Confusion on double integral

Question

If i do closed integral($dxdy$) in Cartesian coordinates and try to find area of a circle by putting $y=\sqrt{(a^2-x^2)}$ then $dy$ becomes $y$ and then its closed integral($ydx$) and the double integral becomes $0$ for closed integral in cartesian coordinates but if I do this in polar coordinates I get the area of the circle. So whats happening here? Why do I get $0$ in Cartesian coordinates but I get the area of the circle in polar coordinates?

Answer 1

$$\int_{-a}^a \int_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}} 1 dy dx$$ is the integral you want to use. $1dy$ integrates to $y$ evaluated between limits, giving ${\sqrt{a^2-x^2}}--{\sqrt{a^2-x^2}} = 2{\sqrt{a^2-x^2}}$, so the new integral becomes $$\int_{-a}^{a}2{\sqrt{a^2-x^2}}dx$$