polar area intergral check

[kpx001]

inside r = 3sin(theta) and outside r = 2 - sin(theta). intergral 0 to pi/2 (3sin(theta))^2 - 2*intergra pi/6 to pi/2 (2 - sin(theta))^2 is this intergral correct to solve this problem?

 

[skeeter]

$\displaystyle 3\sin{\theta} = 2 - \sin{\theta}$

$\displaystyle \sin{\theta} = \frac{1}{2}$

$\displaystyle \theta = \frac{\pi}{6}$
$\displaystyle \theta = \frac{5\pi}{6}$


$\displaystyle A = \frac{1}{2} \int_{\frac{\pi}{6}}^{\frac{5\pi}{6}} (3\sin{\theta})^2 - (2 - \sin{\theta})^2 \, d\theta$

 

[kpx001]

i see how ur way would work, but is my way incorrect? im just doing what makes more sense for me.

 

[galactus]

No, your set up isn't quite correct. Follow skeeter's lead. You can see from the graph.

You could also use $\displaystyle \int_{\frac{\pi}{6}}^{\frac{\pi}{2}}\left[(3sin(t))^{2}-(2-sin(t))^{2}\right]$