For $F=yx^3\hat{i}+y^2\hat{j}$ compute the line integral $\int_C F\cdot dr$ where $C$ is the curve $y = x^2$ for $0\le x\le 1$.
Can this be solved with Green's Theorem? I am aware of the method where everything is written in terms of x, but it seems like Green's theorem should work?
On
Suppose we close the curve $C$ with the line segment $C'$ joining $(1,1)$ to $(0,0)$, parameterized by
$$\hat r(t) = (1-t)\,\hat\imath + (1-t)\,\hat\jmath, \qquad t\in[0,1]$$
By Green's theorem, the integral over the closed path with interior $D$ is
$$\color{green}{I} = \oint_{C\cup C'} \left(yx^3\,\hat\imath + y^2\,\hat\jmath\right)\cdot d\hat r = \iint_D \frac{\partial(y^2)}{\partial x} - \frac{\partial (yx^3)}{\partial y} \, dx\,dy = \int_0^1 \int_{x^2}^x -x^3 \, dy \, dx$$
To recover the line integral over $C$, we subtract from $\color{green}{I}$ the contribution of the integral along $C'$, which is
$$\int_{C'} \left(yx^3\,\hat\imath + y^2\,\hat\jmath\right) \cdot d\hat r = \int_0^1 \left((1-t)^4\,\hat\imath+(1-t)^2\,\hat\jmath\right) \cdot \left(-\hat\imath-\hat\jmath\right) \, dt = - \int_0^1 (1-t)^4+(1-t)^2 \, dt$$
Green's theorem is not really useful here because the curve $y = x^2$ for $0\le x\le 1$ is not closed. If you really wanted to, you could extend the curve $C$ to a piecewise smooth closed curve and then apply Green's theorem; however, you would then still need to calculate separate line integrals and subtract them from the double integral obtained from Green's theorem in order to figure out the original line integral. Try it out for fun if you want to get some extra practice!