AP Calculus BC 9.9 Polar Region Areas Overview

Review polar-region areas by finding intersections, splitting angular bounds and subtracting overlapping regions accurately.

[Maximum number: 3]

Curve C is defined by the polar equation $r(\theta)=2 \sin ^{2} \theta$ for $0 \leq \theta \leq \pi$ . Curve C and the semicircle

$r=\frac{1}{2}$ for $0 \leq \theta \leq \pi$ are shown in the x y-plane.

Figure for Question 9.9 Finding the Area of the Region Bounded by Two Polar Curves question 1 — AP Calculus BC

(Note: Your calculator should be in radian mode.)

Find the area of the region that lies inside curve C but outside the graph of the polar equation

$r=\frac{1}{2}$ . Show the setup for your calculations.

B Find the area of the region that lies inside curve C but outside the graph of the polar equation $r=\frac{1}{2}$ .

Show the setup for your calculations.

For $0 \leq \theta \leq \pi, r(\theta)=\frac{1}{2}$ for $\theta=\theta_{1}=\frac{\pi}{6}=0.523599$ and $\theta=\theta_{2}=\frac{5 \pi}{6}=2.617994$ .	Integrand including $(r(\theta))^{2}$	Point 2 (P2)
$\frac{1}{2} \int_{\theta_{1}}^{\theta_{2}}\left((r(\theta))^{2}-\left(\frac{1}{2}\right)^{2}\right) d \theta$	Integrand	Point 3 (P3)
=2.066769<br>The area is 2.067 (or 2.066).	Answer	Point 4 (P4)
Scoring Notes for Part B

- P2 is earned for a definite integral including $(r(\theta))^{2}$ , such as $\int_{a}^{b}\left((r(\theta))^{2}-\left(\frac{1}{2}\right)^{2}\right) d \theta$ or

$\int_{a}^{b}(r(\theta))^{2} d \theta$ , with or without the differential $d \theta$ .

- P3 is earned for a definite integral (or integrals) with a correct integrand, such as

$\int_{a}^{b}\left((r(\theta))^{2}-\left(\frac{1}{2}\right)^{2}\right) d \theta$ or $\int_{a}^{b}(r(\theta))^{2} d \theta-\int_{c}^{d}\left(\frac{1}{2}\right)^{2} d \theta$ , with or without the differential $d \theta$ .

- The limits $\theta_{1}=\frac{\pi}{6}=0.523599$ and $\theta_{2}=\frac{5 \pi}{6}=2.617994$ and the factor $\frac{1}{2}$ are assessed in P4, not

in P2 or P3.

- P4 is earned for the correct answer, with or without supporting work. A reported answer should be

accurate to three places after the decimal point, rounded or truncated. An inappropriately rounded

answer does not earn the point, unless an earlier point was not earned due to inappropriate rounding.

- Incorrect or unclear communication between the correct integral and the correct answer is treated as

scratch work and is not considered in scoring. For example:

○ $\int_{\pi / 6}^{5 \pi / 6}\left((r(\theta))^{2}-\left(\frac{1}{2}\right)^{2}\right) d \theta=4.133538$ so the area is 2.067.

Note: This response earns P2 and P3 for the integral. It also earns P4 for the correct answer.

○ $\int_{\pi / 6}^{5 \pi / 6}\left((r(\theta))^{2}-\left(\frac{1}{2}\right)^{2}\right) d \theta=2.067$

Note: This response earns P2 and P3 for the integral. It also earns P4 for the correct answer. (In

this instance, incorrect linkage is not considered in scoring.)

- Special case: An indefinite integral with a correct integrand does not earn P2, earns P3, and is

eligible to earn P4 with a correct answer.

- A response of $\int_{\pi / 6}^{\pi / 2}\left((r(\theta))^{2}-\left(\frac{1}{2}\right)^{2}\right) d \theta=2.067$ , using the symmetry of the region, earns P2, P3,

and P4.

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Covered syllabus points

9.9 Finding the Area of the Region Bounded by Two Polar Curves question 1