Use greens theorem to find work done 
Use Green's Theorem to find the work done by the force $\mathbf{F}(x,y)=x(x+y)\mathbf{i}+xy^2\mathbf{j}$ in moving a particle from the origin along the $x$-axis to $(1,0)$, then along the line segment to $(0,1)$, and back to the origin along the $y$-axis.

So I was able to find $\frac{\partial Q}{\partial x} -\frac{\partial P}{\partial y}$ to be $y^2 -x$ and I integrated that with respect to $y$ and $x$ by using $y= 1-x$ as my upper bound and $y=0$ as my lower bound, and $0 < x < 1$ for my $x$ integral. but it came out to $-\frac{7}{36}$, and the answer is $-\frac1{12}$. I'm not sure if I'm doing something fundamentally wrong here or if its a calculation error. I checked it twice. How do I do it correctly?
 A: Yes, you are correct, by Green's Theorem, you should evaluate
$$\int_{x=0}^1\int_{y=0}^{1-x}(y^2-x)dydx=\int_{x=0}^1\left[\frac{y^3}{3}-xy\right]_{y=0}^{1-x}dx=\int_{x=0}^1\left(\frac{(1-x)^3}{3}-x(1-x)\right)dx\\
=\frac{1}{3}\int_{t=0}^1t^3 dt-\int_{x=0}^1 x(1-x)dx.$$
Can you take it from here?
A: \begin{align}\int_0^1 \int_0^{1-x} (y^2-x) \, dy\, dx &= \int_0^1 \left[ \frac{y^3}{3}-xy \right]_{y=0}^{y=1-x}\, dx \\
&=\int_0^1 \frac{(1-x)^3}{3}-x(1-x) \, dx \\
&=  \int_0^1 \frac{(1-x)^3}{3} - x+x^2 \, dx \\
&= \left[ -\frac{(1-x)^4}{12}-\frac{x^2}{2}+\frac{x^3}{3}\right]_{x=0}^{x=1} \\
&=  \left[ -\frac12+\frac13+\frac1{12} \right]\\
&=\left[\frac{-6+4+1}{12} \right]\\
&=-\frac1{12}\end{align}
Your method seems correct but most likely you make a careless mistake somewhere.
A: $$\int_{y=0}^1\int_{x=0}^{1-y}(y^2-x)dxdy = \int_{x=0}^1(y^2x-\frac{x^2}{2})|_{0}^{(1-y)}dy$$
If you expand you get
$$=\frac{1}{2}\int_{y=0}^1(-2y^3+y^2+2y-1)dy$$
$$=\frac{1}{2}\left(\frac{-2y^4}{4}+\frac{y^3}{3}+\frac{2y^2}{2}-y\right)|_{0}^{1}$$
$$=-\frac{1}{12}$$
