Let C be the part of the plane curve defined by $$y^2=x^3-x$$ between $$(\frac{-1}{\sqrt3},\sqrt[4]{\frac4{27}})$$ and $$(0,0)$$ oriented from left to right. How would I calculate
$$\int_{C}y^2\vec{i}+(2xy+4y^3e^{y^4})\vec{j}ds$$
I have already found that the vector field is conservative, I'm just not sure how to proceed from there.
On
${\bf F}(x,y)=(y^2,2xy+4y^3e^{y^4})$ is the vector field. If ${\bf F}$ is conservative then ${\bf F}=\nabla G$ for some scalar field $G(x,y)$. And, $$ \int_C {\bf F}\cdot d{\bf r} = \int_a^b {\bf F}({\bf r}(t))\cdot {\bf r}'(t)dt =G({\bf r}(b))-G({\bf r}(a)) $$ where ${\bf r}(t)$ is any parametrisation of the curve $C$. We have the endpoints ${\bf r}(b)$ and ${\bf r}(a)$ so we only need to find $G$.
${\bf F}=\nabla G$ means that \begin{eqnarray*} \frac{\partial G}{\partial x}&=&y^2 \qquad and \qquad \frac{\partial G}{\partial y}=2xy+4y^3e^{y^4}\\ \implies G&=&xy^2+f(y)+C\qquad and \qquad G=xy^2+e^{y^4}+g(x)+C \end{eqnarray*} for some functions $f(y)$ and $g(x)$. So, take $f(y)=e^{y^4}$ and $g(x)=0$ and we have $G(x,y)=xy^2+e^{y^4}+C$.
Now, $$ \int_C {\bf F}\cdot d{\bf r} = G({\bf r}(b))-G({\bf r}(a)) = G(0,0)-G\left(-\frac{1}{\sqrt{3}},\sqrt{\frac{4}{27}}\right) $$
Since the vector field is conservative it has a potential function. Check that the gradient of
$$f(x,y) = xy^2 + e^{y^4}$$
is indeed the vector field in the problem. Then use the fundamental theorem of line integrals:
$$\int_{\vec{a}}^{\vec{b}}\nabla f \cdot d\vec{r} = f(\vec{b}) - f(\vec{a})$$
to get that the integral equals
$$f(0,0) - f\left(-\frac{1}{\sqrt{3}},\sqrt[4]{\frac{4}{27}}\right) = \frac{11}{9} - e^{\frac{4}{27}} $$