00:01
Hello, so the area we're looking to find here is going to be the area between the points of intersection of these curves is going to imply that three sine theta is going to be equal to 1 plus sine of theta, which is then going to give us that 2 sine of theta is equal to 1.
00:20
Giving us that sine of theta is going to be equal to 1 half.
00:25
So therefore, theta here is going to be equal to pi over 6 and 5 pi over 6.
00:30
Then we're taking the integral here, we can pull out the one -half in front.
00:34
So we did one -half times the integral from pi over six to five pi over six of three side of theta, um, all squared, d theta, and then minus one integral, uh, from again, pi over six to five pi over six of five, over six of one, 1 plus squared d theta.
01:07
So this is then going to give us, we can use our identities here, and we're going to get a 9 4th, and then times the integral from pi over 6, the 5, pi over 6 of 1d theta, and then minus times the interval from pi over 6.
01:38
Um, uh, we can write this as 1 plus, two side of theta and then plus one half times the quantity, um, one minus cosine of d theta.
01:59
All right, so we integrate here, and we get nine -fourths times theta minus one -half sine of two theta, evaluating from pi over six or pi over six...