I can't seem to get a value for the area between these two polar curves: 2+sin(theta) and 3sin(theta)

Area = .5 * Integral of (r^2 * dtheta)

Question

I can't seem to get a value for the area between these two polar curves: 2+sin(theta) and 3sin(theta)

Area = .5 * Integral of (r^2 * dtheta)

anonymous · Answer

I found the intersection at the top of both curves, so I used pi/2 and 5pi/2 for my bounds.

anonymous · Answer

I believe it's $$.5\int\limits_{}^{}topfnc^2-bottomfnc^2 d \theta$$

anonymous · Answer

yes, since there are two equations.  2+sin(theta) is on top, but when I simplify everything and take the integrals, I get 0?!?

abb0t · Answer

$$\sin(\theta)=$$

abb0t · Answer

-2/1

anonymous · Answer

.5 * Integral(4+2sin(theta)-8sin^2(theta) between pi/2 and 5pi/2.   Is this right?

anonymous · Answer

why did you just rearrange that?

abb0t · Answer

you have two circles. One that starts on the origin to 3 and another -1 to 3.

anonymous · Answer

I think it should be 
$$.5\int\limits_{}^{}4+4\sin \theta-8\sin \theta d \theta$$
I'm not certain on your bounds though. I'm not saying they're wrong, just that I don't know.

anonymous · Answer

oh it is 4sin(theta) thanks.

abb0t · Answer

based on the graph, you should see which is on top and which is on the bottom. You can graph them on wolfram alpha by putting r=2+sin(theta) and r=3sin(theta) and you will see

anonymous · Answer

to find the bounds, I set the equations = to each other (find the intersection)
3sinx = 2+ sinx
2sinx = 2
sinx = 1
x = pi/2