The value of $\int_C (y^2\,dx + x^2\,dy)$, where $C$ is the triangle given by $x=0$, $x+y=1$, $y=0$ is

Using Green’s theorem:

$\oint_C (M\,dx+N\,dy)=\iint_D\left(\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}\right)dA$

Here $M=y^2,\; N=x^2$ so $\frac{\partial N}{\partial x}=2x,\; \frac{\partial M}{\partial y}=2y$

Integral $=\iint_D 2(x-y)\,dA$ over triangle with vertices $(0,0),(1,0),(0,1)$

By symmetry $\iint_D x\,dA=\iint_D y\,dA$ so $\iint_D (x-y)\,dA=0$

Hence value $=0$