Evaluate $\int \vec{F} \cdot d\vec{r}$ o the surface $z=4-y^2$ cut off by $x=0$, $z=0$, and $y=x$.
I particularly need help with evaluating the integral on $C_3$. Please see picture
I am not too sure about the way I drew this, especially the part involving $C_3$. Anyway this is what I tried for $C_3$:
$C_3$ is on the line $y=x$, I think, and we are told $z=4-y^2 \implies x=\sqrt{4-z}$. We note that $y=x$ on all point at line $C_3$
$\therefore$ $\vec{r} =\left<x,y,z\right>=\left<\sqrt{4-z},0,z\right>$
Can someone please confirm whether that's the right parametrization and drawing for $C_3$ ? Any helpful suggestions are greatly welcomed and appreciated. -Thanks.
The equation $y=x$ is not the equation of a line, it is the equation of a plane. The intersection of the surface $z=4-y^2$ with the plane surface $y=x$ is the set of points $$\vec{r} = (x,x,4-x^2).$$ This is the parametrization of the curve. The tangent vector is found by differentiating each coordinate, and we obtain $$\frac{d\vec{r}}{dx} = (1,1,-2x).$$ The limit for $x$ can be found by setting $z=0$ in the equation $z=4-x^2$. This is when $4-x^2=0$. Therefore $x=2$ and $x=-2$. The second point is thrown out because it is outside the bounded volume. Thus the parametrization is for $0 \leq x \leq 2$.