Show that closed curve $C$ is contained in surface and solve closed integral of C oriented by the tangent vector given by the parametrization. I do not understand how to show that $C$ is contained in the given surface in the question below. I've tried to prove it, but to no avail. Can someone please show me the proof, so that I can understand how to do a thing like this in the future?
And for question $(b)$, I'd like to know the solution of the integral, which can only be given if $(a)$ has been executed correctly.
EDIT: I'm really stuck, because I don't know which steps to take to even set up the integral. A detailed solution would be amazing. Please help me asap.
Here's the question:

 A: (a) follows readily from the trig identity
$$ \sin 2t = 2 \sin t \cos t $$
Thus it immediately follows that $z = 2xy$. So the space curve defined by $\mathbf{r}$ is on the surface $z=2xy$ because any point on the curve $(\cos t, \sin t, \sin 2t)$ satisfies the surface equation $z = 2xy$.
As for (b), I like to rewrite the expression
$$ \int_C \left(f \, dx + g \, dy + h \, dz \right) $$
like so
$$ \int_C (f,g,h) \cdot d\mathbf{r} $$
where "$\cdot$" is the vector dot product, and $d\mathbf{r} = (dx, dy, dz)$.
If we have the parametrization of $C$ by a vector-valued function $\mathbf{r}(t) = \big(x(t),\ y(t),\ z(t) \big)$, then
$$ \int_C (f,g,h) \cdot d\mathbf{r} = \int_a^b \Big(f(\mathbf{r}(t)),\ g(\mathbf{r}(t)),\ h(\mathbf{r}(t)) \Big) \cdot \frac{d\mathbf{r}}{dt} \, dt $$
where the notation $f(\mathbf{r}(t))$ means replace every $x$, $y$, and $z$ in $f$ with the corresponding component of the parametrization. When you do that, you'll get a vector that needs to be "dotted" with the derivative $d\mathbf{r}/dt$ which is computed by taking the derivative of each component in turn:
$$ \frac{d\mathbf{r}}{dt} = (-\sin t,\ \cos t,\ 2\cos 2t) $$
After "dotting" them, you'll have an ordinary 1D integral of one variable. Can you take it from there?
Without doing the algebra myself, it looks like the final integral you'll get may not be doable by hand. You may need a computer to do it numerically.
