I'm at a loss for ideas on what to do. I'm given two paraboloids and their equations: $z_1 = x^2 + 1$ and $z_2 = 5 - y^2$.
I know their intersection is when $z_1 = z_2$.
This gives me $x^2 + y^2 = 4$, the equation of a circle. Which makes sense, because the projection of the intersection on the xy plane is a circle.
With this, I could parameterize with $x(\theta) = 2\cos \theta $ and $y(\theta) = 2 \sin \theta$, using polar coordinates.
My questions begin here, is $z(t) = 0$? Practically, it doesn't make sense because the intersection of the curves go up and down the $z$ axis, but I don't understand what else it could be, if not just $z$ ranging from $z_2 = 5 - y^2$ to $z_1 = x^2 + 1$.
In either case, my next question, is regarding calculating closed curve integral of a vector field $\overrightarrow{F}$ around this intersection. My problem with this, is that to find the normal vector $\overrightarrow{n}$ from $\int\int_S \nabla \times \overrightarrow{F} \cdot \overrightarrow{n} dS$, I need to calculate the cross product of the partial derivatives of my parametrized intersection equation.
Suppose that I use $z(z) = z$, then my parametrized equation is $R(\theta, z) = 2\cos \theta \overrightarrow{i} + 2\sin \theta\overrightarrow{j} + z \overrightarrow{k}$. I'll spare you the details, but ultimately I'm left evaluating an intergral with a cosine and sine ranging from $0$ to $2\pi$, which leaves me with a big ZERO. I'm sure there is something wrong somewhere, and I cannot figure out what.
Hints on potential solution tracks would be greatly appreciated.