divergence-theoremintegrationmultiple integralmultivariable-calculus
Let $\vec{F}=(x+2y)e^zi+(ye^z+x^2)j+y^2zk$ and let $S$ be the surface $x^2+y^2+z=1$, $z\geq 0$. If $\hat{n}$ is a unit normal to $S$ and $$\left|\iint_S(\nabla\times \vec{F})\cdot \hat{n}\, dS\right|=\alpha\pi.$$ Then $\alpha=?$
We can't apply Gauss divergence theorem here since the surface S is not closed. How to proceed in this question then? Please help.
While Daniel Fischer's answer probably is more direct, I find this way of showing that the circulation over a closed surface is zero (using Stokes theorem) more intuitive:
Here I split the closed surface $S$ into two surfaces $S_1$ and $S_2$ with a shared boundary, the curve $C_1$. If we apply Stokes theorem to the two surfaces separately we get:
$$\int{\int_{S_1} (\nabla \times \bar F) d \bar S} = \int_{C_1} {\bar F \cdot d \bar r}$$
and
$$\int{\int_{S_2} (\nabla \times \bar F) d \bar S} = - \int_{C_1} {\bar F \cdot d \bar r}$$
notice the negative sign before the right hand integral in the second equation. This is because the curve $C_1$ runs in the opposite direction to the normals of $S_2$ compared to $S_1$.
Combing the two surface integrals to get the integral over the entire surface, $S = S_1 \cup S_2$:
$$ \begin{eqnarray} \int{\int_{S} (\nabla \times \bar F) d \bar S} &=& \int{\int_{S_1} (\nabla \times \bar F) d \bar S} + \int{\int_{S_2} (\nabla \times \bar F) d \bar S} \\ \int{\int_{S} (\nabla \times \bar F) d \bar S} &=& \int_{C_1} {\bar F \cdot d \bar r} - \int_{C_1} {\bar F \cdot d \bar r} \\ &=& 0 \end{eqnarray} $$
At the bottom of the cylinder, $z=0$. Since $F_z(x,y,0)=0$ we find
$$\begin{align} \int_{S_1} \vec F \cdot \hat n\, dS&=\int_{S_1}\vec F(x,y,0)\cdot \hat z\,dS\\\\ &=0 \end{align}$$
and there is no flux contributed.
At the top of the cylinder, $z=5$. Since $F_z(x,y,5)=15$ we find
$$\begin{align} \int_{S_2} \vec F \cdot \hat n\, dS&=\int_{S_2}\vec F(x,y,5)\cdot \hat z\,dS\\\\ &=\int_0^{2\pi}\int_0^2 (15)\,\rho\,d\rho\,d\phi\\\\ &=60\pi \end{align}$$
Putting it all together, we find
$$\begin{align} \int_S \vec F\cdot \hat n \,dS&=\int_V \nabla \cdot \vec F\,dV-\int_{S_2} \vec F \cdot \hat n\, dS-\int_{S_1} \vec F \cdot \hat n\, dS\\\\ &=120\pi-60\pi-0\\\\ &=60\pi \end{align}$$
Best Answer
Notice that the boundary of the surface is the curve $\{(x,y,z):x^2+y^2=1\cap z=0\}$ By Stokes' theorem if two surfaces share the same boundary then the integral of the curl on both surfaces will be identical. I.e.
$$\iint\limits_S (\nabla \times F)\cdot dS = \iint\limits_{x^2+y^2\leq 1 \:\cap\:z=0} (\nabla \times F)\cdot dS$$
with both oriented either upwards or downwards.
Why does this make life easier? For starters, the Jacobian between the $z=0$ plane and the usual $xy$ coordinates is $1$ (the Jacobian of anything from itself to itself is $1$) and the normal vector only points in the $z$ direction, meaning we don't have to even compute the whole curl, only the $z$ component, which is
$$\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 2x-2e^z$$
This gives us the following equality
$$\iint\limits_{x^2+y^2\leq 1 \:\cap\:z=0} (\nabla \times F)\cdot dS = \iint\limits_{x^2+y^2\leq 1}2x-2e^0\:dA = \iint\limits_{x^2+y^2\leq 1}2x\:dA - \iint\limits_{x^2+y^2\leq 1}2\:dA$$
$2x$ is an odd function, so its integral will vanish on the disk by $x$ symmetry. The only integral left is a constant, which just gives us the area of the surface times that constant:
$$\iint\limits_{x^2+y^2\leq 1 \:\cap\:z=0} (\nabla \times F)\cdot dS = -2\pi$$
Thus $\alpha =2$