Question

I'm trying to determine what the limits of integration ought to be for r=8-4sin(θ). Since it's a cardioid, I *think* that I could say from (pi/2) to (3pi/2), and then double that integral to get the area bounded by the function. Is that the right approach?

Answer 1

Lol I was trying to draw a picture of it.
But failing miserably.

Yah I think you've got the right approach.
Personally I would go with -pi/2 to pi/2 for the bounds
But I guess it doesn't really matter.

Answer 2

Ok, so I have \[r = 8=4\sin(\theta)\] and \[A = \frac{1}{2}\int\limits_{\frac{-\pi}{2}}^{\frac{\pi}{2}}[8-4\sin(\theta)]^{2} d\theta\]

Answer 3

Is there a formula for Area in polar? I always forget it.
I always need to start from here:\[\Large\rm \int\limits_{\theta}~\int\limits_{r=0}^{8-4\sin \theta}r~dr~d \theta\]Integrating r gives us:\[\Large\rm \frac{1}{2}\int\limits_{\theta}r^2|_{r=0}^{8-4\sin \theta}d \theta\]And then plugging in the bounds and stuff.. ah yes!
Mmmm ok great. Your setup looks good so far!

Answer 4

Cool, let's see if I get it this time...

Answer 5

Actually, I hadn't seen that formula for polar area, but it looks a lot simpler than the way I was going. I'll try both.

Answer 6

Ugh. I got 36pi, but that doesn't seem to be it.

Answer 7

Ha! I forgot to multiply by two at the end! I think that may have been all I was missing to begin with. Thanks for your help.

Answer 8

Ohhh the doulbing thing! Ah yes!

Answer 9

There was a small little mistake early on actually.
Expanding out the square should have given you -4sin(theta) for the middle term.
Since that ended up zero'ing out, it didn't matter though :)

Answer 10

Cool, one assignment to go, and then time to review for the final. :D

Answer 11

One *problem*, I mean.

Answer 12

oo nice.
Calc 2 or something?

Answer 13

Yeah, I'm working on a physics degree.