Line integral segment of parabola Suppose
$$
\vec{F} = \nabla f(x,y) = 6y \sin (xy) \vec{i} + 6x \sin (xy) \vec{j},
$$
and C is the segment of the parabola $y = 5 x^2$ from the point $(2,20)$ to $(6,180)$. Then, what is
$$\int_C \vec{F} \cdot d\vec{r}\ ?$$ 

I can not fnd the right answer, tried many ways to do it! could not get so far on this
Parameterize the curve $C = C(t)$, say letting $t \in [2,6]$ and then $x(t) = t$ and $y(t) = 5t^2$, so
$$
d\vec{r} = d \pmatrix{t \\ 5t^2} = \pmatrix{1 \\ 10t} dt
$$
So now compute
$$
\int_{t=2}^{t=6} 6\sin(xy) \pmatrix{y \\ x} \cdot \pmatrix{1 \\ 10t} dt,
$$
plugging in the correct $x = x(t), y=y(t)$ as  we defined above.

Can you take it from here?
Parameterize the curve $C = C(t)$, say letting $t \in [2,6]$ and then $x(t) = t$ and $y(t) = 5t^2$, so
$$
d\vec{r} = d \pmatrix{t \\ 5t^2} = \pmatrix{1 \\ 10t} dt
$$
So now compute
$$
\int_{t=2}^{t=6} 6\sin(xy) \pmatrix{y \\ x} \cdot \pmatrix{1 \\ 10t} dt,
$$
plugging in the correct $x = x(t), y=y(t)$ as  we defined above.

Can you take it from here?