(θ) and inside
the region enclosed by r = 1 + cos(θ), so when we describe the region using
polar coordinates, we have 1 − cos(θ) ≤ r ≤ 1 + cos(θ).
What are the bounds on θ? At the points where the two curves intersect, we
have 1 − cos(θ) = 1 + cos(θ), so cos(θ) = 0. The general solution to this is
θ = π2 + kπ, where k is an integer, so the bounds on θ will be − π2 ≤ θ ≤ π2 .
Therefore the area of the region R is
Z
ZZ
π/2
Z
1+cos(θ)
1dA =
r dr dθ
−π/2
R
Z
=
1−cos(θ)
r=1+cos(θ)
π/2
1 2 −π/2
1
=
2
Z
1
2
Z
=
r
dθ
2 r=1−cos(θ)
π/2
[(1 + cos(θ))2 − (1 − cos(θ))2 ] dθ
−π/2
π/2
4 cos(θ) dθ
−π/2
θ=π/2
= 2 sin(θ)|θ=−π/2
= 4.
Quiz 8
Name:
SOLUTIONS
Maths 114 - Calculus II
November 11, 2010
Note: In order to receive full credit, you must show work that justifies your answer.
1. Find the volume of the solid under the surface z = 2xy and above the triangle
with vertices (0, 0), (2, 0) and (0, 2).
The triangle is bounded by the coordinate axes and the line y = 2 − x, so we
can describe the triangle as being all points (x, y) such that 0 ≤ x ≤ 2 and
0 ≤ y ≤ 2 − x (of course, we could also describe it as being all points (x, y) such
that 0 ≤ y ≤ 2 and 0 ≤ x ≤ 2 − y). The volume of the solid is the double
integral of the function f (x, y) = 2xy over the triangle:
ZZ
Z 2 Z 2−x
f (x, y)dA =
2xy dy dx
R
0
0
Z 2
y=2−x
=
xy 2 y=0 dx
Z0 2
=
x(2 − x)2 dx
Z0 2
=
4x − 4x2 + x3 dx
0
x=2
4 3 1 4 2
= 2x − x + x 3
4
x=0
32
+4−0
= 8−
3
4
=
.
3
GRADING NOTE: Since I said the quiz would only cover section 16.4 and this question isn’t
designed to be solved using polar coordinates, I’ve decided to make your total score for this
quiz equal to your score for question 2 plus the maximum of your scores for both questions.
2. Draw the graph of r = cos(3θ) and calculate the area of the region it bounds.
The graph is a ‘three leaved rose’ (see the solution to exercise 11.4 13 on pg A98
of the textbook). The total area is three times the area of one ‘petal’, and the
bounds on θ for the petal which lies along the positive x-axis are given by θ
such that cos(3θ) = 0, so 3θ = ±π/2, i.e. θ = ±π/6.
For a fixed value of θ in the interval [−π/6, π/6], the points inside the region
have r between zero and cos(3θ). Therefore the area is
ZZ
Z
π/6
cos(3θ)
Z
r dr dθ
1dA =
−π/6
R
Z
π/6
=
0
r=cos(3θ)
1 2 r
dθ
2 r=0
−π/6
Z π/6
=
1
2
Z
cos2 (3θ) dθ
−π/6
π/6
cos2 (3θ) dθ
=
(even function on symmetric interval)
0
Z
π/6
1
[cos(6θ) + 1] dθ (double angle formula)
2
0
θ=π/6
1
1 =
sin(6θ) + θ 12
2 θ=0
1
1 π
=
sin(π) +
−0
12
2 6
π
.
=
12
=
The total area is thus 3(π/12) = π/4.