Calculating flux, question about orientation and sign of integral

Question

Let $$B=\{(x,y,z) \in R^3 : x^2+y^2=4z\}$$

$$L=\{(x,y,z) \in R^3 : x=2, y=0\}$$

Let $P$ be a fragment of $B$ of points such that distance from them to $L$ is less than 1. $P$ is oriented as follows: in any point, normal vector giving positive side has positive $z$ component.

Calculate flux over vector field $F(x,y,z)=(\frac{-y}{z},\frac{x}{z}, \frac{y}{x-z})$ through $P$ from negative to positive side.

So I calculated that $P=\{(x,y,z) \in R^3 : x^2+y^2=4z, (x-2)^2+y^2<1\}$ and that normal vector giving positive side points inwards. I considered parametrization $f(u,v)=(u,v,\frac{u^2+v^2}{4})$ from $\operatorname{Ball} K=\{(u-2)^2+v^2<1\}$ to $P$.

So:

$$\operatorname{Flux} = \int_P\frac{-y}{z}dy \wedge dz -\frac{x}{z} dx \wedge dz + \frac{y}{x-z} dx \wedge dy=(\color{red}{\pm1})\int_K \frac{-v}{\frac{u^2+v^2}{4}}(\frac{-u}{2})-\frac{u}{v}(\frac{v}{2})+\frac{v}{u-\frac{u^2+v^2}{4}}\qquad (1)$$

and I am not sure about the sign after equality (in red). I think it should be negative because normal point inwards, can you explain it and check if everything before is correct?

Answer 1

My previous answer was too involved and came to the wrong conclusion for some reason.

The answer is you need the plus sign.

The surface $B$ is the boundary of an elliptic paraboloid. I gather that $z\ge 0\,.$ The inward pointing unit normal vector at $B$ is $$\tag{1} \mathbf{n}=\frac{1}{\sqrt{x^2+y^2+4}}\begin{pmatrix}-x\\-y\\2\end{pmatrix}\,. $$ Following Ted Shifrin's comment: The normal vector is $$\tag{2} \frac{\partial f}{\partial u}\times \frac{\partial f}{\partial v}=\begin{pmatrix}1\\0\\\frac{u}{2}\end{pmatrix}\times \begin{pmatrix}0\\1\\\frac{v}{2}\end{pmatrix}=\begin{pmatrix}-\frac{u}{2}\\-\frac{v}{2}\\\color{red}{+}1\end{pmatrix} $$ when surface parametrized by $f(u,v)=(u,v,(u^2+v^2)/4)^\top\,.$ The normal vector (2) has positive $z$-component and points inward like $\mathbf{n}$ does.

The flux of $F$ through $P$ in the direction of those normal vectors is $$ \int_PF\cdot \mathbf{n}\,dS= \int_K F\cdot\Big\{\frac{\partial f}{\partial u}\times \frac{\partial f}{\partial v}\Big\}\,du\,dv\,.\tag{3} $$ (The reason why the cross product does not need to be normalized is discussed in this answer.)

To bring (3) into a notation using differential forms we write $f=(x,y,z)^\top\,.$ Then the components of $\{\frac{\partial f}{\partial u}\times \frac{\partial f}{\partial v}\}\,du\,dv $ are \begin{align} \Big\{\frac{\partial y}{\partial u}\frac{\partial z}{\partial v}-\frac{\partial z}{\partial u}\frac{\partial y}{\partial v}\Big\}\,du\,dv &=dy\wedge dz\,,\\ \Big\{\frac{\partial z}{\partial u}\frac{\partial x}{\partial v}-\frac{\partial x}{\partial u}\frac{\partial z}{\partial v}\Big\}\,du\,dv &=dz\wedge dx\,,\\ \Big\{\frac{\partial x}{\partial u}\frac{\partial y}{\partial v}-\frac{\partial y}{\partial u}\frac{\partial x}{\partial v}\Big\}\,du\,dv &=dx\wedge dy\,. \end{align} Therefore with the desired inward orientation of $\mathbf{n}$ the flux is $$ \int_PF\cdot \mathbf{n}\,dS=\int_PF_x\,dy\wedge dz+F_y\,dz\wedge dx+F_z\,dx\wedge dy\,. $$ In your case this is calculated by the RHS of (3) and using (2) as $$ \int_K\frac{-v}{\frac{u^2+v^2}{4}}(\frac{-u}{2})-\frac{u}{\color{red}{\frac{u^2+v^2}{4}}}(\frac{v}{2})+\frac{v}{u-\frac{u^2+v^2}{4}}\,du\,dv $$ where I fixed an error of yours.