# Thread: Area in Polar Coordinates

1. ## Area in Polar Coordinates

So, I've tried this problem about 10 times, and I'm not quite sure what I'm doing wrong.

Find the area inside one leaf of the rose
r= 4sin(5θ)

I know the graph looks like a 5 leaf rose.
I used the equation A= integral (1/2r^2)dθ

Maybe I'm bounding my integral wrong?? I have it from 0 to pi/5
Or maybe my antiderivative is wrong??? I have 4θ - (4/10)(sin10θ)

Any help would be greatly appreciated! This question is really starting to frustrate me!

2. As you done, you can use the limits 0 to Pi/5, then since there is symmetry, you can multiply by 5.

$\frac{5}{2}\int_{0}^{\frac{\pi}{5}}\left(4sin(5\th eta)\right)^{2}d{\theta}=40\int_{0}^{\frac{\pi}{5} }sin^{2}(5\theta)d{\theta}$

When integrating a $sin^{2}({\theta})$, a good identity to use is

$sin^{2}(\theta)=\frac{1}{2}(1-cos(2{\theta}))$.

$=20\int_{0}^{\frac{\pi}{5}}\left(1-cos(10\theta)\right)d{\theta}$

$=20{\theta}-2sin(10{\theta})$

$20(\frac{\pi}{5})-2sin(10(\frac{\pi}{5}))=4{\pi}$