Written Homework 10
MAC 2312
July 28, 2015
Question #1 Rewrite the following polar equation into a cartesian equation. Graph
the polar equation. [Assigned July 20, 2015, due in class July 21, 2015]
r = 6 sin(θ)
We first multiply through by r and then apply our transformation rules:
r2 = 6r sin(θ) ⇒ x2 + y 2 = 6y ⇒ x2 + y 2 − 6y = 0 ⇒ x2 + (y − 3)2 = 9
6
5
4
3
2
1
−3 −2 −1
−1
1
2
3
Question #2 Sketch the polar curve and find the area inside the curve.
[Assigned July 21, 2015, due in class July 23, 2015]
r = 2 + 2 cos(θ)
3
2
1
−1
−1
1
2
3
4
5
−2
−3
Z π
Z π
1
2
A=2
(2 + 2 cos(θ)) dθ =
4 + 8 cos(θ) + 4 cos2 (θ) dθ
2 0
0
π
π
Z π
1
1
(4θ + 8 sin(θ)) + 4
(1 + cos(2θ)) = (4π) + 2 θ + sin(θ) 2
0 2
0
0
= 4π + 2π = 6π
Question #3 Sketch the polar curve and find the area inside the curve.
[Assigned July 21, 2015, due in class July 23, 2015]
r = sin(3θ)
1
−1
1
−1
We will calculate the area of one loop and multiply by three:
!
Z π/3
Z
1
3 π/3 1
2
3
sin (3θ) dθ =
(1 − cos(6θ)) dθ
2 0
2 0 2
π/3
1
π
3
3 π θ − sin(6θ) =
=
=
4
6
4 3
4
0
Question #4 Sketch the following polar graphs. Find the area inside the first curve
but outside the second curve.
[Assigned July 23, 2015, due in class July 24, 2015]
r1 = 2
r2 = 2 sin(2θ)
2
1
−2
−1
1
2
−1
−2
To find the shaded area, we will take the area of the circle that we know
is 2π, and subtract off the area of the rose:
!
Z π/1
Z π/2 1
1
2
Arose = 4
(2 sin(2θ)) dθ = 2
4
(1 − cos(4θ)) dθ
2 0
2
0
π/4
π 1
= 4 θ − sin(4θ) = 4
=π
4
4
0
Thus, the total area of the shaded region is 2π − π = π.
Question #5 Sketch the following polar graphs. Find the area inside both curves.
[Assigned July 24, 2015, due in class July 27, 2015]
r2 = 1 − cos(θ)
r1 = 1 + sin(θ)
2
1
−2
−1
1
2
−1
−2
We can compute half of this area and then double it. We will need to
find where the two curves intersect:
3π 7π
,
1 + sin(θ) = 1 − cos(θ) ⇒ sin(θ) = − cos(θ) ⇒ θ =
4 4
We can now set up the integral to compute the area:
! Z
Z 7π/4
7π/4
1
2
2
(1 + sin(θ)) dθ =
1 + 2 sin(θ) + sin2 (θ) dθ
2 3pi/4
3pi/4
7π/4
1
3θ
1
3
+ 2 sin(θ) − cos(2θ) dθ =
− 2 cos(θ) − sin(2θ) =
2
2
2
4
3pi/4
3pi/4
√
√
21π √
1
1
3π
9π √
12π
=
− 2+
+ 2+
−2 2=
−2 2
−
=
8
4
8
4
8
2
Z
7π/4 Question #6 Sketch the following polar graphs. Find the area inside the first curve
but outside the second curve.
[Assigned July 24, 2015, due in class July 27, 2015]
r1 = 2 cos(θ)
r2 = 2 cos(3θ)
2
1
−2
−1
1
2
−1
−2
To find the shaded area, we will simply find the area of one loop and
subtract it from the area of the circle.
Aloop = 2
1
2
Z
!
π/6
2
(2 cos(3θ)) dθ
Z
=
0
0
π/6
1
4
(1 + cos(6θ)) dθ
2
π/6
π π
1
= 2 θ + sin(6θ) = 2
=
6
6
3
0
Thus, the area of the shaded region is π −
π
2π
=
.
3
3
Question #7 Sketch the following polar graphs. Find the area inside the first curve
but outside the second curve.
[Assigned July 24, 2015, due in class July 27, 2015]
r1 = 2 cos(θ)
r2 = 2(1 + 2 cos(θ))
6
4
2
−6
−4
−2
2
4
6
−2
−4
−6
To find the area of the shaded region, we can take the area of the circle
and subtract away the area of the inner loop.
Aloop = 2
1
2
Z
4π/3
!
2
(2 (1 + 2 cos(θ))) dθ
2π/3
Z
4π/3
4 1 + 4 cos(θ) + 4 cos2 (θ) dθ
=
2π/3
4π/3
=4
1 + 4 cos(θ) + 2 (1 + cos(2θ)) dθ = 4 (3θ + 4 sin(θ) + sin(2θ)) 2π/3
2π/3
√ !
√ !
√
√
√
3
3
− 4 2π + 2 3 −
= 8π + 4 3
= 4 4π − 2 3 +
2
2
Z
4π/3