I'm trying to compute the line integral $$ \int_{C(0,4)^+} (x^2+y^3)dx + (y^4+x^3)dy, $$ where $C(0,4)$ is the circle, positively oriented, with center 0 and radius 4.
So far, I've tried using Stokes theorem: $\int_{\partial M} \omega = \int_M d\omega$. So $C(0,4)$ is the boundary of the closed ball $B(0,4)$. We compute: $$ d((x^2+y^3)dx + (y^4+x^3)dy)) = d(x^2+y^3) \wedge dx + d(y^4+x^3) \wedge dx = (3x^2-3y^2) dx \wedge dy $$ So we need to compute the integral $$ \int_{B(0,4)^+} (3x^2-3y^2) dx \wedge dy. $$ Here is where I'm stuck. I tried to use a coordinate patch $\alpha : (0,2\pi) \times (0,4) \to B(0,4) : (\theta,r) \mapsto (r \cos \theta, r \sin \theta)$, and compute $\alpha^\ast(\omega)$, but I'm not sure whether I'm on the right track since the expressions become rather complicated and nothing cancels out since we have $x^2-y^2$, and for polar coordinates $x^2+y^2$ would be convenient. Any tips?
I think this is an application of Green's Theorem, which is a specific case of Stokes' Theorem. Namely, $$ \int_{\partial D} P \mathrm{d}x + Q \mathrm{d}y = \iint_{D} \frac{\partial Q}{ \partial x } - \frac{\partial P}{ \partial y } \mathrm{d}S $$ In your case, $$ \int_{C(0,4)^{+}} (x^{2} + y^{3}) \mathrm{d}x + (y^{4} + x^{3}) \mathrm{d}y = \iint_{D(0,4)} 3x^{2} - 3y^{2} \mathrm{d}S $$ Then using the polar coordinates (and multiplying by the Jacobian Term), we get $$ \iint_{D(0,4)} 3x^{2} - 3y^{2} \mathrm{d}S = \int_{0}^{2\pi} \int_{0}^{4} 3r^{3}(\cos^{2}(\theta) - \sin^{2}(\theta)) \mathrm{d}r \mathrm{d}\theta = 192 \int_{0}^{2\pi} \cos^{2}(\theta) - \sin^{2}(\theta) \mathrm{d}\theta $$ $$ = 192 \int_{0}^{2\pi} \cos(2 \theta) \mathrm{d} \theta = 96 \sin(2\theta) \mid_{0}^{2\pi} = 0 $$