Question

What do I do when I get a negative are under a polar region?

Answer 1

This question: Find the area of the region inside\[r=3sin theta\] and outside \[r=1+\sin \theta\] I solved for the area using \[\frac{ A }{ 2 } = \frac{ 1 }{ 2 }(\int\limits_{\pi/6}^{\pi/2}(3\sin \theta)^2 d \theta - \int\limits_{\pi/6}^{\pi/2}(`+\sin \theta)^2 d \theta\]
I made use of symmetry to shorten my limits from pi/6 -> 5pi/6 to two pi/6 -> pi/2
My final answer, however, was -pi. Could it be that i'm mixing up the subtraction?

Answer 2

https://www.desmos.com/calculator/g1cijxxjy8
Here's an illustration of the situation.
RED is the inside, BLUE is the outside, the 2 lines are the limiters, and they're symmetric about the y axis.

Answer 3

http://www.wolframalpha.com/input/?i=integrate+%283+sin%28x%29%29%5E2+-+%281%2Bsin%28x%29%29%5E2+dx+from+pi%2F6+to+pi%2F2

Answer 4

So my computation is just...wrong? Alright. I'll check, thanks.

Answer 5

I simply subtracted limiter a from limiter b in both somehow instead of b from a... Well thanks!