I want to evaluate $$\int_R(-x^2y \ dx + xy^2 \ dy)$$ where R is the closed curve of the semicircle $x^2 + y^2 = a^2, y>0$ along the segment $(-a,a)$ of the x-axis, by first expressing it in the form of $$\int\int_D\frac{\partial Q}{\partial X} - \frac{\partial P}{\partial y}dx\ dy = \int\int y^2 - x^2 dx \ dy$$
I have differentiated the equation above to get the form below, and I believe the limits should be :$$\int^a_{-a} \int^{\sqrt{a^2-y^2}}_0 y^2 + x^2 dy \ dx$$
Is this the correct way? But If I integrate in this manner, I get an expression like $$[\frac{y^3}{3} + x^2y]$$ which when I substitute the limits in, I cannot integrate further. Is this the right way to do it?
Here is what I get after continuing the integration:
$$\int^a_{-a} (a^2-x^2)^\frac{1}{2} (\frac{a^2-x^2}{3} + x^2) \ dx$$
This seems not possible to be integrated because even after doing integration by-parts, there will always be that power of half that I cannot remove....
By Green's theorem $$ \oint_{R}-x^2 y\,dx + xy^2\,dy = \iint_{D}\left(x^2+y^2\right)\,dx\,dy $$ where $D$ is the region given by $x^2+y^2\leq a^2$ and $y\geq 0$.
By switching to polar coordinates the RHS turns into $$ \int_{0}^{\pi}\int_{0}^{a}\rho^3\,d\rho\,d\theta=\frac{\pi}{4}a^4.$$