Question: Test Green's Formula in the case where $M = x^2 - y^2$ and $N=2xy$ and $\Omega$ is the triangle with vertices $(0, 0), (1,1), (2, 0)$.
My attempt: $$\iint_\Omega \Big(\dfrac{\partial M}{\partial x} - \dfrac{\partial N}{\partial y} \Big) \,dx\,dy = \iint_\Omega (2x - 2x) \,dx\,dy = 0,$$ So must the following integral be zero as well: $$\int_\Gamma (x^2-y^2) \,dy + 2xy \,dx = 0+ \int_{\frac13}^{\frac23} (-4+12t)(-3) \,dt + 0 + \int_{0}^{\frac13} 2(9t^2)(3) \,dt + \int_{\frac13}^{\frac23} 3(2)(3t)(2-3t) \,dt +0= \dfrac{84}{9} \ne 0 \ !!$$ Where the first three integral is for ... dy and the last three is for ... dx. Where am I doing wrong?
I used $(3t,3t)$ for from $(0,0)$ to $(1,1)$; $(3t,2-3t)$ for from $(1,1)$ to $(2,0)$; and, $(6-6t,0)$ for from $(2,0)$ to $(0,0)$.
By the book the Green's Formula is $$\int_\Gamma (Mdy+Ndx)=\iint_{\Omega}\left(\frac{\partial M}{\partial x}-\frac{\partial N}{\partial y}\right)dx\,dy.$$
I calculated the line integrals clockwise but the difference is a minus in final result; so no big deal!
You can parametrize the boundary of your triangle by $\gamma_1(t) =(2t,0)$, $\gamma_2(t)=(2-t,t)$ and $\gamma_3(t) = (1-t,1-t)$ over $[0,1]$. Then the pullback of $\omega$ under these three functions are
$$\omega_1 = 0,\quad \omega_2 = 2(t^2-4t+2)dt,\quad \omega_3= -2(t-1)^2 dt.$$
But note that $\omega_2 = 2((t-1)^2 - (2t-1))$ so that the total pullback is the linear function $\omega_1+\omega_2+\omega_3 = -2(2t-1)$ which integrates to $0$ over $[0,1]$.