Let S be the paraboloid $z = 5x^2 + 3y^2$ in $\mathbb{R}^3$ lying over the region $R$ in the $xy$-plane bounded by the lines $x+y=3$ and the coordinate axes. Suppose that the orientation of $S$ is such that the normal ${\bf n}$ is pointing upwards. Let $C$ be the boundary of $S$. Consider the vector field ${\bf F}=(x,z,x+y)$.
I'm having trouble parametrizing the boundary and keep getting confused. A little help would be great.
The line $x+y=3$ meets the $x$-axis at $(3,0)$ and the $y$-axis at $(0,3)$.
The chord from $(0,0)$ to $(3,0)$ is parametrised by $(3t,0)$ where $0 \le t \le 1$ and the chord from $(0,0)$ to $(0,3)$ is parametrised by $(0,3t)$ where, again, $0 \le t \le 1$. You can parametrise the chord between $(0,3)$ and $(3,0)$, where $0 \le t \le 1$, by $$(0,3)\cdot (1-t)+(3,0) \cdot t = (3t,3(1-t)).$$
You can now "lift" these from the plane up onto the parabaloid. If $x=3t$ and $y=0$ then $z=45t^2$. If $x=0$ and $y=3t$ then $z=27t^2$. If $x=3t$ and $y=3(t-1)$ then $$z = 5(3t)^2 + 3\left[3(t-1)\right]^2 = \ ?$$