Compute the line integral $\int_{\gamma}F\cdot ds$, when $F(x,y) = (x,y^2)$ and $\gamma \subset \mathbf{R}^2$ is the curve of the parabola $x=3y^2$ starting at $(0,0)$ and ending at $(3,1)$.
I tried to parameterize $\gamma$ as follows $\gamma(t)=(3t^2,t)$ and then computing
$$\int_{0}^{3}F(\gamma(t)) \|\gamma'(t)\| \ dt = \int_{0}^{3}F(3t^2,t) \|(6t,1)\| \ dt = \int_{0}^{3}(3t^2,t^2) \cdot(36t^2,1) \ dt$$
but I'm quite certain this is not the correct integral. What might I be missing here?
The vector field $F(x,y) = (x,y^2)$ is conservative as it is gradient of,
$f(x,y) = \displaystyle (\frac{x^2}{2} + \frac{y^3}{3}) $
So your line integral is path independent and can simply be between the end points which are $(0, 0)$ and $(3, 1)$.
$\displaystyle \int_C \vec{F} \cdot dS = \int_C \nabla f \cdot dS = f(3,1) - f(0, 0)$
On your parametrization, it should be
$\displaystyle \int_{0}^{1} (3t^2,t^2) \cdot (6t,1) \ dt = \int_{0}^{1} (18t^3 + t^2) \ dt = \frac{29}{6}$