23

The definition of radian measure is given by $$\text{radian measure} = \frac{\text{arc length}}{\text{radius of circle}}\quad \text{or}\quad \theta = \frac{s}{r}.$$ We've traveled (arc length) $s=2\pi$ units along a circle with radius $r=6$. The radian measure is $$\theta = \frac{s}{r} = \frac{2\pi}{6} = \frac{\pi}{3}.$$

Note that $s = r\theta$. In particular, note that if the angle we traverse is a little angle $d\theta$, then the little distance traveled is $$ds = rd\theta.$$

Note that $dx = 3du+2dv$ and $dy=-1du+4dv$. We can write this as $$(dx,dy)=(3,-1)du+(2,4)dv.$$ The area of the parallelogram formed by the vectors $(3,-1)$ and $(2,4)$ is $$A=|12+2|=14.$$ The Jacobian of the transformation is 14, so areas in the $uv$-plane are multiplied by 14 to obtain areas in the $xy$-plane. In terms of integrals, we write $$A_{xy} = \iint_{R_{xy}}dxdy = \iint_{R_{uv}}14 dudv.$$ The 14 is our area stretch factor that appears when we change from one coordinate system to another.

Note that $dx = 2du+3dv$ and $dy=4du+5dv$. We can write this as $$(dx,dy)=(2,4)du+(3,5)dv.$$ The area of the parallelogram formed by the two vectors above is $A=|2\cdot 5-3\cdot 4|=2$. This means $$dA_{xy} = 2 dudv,$$ and so the area of the transformed region is twice the original area, hence $$A_{xy} = 5\cdot 2 = 10.$$

We'll draw it together. To find the point of intersection, we need $2-2\cos\theta = 2\cos\theta$ which means $2 = 4\cos\theta$ or $\cos\theta = \frac{1}{2}$. This occurs at $\theta = \pi/3$, which gives $r=2\cos(\pi/3) =1$.

Notice that rays starting from the origin, and heading out the curves, hit the cardioid for $0\leq \theta\leq \pi/3$, and then swap to hitting the circle for $\pi/3\leq \theta\leq \pi/2$. This means we'll need two different integrals. The solution is $$ \int_{0}^{\pi/3}\int_{0}^{2-2\cos\theta} r dr d\theta + \int_{\pi/3}^{\pi/2}\int_{0}^{2\cos\theta} r dr d\theta. $$