Question

Find the area inside the polar circle r=-4sin(t) and outside the circle r=2.

Answer 1

I know I have to use the following equation, but I'm not really sure how to find a and b.

\[\int\limits_{a}^{b}\frac{ 1 }{ 2 }(f(t)^2 - g(t)^2)\]

Answer 2

\[\int\limits_{a}^{b}\frac{ 1 }{ 2 } ((-4\sin \theta)^2-2^2)d \theta\]
=\[\int\limits_{a}^{b}\frac{ 1 }{ 2 } ((16\sin ^2 \theta)-4)d \theta\]

Answer 3

=\[ -2 \theta + \int\limits_{a}^{b}\frac{ 1 }{ 2 } (16\sin^2 \theta)d \theta\]

Answer 4

=\[ -2 \theta + \int\limits_{a}^{b}\frac{ 1 }{ 2 } (16 \frac{(1-cos2 \theta)}{2})d \theta\]

Answer 5

Well, a and b are what it takes to go around the circle once, amiright?

http://www.wolframalpha.com/input/?i=polar+graph+r%3D-4sin%28t%29

So, I suppose two possible a and b values are 0 and 2pi.

Answer 6

equate the two circles to get the points of intersection
\[-4\sin t = 2\]

Answer 7

\[\sin t = -1/2\]\[t=7\pi/6,11\pi/6\]

Answer 8

Oh, whoops, I misinterpreted the problem. Ignore me.

http://www.wolframalpha.com/input/?i=plot+polar+r%3D-4sin%28t%29%2C+r%3D2

Answer 9

Hm. I'll give that a shot - thank you.

Answer 10

\[A=\int\limits_{7\pi/6}^{11\pi/6}\frac{ 1 }{ 2 }\left[ \left( -4 \sin t \right)^2-\left( 2 \right)^2\right] dt\]

Answer 11

So in the long run I keep coming up with the following as my solved integral, but I think I'm doing something wrong somewhere. I"m really excellent at making basic algebra errors, so I think I'm probably dropping a 1/2 somewhere...

\[2\theta - 2\sin2\theta \]

Answer 12

\[16\sin^2 \theta - 4=8(2\sin^2 \theta)-4=8(1- \cos 2\theta)-4=4-4\cos 2\theta\]

Answer 13

I figured it out! Thanks so much for all your patience.

Answer 14

yw.