I need to find the work from $(0,0)\to(1,0)\to(1,1)$ of the following vector field:$\vec F(x,y)=xy^2\hat i+yx^2\hat j$
My attempt:
$$\oint_{c}\vec F d\vec r=\int_{(0,0)\to (1,0)}\bigg(xy^2\; dx +yx^2 \; dy\bigg)+\int_{(1,1)\nwarrow(1,0)}\bigg(xy^2\; dx+yx^2\; dy\bigg)$$
Is it correct so far? if yes so how can I proceed from this point?
METHOD 1:
We can simplify the problem by noting that $\nabla \times \vec F=0$. Therefore, $\vec F$ is therefore conservative on any connected domain. By Stokes' Theorem, the integral of $\vec F$ over any contour $C$ that bounds a connected domain $S$ is
$$\oint_C \vec F\cdot d\vec \ell =\int_S \nabla \times \vec F\cdot \hat ndS$$
Thus, the integral of interest is equal to
$$\int_{C_1+C_2}\vec F\cdot d\vec \ell=-\int_{(1,1)\,\text{to}\,(0,0)}(xy^2dx+x^2ydy)=\int_0^1 2t^3=\frac12$$
METHOD 2:
If we choose to evaluate the line integral directly, that is without deforming the contour as in Method 1, then we proceed as follows.
$$\int_{C_1+C_2}\vec F\cdot d\vec \ell=\int_{C_1}\vec F\cdot d\vec \ell+\int_{C_2}\vec F\cdot d\vec \ell$$
where
$$\int_{C_1}\vec F\cdot d\vec \ell=\int_0^1 (x(0)^2)dx =0 \tag 1$$
and
$$\int_{C_2}\vec F\cdot d\vec \ell=\int_0^1 (1)^2y\,dy=\frac12 \tag 2$$
Putting together $(1)$ and $(2)$ we obtain
$$\int_{C_1+C_2}\vec F\cdot d\vec \ell=\frac12$$
as expected!