### Home > APCALC > Chapter 12 > Lesson 12.1.3 > Problem12-37

12-37.

Multiple Choice: To calculate the area of the region in the first quadrant outside the circle $r = 4\cos(θ)$ and inside the lemniscate $r^2 = 8\sin(2θ)$, use the integral:

1. $\frac { 1 } { 2 } \int _ { \pi / 4 } ^ { \pi / 2 } ( 16 \operatorname { cos } ^ { 2 } ( \theta ) - 8 \operatorname { sin } ^ { 2 } ( 2 \theta ) ) d \theta$

1. $\frac { 1 } { 2 } \int _ { \pi / 4 } ^ { \pi / 2 } ( 8 \operatorname { sin } ^ { 2 } ( 2 \theta ) - 16 \operatorname { cos } ^ { 2 } ( \theta ) ) d \theta$

1. $\frac { 1 } { 2 } \int _ { 0 } ^ { \pi / 4 } ( 16 \operatorname { cos } ^ { 2 } ( \theta ) - 8 \operatorname { sin } ^ { 2 } ( 2 \theta ) ) d \theta$

1. $\frac { 1 } { 2 } \int _ { 0 } ^ { \pi / 4 } ( 8 \operatorname { sin } ^ { 2 } ( 2 \theta ) - 16 \operatorname { cos } ^ { 2 } ( \theta ) ) d \theta$

1. $\int _ { 0 } ^ { \pi / 4 } ( 4 \operatorname { cos } ( \theta ) - 2 \sqrt { 2 } \operatorname { sin } ( 2 \theta ) ) d \theta$

Determine where the functions intersect for the bounds of integration. Remember, this is only in the first quadrant.

$16\cos^2(\theta)=8\sin(2\theta)$

$16\cos^2(\theta)=16\sin(\theta)\cos(\theta)$

$\cos(\theta)=0\text{ or }\sin(\theta)=\cos(\theta)$

Graph the functions to determine which is the "upper" function and which is the "lower" function.

Use the area inside polar curves formula:

$\int_a^b\frac{1}{2}(R^2-r^2)d\theta$