# Find the work done by the force field $F(x,y)=(x^2+xy)\bar{i}+(xy^2)\bar{j}$ along a path

Find the work done by the force field $$F(x,y)=(x^2+xy)\bar{i}+(xy^2)\bar{j}$$ when a particle moves from the origin, along the $$x$$ axis to the $$(1,0)$$, then on the line segment that joins the $$(1,0)$$ with the $$(0,1)$$ and finally returns along the $$y$$ axis to origin.

I have thought to do the following:

By Green's theorem we have to

$$\int_CF\cdot rdr=\int\int_D(y^2-x)dA=\int_{0}^{1}\int_{0}^{1}(y^2-x)dxdy=\int_{0}^{1}[y^2x-x^2/2]_{0}^{1}dy=\int_{0}^{1}(y^2-1/2)dy=y^3/3-y/2]_{0}^{1}=1/3-1/2=-1/6$$

Is this fine? Thank you.

• You're integrating over a square; the integral over the triangle enclosed by the path is $\int_0^1 \int_0^{1 - x} (y^2 - x) dy dx$ (also, $F \cdot r dr$ should be $\mathbf F \cdot d\mathbf r$). – Maxim Nov 18 '18 at 19:59

Just perform the three integrals:

I: $$\int\limits_{x=0}^1 (x^2 + x y)\ dx = \int\limits_{x=0}^1 x^2\ dx = {1 \over 3}$$

where along the first path $$y=0$$.

II: $$\int\limits_{x=1}^0 x^2 + x y\ dx + \int\limits_{y=0}^1 x y^2\ dy$$

where along the second path $$y = 1-x$$ so

$$\int\limits_{x=1}^0 x^2 + x (1 - x)\ dx + \int\limits_{y=0}^1 (1-y) y^2\ dy = -{5 \over 12}$$

III: $$\int\limits_{y=1}^0 x y^2\ dy = 0$$

where along the third path $$x=0$$:

Total = $$-{1 \over 12}$$