# Thread: Double Integration in Polar Coordinates

1. ## Double Integration in Polar Coordinates

Ok, so I have tried this problem like 4 times and have not gotten the correct answer yet. The instructions say, "Use a double integral to find the area of the shaded region."

11. r=2(1-cosθ)
Then there is a graph with the entire limacon filled in.

This is what I have done but I haven't gotten the correct answer.

∫∫rdrdθ

∫ (1/2 r^2 |(0 to 2(1-cosθ)) )dθ

∫ (1/2 *4 (sin^2(θ)))dθ

∫ (2 sin^2(θ))dθ

∫ (2*-1/2 (cos(2θ)-1))dθ

∫ (-cos(2θ) + 1)dθ

u=2θ
du=2dθ

-1/2 ∫ (cosu)du + ∫1dθ

-1/2 (sin2θ) | (0 to 2π) + θ | (0 to 2π)

0 - 0 + 2π - 0 =

Why is this not correct? It is supposed to be 6π. Where have I made a mistake?

2. Originally Posted by dude15129
Ok, so I have tried this problem like 4 times and have not gotten the correct answer yet. The instructions say, "Use a double integral to find the area of the shaded region."

11. r=2(1-cosθ)
Then there is a graph with the entire limacon filled in.

This is what I have done but I haven't gotten the correct answer.

∫∫rdrdθ

∫ (1/2 r^2 |(0 to 2(1-cosθ)) )dθ

∫ (1/2 *4 (sin^2(θ)))dθ This is where your mistake is: $\displaystyle \color{red}(1-\cos\vartheta)^2\neq \sin^2\vartheta$!!!

∫ (2 sin^2(θ))dθ

∫ (2*-1/2 (cos(2θ)-1))dθ

∫ (-cos(2θ) + 1)dθ

u=2θ
du=2dθ

-1/2 ∫ (cosu)du + ∫1dθ

-1/2 (sin2θ) | (0 to 2π) + θ | (0 to 2π)

0 - 0 + 2π - 0 =

Why is this not correct? It is supposed to be 6π. Where have I made a mistake?
--Chris