Question

Find the area of the region that lies inside the first curve and outside the second curve

Answer 1

\[r= 12 \sin \theta , r=6\]

Answer 2

i know that \[A= \frac{ 1 }{ 2 }\int\limits_{a}^{b}(r:1st)^2-(r:2nd)^2 dtheta\] but im not sure how to find a and b on the integral

Answer 3

Do i set the 2 equal to each other and solve for theta to find a and b?

Answer 4

thats the area you want?

Answer 5

first solve
$$\Large {
r= 12 \sin \theta , r=6\\
6 = 12 \sin \theta \\
\frac 1 2 = \sin \theta
\\ \theta = \pi /6 ~, ~5 \pi / 6
}
$$

Answer 6

it wouldn be just neg and pos pi/6

Answer 7

idk, this is really hard!

Answer 8

$$
\Large {
\frac12\int_{\pi/6}^{5\pi/6}\frac12 [(12 \sin \theta)^2 - 6^2]~ d\theta



}
$$

Answer 9

sin theta = 1/2 , the solution comes from the unit circle .

Answer 10

ok