Home > CALC > Chapter 12 > Lesson 12.1.3 > Problem12-40

12-40.

Multiple Choice: To find the area of the region in the first quadrant outside the circle $r = 4\operatorname{cos} θ$ and inside the lemniscate $r^2 = 8\operatorname{sin} 2θ$ , 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.
$16\cos^2(\theta)=8\sin(2\theta)$
$16\cos^2(\theta)=16\sin(\theta)\cos(\theta)$
Graph the functions to determine which is the "upper" function and which is the "lower" function.
Use the area inside polar curves formula