Let $F=(x,y):= (6x+2y^2, 4xy+3y^2)$. Calculate the line integral $\int_\gamma$, where $\gamma$ is the ellipse $x^2+2y^2=1$ that connects the points $(-1,0)$ and $(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}})$ (in the positive direction).
I have tried to parametrize as follows:
$x=cos(t)$ and $y=\frac{1}{\sqrt{2}}sin(t)$, with $\frac{\pi}{4} \leq t \leq \pi$.
..thinking that the point $(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}})$ must represent $t=\frac{\pi}{4}$
but this gives entirely the wrong answer.
Is this a doable approach? Should the span of $t$ perhaps be determined in another way?
With the given parameterization of $\gamma$, note that as $t$ varies from, say, $t=0$ to $t=\pi$, we start on the positive $x$-axis at $(1,0)$ when $t=0$; we pass through the positive $y$-axis at $\left(0,\frac1{\sqrt2}\right)$ when $t=\frac\pi2$; and we stop on the negative $x$-axis at $(-1,0)$ when $t=\pi$. Notice that we are traversing the ellipse in the counterclockwise direction, so the curve is positively oriented.
We know $t=\pi$ corresponds to $(-1,0)$, so now we just need to find $t>\pi$ such that $(x(t),y(t)) = \left(\frac1{\sqrt3},\frac1{\sqrt3}\right)$. Solve:
$$\begin{cases}\cos(t) = \frac1{\sqrt3} \\ \frac1{\sqrt2}\sin(t)=\frac1{\sqrt3}\end{cases} \implies \tan(t) = \sqrt2 \implies t = \arctan(\sqrt2)+k\pi$$
where $k$ is an integer. We need to choose the smallest $k$ that brings us to the right endpoint and make sure we don't do another lap. (At any rate, nothing about the question suggests we should traverse it completely.) In this case, it's $k=2$, so the line integral should be taken over the interval $[\pi,\arctan(\sqrt2)+2\pi]$.
Why $k=2$? It may help to picture the unit circle; both $\cos(t)$ and $\sin(t)$ are positive, so the point $\left(\frac1{\sqrt3},\frac1{\sqrt3}\right)$ lies somewhere in the first quadrant, making an angle $t$ with the positive $x$-axis. If we add $\pi$ to this angle, we end up in the third quadrant with both $\sin$ and $\cos$ negative. Adding another $\pi$ will return us to the first quadrant.
Now substituting everything into the integral gives
$$\begin{align*} I &= \int_\gamma \vec F(x,y) \cdot d\vec r \\[1ex] &= \int_\gamma (6x(t)+2y(t)^2,4x(t)y(t)+3y(t)^2) \cdot \left(\frac{dx}{dt},\frac{dy}{dt}\right) \, dt \\[1ex] &= \int_\pi^{\arctan(\sqrt2)+2\pi} \left(6\cos(t)+\sin^2(t), 2\sqrt2\cos(t) \sin(t)+\frac32\sin^2(t)\right) \cdot \left(-\sin(t),\frac1{\sqrt2}\cos(t)\right) \, dt \\[1ex] &= \frac18 \int_\pi^{\arctan(\sqrt2)+2\pi} \sin(t) \left(4-48\cos(t)+12\cos(2t)+3\sqrt2\sin(2t)\right) \, dt \end{align*}$$
The remaining integration is elementary with the help of these trig identities.