Hello, elizsimca!
Set up the integral to find the area shared by the polar curves given by:
. . ![r\:=\:\sin2\theta,\;\;r\:=\:\frac{1}{2},\;\;\left[0,\,\frac{\pi}{2}\right]](http://latex.codecogs.com/png.latex?r\:=\:\sin2\theta,\;\;r\:=\:\frac{1}{2},\;\;\left[0,\,\frac{\pi}{2}\right])
I found the points of intersection: . 
. . . . Good! The first curve is a 4-leaf rose curve.
The petals at one unit long and are aligned along the ±45° lines.
The second is a circle centered at the origin with radius ½.
We are in Quadrant 1.
We want the area of that petal which is inside the circle. Code:
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* * /
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|:::/:::::*
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- * - - - - * - - - -
| I'm afraid it takes three integrals . . .
![A \;=\;\frac{1}{2}\int^{\frac{\pi}{12}}_0 \sin^2\!2\theta\,d\theta + \frac{1}{2}\int^{\frac{5\pi}{12}}_{\frac{\pi}{12}} \left[\left(\frac{1}{2}\right)^2 - \sin^2\!2\theta\right]\,d\theta + \frac{1}{2}\int^{\frac{\pi}{2}}_{\frac{5\pi}{12}}\ sin^2\!2\theta\,d\theta](http://latex.codecogs.com/png.latex?A \;=\;\frac{1}{2}\int^{\frac{\pi}{12}}_0 \sin^2\!2\theta\,d\theta + \frac{1}{2}\int^{\frac{5\pi}{12}}_{\frac{\pi}{12}} \left[\left(\frac{1}{2}\right)^2 - \sin^2\!2\theta\right]\,d\theta + \frac{1}{2}\int^{\frac{\pi}{2}}_{\frac{5\pi}{12}}\ sin^2\!2\theta\,d\theta)
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Originally Posted by elizsimca 
