## area

Find the area of the region bounded by: $$r= 6-2sin(\theta)$$

here's what i did:

$$6-2sin(\theta) = 0$$
$$sin(\theta) = 1/3$$

so the bounds are from arcsin(-1/3) to arcsin(1/3) right?

my integral:
$$\int_{-.339}^{.339} 1/2*(6-2sin(\theta))^2$$

i get a answer of 0.6851040673*10^11, and it's wrong. all my steps seems to be correct, i cant figure out the problem.
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 What is the answer that you should have got ??? marlon
 $$r= 2sin(\theta)$$ is an ellipse so $$r= 6-2sin(\theta)$$ is just shifting and stretching it. Therefore the bounds on $$\theta$$ are $$0 \leq \theta \leq 2 \pi$$

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## area

I agree.It's a shifted & stretched ellipse.Pay attention with the numbers...You can't get a big value for the area.It's ~100...

Daniel.

$38\pi$ to be exact.

 Quote by asrodan $$r= 2sin(\theta)$$ is an ellipse so $$r= 6-2sin(\theta)$$ is just shifting and stretching it. Therefore the bounds on $$\theta$$ are $$0 \leq \theta \leq 2 \pi$$
can you explain it to me agian? i dont really understand it that well. are you saying if it's an ellipse, the bounds will always be from 0 ->2pi?
 Blog Entries: 9 Recognitions: Homework Help Science Advisor Yes,it's like for a circle,or for any closed curve enclosing the origin inside it... Daniel.
 ah, i get it now. thank you

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