Using a double integral, find the area inside the petals of the rose curve $r=2\sin2\theta$ and outside the circle $r=1$.
I have done work on this and am thinking that the period would be pi, therefore that being the upper limit of integration. I am guessing that the lower one is $0$.
I ended up getting the answer $\pi$. That was for the rose curve part. Does anyone know where to start with the circle part? Is my lower limit of integration correct? I started my integral by using only the letter r.
The curve $r=2\sin2\theta$ is symmetrical by the $x$-axis and the $y$-axis. Therefore, the area you want to obtain is $8A$, since the area cut by $x$-axis and $y=x$ as $A$.
Then let's get the value of $A$. Since the first quadrant is $0\le\theta\le\frac\pi4$. and the solution of $1\le2\sin2\theta$ is $\frac\pi{12}\le\theta\le\frac{\pi}{4}$.
So $A=\int_{\pi/12}^{\pi/4}\frac12(r_1^2-r_2^2)d\theta$.($r_1=2\sin2\theta, r_2=1$). Then...
$$ A=\int_{\frac\pi{12}}^{\frac\pi4}\frac12\left((2\sin2\theta)^2-1^2\right)d\theta =\frac12\int_{\frac\pi{12}}^{\frac\pi4}(4\sin^22\theta-1)d\theta =\frac12\int_{\frac\pi{12}}^{\frac\pi4}(2(1-\cos4\theta)-1)d\theta\\ =\frac12\int_{\frac\pi{12}}^{\frac\pi4}(1-2\cos4\theta)d\theta =\left[\theta-\frac{\sin4\theta}2\right]_{\frac\pi{12}}^{\frac\pi4} =\frac{2\pi+3\sqrt3}{12}. $$
So, the answer is $8A=\frac{4\pi+6\sqrt3}3$.