area

[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 cant 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 dont 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