Calculate the Flux through S where $\vec F=y\vec j-z\vec k, \text{enclosed by}\quad y=x^2+z^2, \space y\in[0,1], \quad x^2+z^2 =1$

Question

I am struggling with Flux integrals and would like someone to check what I am not doing right.

I am to evaluate

$$\int\int_S \vec F \cdot d\vec S $$ where $\vec F=y\vec j-z\vec k, \text{enclosed by}\space S : y=x^2+z^2, \space y\in[0,1], \quad x^2+z^2 =1$

So I am thinking that I can use

$$\int\int_S \vec F \cdot (\vec r_x \times\vec r_z)d S $$

where the parametarization of $S$ is obtained by $$x=x, y=x^2+z^2, z=z \Rightarrow \vec r(x,z)=x\vec i+(x^2+z^2)\vec j +z\vec k $$

I believe that this leads to

$$\vec F \cdot (\vec r_x \times\vec r_z) = -y-2z^2$$

and all I have to do is evaluate the double integral

$$-\int \int _D y+2z^2 dA$$

Using the cylindrical method I have

$$-\int_0^{2\pi} \int_0^{1} r^3(1+2\cos^2\theta) dr d\theta = -\pi $$

but the answer is supposedly zero.

Either I made some calculation mistake or there is something that I fundamentally got wrong . . . may I have some assistance?

Answer 1

$\vec F=y\vec j-z\vec k, \text{enclosed by} \ y=x^2+z^2, \ x^2+z^2 =1, \ y\in[0,1]$

It is easier to see this by divergence theorem.

The surface is bound by paraboloid $y = x^2+z^2$ and cylinder $x^2+z^2 = 1$ for $y \in (0,1)$. As the surface is open at $y=0$, we first close the surface by adding a disc at $y = 0$ as applying divergence theorem requires closed surface. We can separately find flux through the disc and subtract it later (which by the way is also zero but that is for later). Now applying divergence theorem,

$div{\vec{F}} = \nabla \cdot \vec{F} = \frac{\partial{(y)}}{{\partial y}} + \frac{\partial{(-z)}}{{\partial z}} = 0$

As the divergence of the vector field is $0$, the flux through the closed surface is zero. But we now need to subtract the flux through the disc at $y = 0$ which we added and is not part of the original surface.

The outward unit normal vector to the disc is $(0,-1,0)$ so $\vec{F} \cdot \hat{n} = -y \ $ but as $y = 0$, the flux through this surface is also zero.

So the surface integral is zero.

EDIT: On your working,

First visualizing the region between $y=0$ and $y=1$, its inner surface is given by the equation of the paraboloid and outer surface is given by the equation of the cylinder. Now when you are finding the flux through the paraboloid surface, please note that out of surface here would mean pointing generally upward in $y$ direction. So while your working seems fine, you have taken normal vector pointing in downward $y$ direction which is not correct for this question.

If I parametrize it in cylindrical coordinates,

$r(\rho, \theta) = (\rho \cos\theta, \rho^2, \rho\sin\theta)$

$r'_{\rho} = (\cos\theta, 2\rho, \sin\theta)$

$r'_{\theta} = (-\rho\sin\theta, 0, \rho\cos\theta)$

$r'_{\theta} \times r'_{\rho} = (-2\rho^2\cos\theta, \rho, - 2\rho^2\sin\theta)$ (keeping $y$ component positive)

$I_1 = \displaystyle \int_0^{2\pi}\int_0^1 (0, \rho^2, -\rho\sin\theta) \cdot (-2\rho^2\cos\theta, \rho, - 2\rho^2\sin\theta) \ d\rho \ d\theta = \pi$

Now parametrizing cylindrical surface as

$r(y, \theta) = (\cos\theta, y, \sin\theta)$

you will get $r'_{y} \times r'_{\theta} = (\cos\theta, 0, \sin\theta)$

So $I_2 = \displaystyle \int_0^{2\pi}\int_0^1 (0, y, -\sin\theta) \cdot (\cos\theta, 0, \sin\theta) \ dy \ d\theta = - \pi$

$I_1 + I_2 = 0$.