Surface integral over the surface of the cone $z=1-\sqrt{x^2+y^2}$ lying above the $xy$-plane and normal making an acute angle with $\vec k$

Question

Let $\vec F=(x^2+y-4,3xy,2xz+z^2)$ and $S$ be the surface of the cone $z=1-\sqrt{x^2+y^2}$ lying above the $xy$-plane and $\vec n$ is the unit normal to $S$ making an acute angle with $\vec k$ , then how to evaluate $\iint_S(\nabla \times \vec F).\vec n dS$ ? If I use Stokes theorem , is the boundary line going to be the circle $x^2+y^2=1$ ? What changes does the normal making acute angle with the unit vector along $z$-axis make ? Please help . Thanks in advance

Answer 1

At any point on the surface you can define your normal in two mutually opposite directions. Your cone points upwards and therefore if you choose the outward normal, it'll make an acute angle with $\hat k$ (the inward normal makes an obtuse angle with $\hat k$).

You're correct about the boundary so your integral becomes (in counter-clockwise direction) $$ \int_{x^2+y^2=1}(x^2+y-4)dx+3xydy $$ which upon using polar coordinates $x=\cos\theta,y=\sin\theta,\theta\in[0,2\pi)$ becomes tractable.

If instead inward normal were used, the integral would change by a factor of $-1$ as the dot product would change by the same factor.