# Convert integration to polar and solve

Evaluate the iterated integral $$\int_{-1}^1\int_0^{\sqrt{3+2y-y^2}}\cos\left(x^2+(y-1)^2\right)\,dy\,dx$$

Confused on how $$y=0$$ and $$y=\sqrt{3+2y-y^2}$$. Is this a typo or am I missing something?

• Use $x=r\cos(\theta)$ and $y=1+r\sin(\theta)$ – DINEDINE Mar 16 '19 at 21:13
• It looks to me that they switched the $dy$ and $dx$ when they typed the question. – user458276 Mar 16 '19 at 21:19
• ok so it should be x = 0, x = sqrt(3+2y−y^2) ? – MasterYoshi Mar 16 '19 at 21:22

What are you confused about? As far as I can see the limits for the integral wrt $$y,$$ being $$0$$ and $$\frac {1+\sqrt 7}{2}$$ from the given conditions, are consistent with everything else.

• Im a bit lost on how to find the theta and r. i cant seem to solve the equation by substituting in $rcostheta$ and $rsintheta$ in $x^2=3+2y-y^2$ – MasterYoshi Mar 16 '19 at 22:25