I have to evaluate $$\int\int_{S}\vec{F}\cdot\vec{N}d\sigma$$ with $$\vec{F}(x,y,z)=-x\vec{i}-y\vec{j}+z^{2}\vec{k}$$ where $S$ is the part of the cone $$z=\sqrt{x^{2}+y^{2}}$$ between the planes $z=1$ and $z=2$ in a way that the orientation of the surface is such that its normal vector $\vec{N}$ satisfies $\vec{N}\cdot\vec{k}<0$.
My question is: how do I work with this $\vec{N}\cdot\vec{k}<0$? I mean: what does it change in my calculus?
There's a symmetry in this problem that allows you to make some simplifications. Consider a point on the cone in cylindrical coordinates $(r,\theta,z)$ where the normal vector is $(a, b, c)$. Then at the point $(r, \theta + \pi, z)$ has the normal vector $(-a, -b, c)$. This can be visually explained by the fact that if you add the normal vectors of a point and its corresponding point halfway around the ring on the cone, you get a vector that points solely in the $-\hat{k}$ direction.
Now, given the horizontal cuts of the cone that you have to evaluate the integral across, there is a one-to-one correspondence of these points: that is, $(r, \theta, z)$ is on this surface if and only if the point $(r, \theta + \pi, z)$ is also on this surface. So the $x$ and $y$ components of the vector field will cancel, and therefore the integral will be the same if we work with the vector field $\vec{G}(x, y, z) = z^2 \hat{k}$.
We then must find the surface normals and a parameterization of the surface and evaluate as normal. The problem of finding the surface normal can be treated as finding the gradient of a function $H(x, y, z) = \sqrt{x^2+y^2} - z$: $$\nabla H(x,y,z) = (\frac{x}{\sqrt{x^2 + y^2}}, \frac{y}{\sqrt{x^2 + y^2}} , -1).$$ We can verify that these are oriented correctly, as $\vec{N} \cdot \hat{k}= \nabla H \cdot \hat{k} = -1 < 0$, as desired. We normalize the normal vector, and the integral then becomes $$\frac{1}{\sqrt{2}} \displaystyle \int_S \vec{F} \cdot \vec{N} d\sigma = \frac{1}{\sqrt{2}} \int_S \vec{G} \cdot \vec{N} d \sigma = \frac{1}{\sqrt{2}} \int_S -z^2 d\sigma$$ which is easy to evaluate by taking a parameterization of the surface $\vec{\sigma}(\theta, z) = (z \cos\theta, z \sin\theta, z)$ with $0 \leq \theta < 2\pi$ and $1 \leq z \leq 2$. The integral then becomes $$\frac{1}{\sqrt{2}} \int_1^2 \int_0^{2\pi} -z^2 ||\vec{\sigma}_z \times \vec{\sigma}_\theta||d\theta dz.$$ Hope this helped!