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.

  • $\begingroup$ 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 $\mathbf F \cdot d\mathbf r$). $\endgroup$ – 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 = $-{1 \over 12}$


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