calculus - Double integral conversions to Polar

Problem 1

Convert to polar form and solve

$$\int^{2}_{0}\int_{0}^{\sqrt{2x-x^2}}((x-1)^2+y^2)^{5/2}\text{ dy dx}$$

$$x^2+y^2=2x, r^2=2rcos\theta , r=2cos\theta$$

$$\int^{\pi/2}_{0}\int^{2\cos\theta}_{0}(r^2-2(rcos\theta)+1)^{5/2}rdrd\theta$$

Problem 2

Convert to polar form and solve

$$\int^{1}_{-1}\int_{0}^{\sqrt{3+2y-y^2}}cos(x^2+(y-1)^2)\text{ dx dy}$$

$$x^2=3+2y-y^2 --> x^2+y^2 = 3+2y -->r^2 =3+2rsin\theta$$

Is the first problem the right integration to follow in polar?
Im stuck at number 2 in finding the bounds..

Answer

In the case of problem 1, the natural substitution is $x=1+r\cos\theta$ and $y=r\sin\theta$, thereby getting$$\int_0^\pi\int_0^1r\times r^5\,\mathrm dr\,\mathrm d\theta.$$
The same idea applies to problem 2: do $x=r\cos\theta$ and $y=1+r\sin\theta$.