Could someone help me to evaluate $$\int_C (x+y)^2dx - (x^2+y^2)dy$$ where C is the positively oriented triangle with vertices at (1,1), (3,2) and (2,5)?
I tried to use Green's theorem to solve it and I got $\iint_D (-4x+2y) dA$ and I have no idea that what should I do next step.
Thanks in advance.
First of all, your integrand is not correct.
$Pdx + Qdy = (x+y)^2dx - (x^2+y^2)dy$
So, $P = (x+y)^2, Q = -(x^2+y^2)$
$\frac{\partial Q}{\partial x} = -2x, \frac{\partial P}{\partial y} = 2x + 2y$
$\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = -4x - 2y$
Now your region is positively oriented triangle with vertices $(1, 1), (3,2)$ and $(2, 5)$. Here is a sketch of your region -
Taking two points at a time, you can find the equations of lines that form this triangle. It is clear from the sketch that you have to either do a change of variable or just split your integral into two parts.
For integrating first over $y$ and then over $x$, your bounds for two integrals become $\frac{x + 1} {2} \leq y \leq (4x-3), 1 \leq x \leq 2$ and $\frac{x + 1} {2} \leq y \leq (11-3x), 2 \leq x \leq 3$