Calculus- Early Transcendentals, 2021a

408 Polar Coordinates, Parametric Equations
Solution. We look for points of intersection.
0 = √ cosθ
0 = cosθ
This has solutions θ = π/2and3π/2, but we must select our bounds for integration correctly. Since cosθ
is negative on (π/2,3π/3), √ cosθ is undefined. Note that one full revolution is 2π; that is, the function
is 2π periodic. If we were to integrate on [3π/2,π/2] we would get the same answer, and can avoid a
scenario in which √ cosθ is undefined. Equivalently, we can integrate on [−π/2,π/2].
∫ b
a
1
2 f (θ)2 dθ =
∫ π/2
−π/2
1
2
(√
cosθ
) 2
dθ =
1
2
∫ π/2
cosθ dθ = 1 ( )∣ ∣∣
π/2
sinθ
−π/2 2
= 1
−π/2
♣
.
.
.
. .
Figure 11.7: Approximating area by sectors of circles.
Example 11.11: Area Between Circles
Find the area between the circles r = 2 and r = 4sinθ, as shown in figure 11.8.
Solution. The two curves intersect where 2 = 4sinθ, orsinθ = 1/2, so θ = π/6 or5π/6. The area we
want is then
∫
1 5π/6
16sin 2 θ − 4 dθ = 4 2 π/6
3 π + 2√ 3.
♣
Figure 11.8: An area between curves.