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.
Best Answer
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 = $-{1 \over 12}$