 In this video, we provide the solution to question number five for practice exam number four for math 12 20 In which case we have to find the area of the polar the area enclosed by the polar curve r equals 3 plus 2 sine of theta This is the this is this this function here is a limouson. It's gonna be 2 pi periodic So the area that we want to compute is gonna equal the integral from 0 to 2 pi All right one half the radius squared. So we're gonna have 3 plus 2 sine Theta squared d theta. I do want to square this thing out You can also use symmetry to help you out here if you wanted to because after all with this limouson It is going to be since it's a sign here. It is going to be symmetric with respect to the y-axis So if you wanted to you could change these bounds here you could instead Well, you could integrate this thing from negative pi halves to pi halves That would give you half of it, but that's exactly half of it here So the two of them can cancel in that situation So then as you're integrating from negative pi halves to pi halves Foil this thing out. You're gonna end up with a 9 plus a 12 sine theta plus 4 sine squared Theta Which I should mention that as you're integrating from negative pi halves to pi halves That's a symmetric interval. So if you integrate at if you integrate at odd function across the symmetric interval It's just gonna cancel out So this sign is actually gonna cancel out as an odd function And then everything else that didn't cancel out the 9 and the 4 sine squared is actually an even function So this actually doubles up the area again So I can rewrite this as 2 times the integral from 0 to pi halves. That's kind of nice I'm gonna have to apply the half angle identity on this 4 sine squared there So which case we're gonna get a 9 plus 4 times 1 half Which will give us a 2 and then you're gonna get 1 minus cosine of 2 theta D theta so then we get the 9 plus 2 is gonna give us 11 11 Minus 2 times cosine of 2 theta d theta. We're now ready to integrate this thing So doing this down below if we take the anti-derivative here 11 becomes 11 theta And then we're going to get a minus Sine of 2 theta as you go from 0 to pi halves Notice that when you plug in 0, this will disappear. So we'll sign When you plug in the pi halves, you're gonna get sine of pi, which is also 0 so that disappears entirely So this thing that's gonna become 2 times 11 times pi halves The 2's cancel and we see that the area here is gonna be 11 pi and so the correct choice was option A