Find the area of the region bounded

[ILoveBaseball]

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 can't figure out the problem.

 

[marlon]

What is the answer that you should have got ?

marlon

 

[asrodan]

$$r= 2sin(\theta)$$ is an ellipse so $$r= 6-2sin(\theta)$$ is just shifting and stretching it.

Therefore the bounds on $$\theta$$ are $$0 \leq \theta \leq 2 \pi$$

 

[dextercioby]

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.

 

[ILoveBaseball]

$$r= 2sin(\theta)$$ 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 don't really understand it that well. are you saying if it's an ellipse, the bounds will always be from 0 ->2pi?

 

[dextercioby]

Yes,it's like for a circle,or for any closed curve enclosing the origin inside it...

Daniel.

 

[ILoveBaseball]

ah, i get it now. thank you