# 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?

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?

• well apparently not, I tried and got it wrong. but at least I got the method right. I did end up with that integral on the right side though. – 2316354654 Dec 5 '17 at 7:44
• In the first integral, you may let $t=1-x$. – Robert Z Dec 5 '17 at 7:48
• what I did was expand everything before integrating and ended up wtih 3 variables. It was probably a simple arithmetic error – 2316354654 Dec 5 '17 at 7:50
• Yes, expanding the cube could be the source of your error. – Robert Z Dec 5 '17 at 7:56
• i just found my arithmetic error. now i got -1/12. you were right, it was the cube expansion. – 2316354654 Dec 5 '17 at 7:59

\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.

$$\int_{y=0}^1\int_{x=0}^{1-y}(y^2-x)dxdy = \int_{x=0}^1(y^2x-\frac{x^2}{2})|_{0}^{(1-y)}dy$$

If you expand you get $$=\frac{1}{2}\int_{y=0}^1(-2y^3+y^2+2y-1)dy$$

$$=\frac{1}{2}\left(\frac{-2y^4}{4}+\frac{y^3}{3}+\frac{2y^2}{2}-y\right)|_{0}^{1}$$

$$=-\frac{1}{12}$$