KendraSan
Nov15-11, 10:50 PM
Find the area between a large loop and the enclosed small loop of the given curve.
r = 4 + 8cos(3θ)
ok so i did 2* ∫ 1/2 r^2 dθ from 0 to 2∏/3 and then subtracted the same integrand from 2∏/3 to ∏. This is what I've been taught to do; however I keep getting 16∏, and the answer is supposed to be 16(∏/3+sqrt(3)).
My steps:
2*(1/2) [(∫ from 0 to 2∏/3) 16+64cos(3θ)+64cos^2(3θ) - (∫from 2∏/3 to ∏) 16+64cos(3θ)+64cos^2(3θ)]
[(∫ from 0 to 2∏/3) 16+64cos(3θ)+32+32cos(6θ) - (∫from 2∏/3 to ∏) 16+64cos(3θ)+32+32cos(6θ)]
(∫ from 0 to 2∏/3) 48+64cos(3θ)+32cos(6θ) - (∫from 2∏/3 to ∏) 48+64cos(3θ)+32cos(6θ)
(∫ from 0 to 2∏/3) 16(3+4cos(3θ)+2cos(6θ)) - (∫from 2∏/3 to ∏) 16(3+4cos(3θ)+2cos(6θ))
which is:
[16(3θ + 4sin(3θ)/3 + 2sin(6θ)/6) from 0 to 2∏/3] -[16(3θ + 4sin(3θ)/3 + 2sin(6θ)/6) from 2∏/3 to ∏]
and then for my final answer i get 16∏
where did i go wrong?
do i have the wrong bounds?
r = 4 + 8cos(3θ)
ok so i did 2* ∫ 1/2 r^2 dθ from 0 to 2∏/3 and then subtracted the same integrand from 2∏/3 to ∏. This is what I've been taught to do; however I keep getting 16∏, and the answer is supposed to be 16(∏/3+sqrt(3)).
My steps:
2*(1/2) [(∫ from 0 to 2∏/3) 16+64cos(3θ)+64cos^2(3θ) - (∫from 2∏/3 to ∏) 16+64cos(3θ)+64cos^2(3θ)]
[(∫ from 0 to 2∏/3) 16+64cos(3θ)+32+32cos(6θ) - (∫from 2∏/3 to ∏) 16+64cos(3θ)+32+32cos(6θ)]
(∫ from 0 to 2∏/3) 48+64cos(3θ)+32cos(6θ) - (∫from 2∏/3 to ∏) 48+64cos(3θ)+32cos(6θ)
(∫ from 0 to 2∏/3) 16(3+4cos(3θ)+2cos(6θ)) - (∫from 2∏/3 to ∏) 16(3+4cos(3θ)+2cos(6θ))
which is:
[16(3θ + 4sin(3θ)/3 + 2sin(6θ)/6) from 0 to 2∏/3] -[16(3θ + 4sin(3θ)/3 + 2sin(6θ)/6) from 2∏/3 to ∏]
and then for my final answer i get 16∏
where did i go wrong?
do i have the wrong bounds?