- #1
ILoveBaseball
- 30
- 0
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 can't figure out the problem.
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 can't figure out the problem.