Find the area of the region bounded by: [tex]r= 6-2sin(\theta)[/tex] here's what i did: [tex] 6-2sin(\theta) = 0[/tex] [tex] sin(\theta) = 1/3[/tex] so the bounds are from arcsin(-1/3) to arcsin(1/3) right? my integral: [tex]\int_{-.339}^{.339} 1/2*(6-2sin(\theta))^2[/tex] 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.
[tex]r= 2sin(\theta)[/tex] is an ellipse so [tex]r= 6-2sin(\theta)[/tex] is just shifting and stretching it. Therefore the bounds on [tex] \theta [/tex] are [tex] 0 \leq \theta \leq 2 \pi [/tex]
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. [itex] 38\pi [/itex] to be exact.
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?