Confirming divergence theorm example

[namu]

Hello, I am having trouble confirming that the flux integral is equal to the divergence over a volume. I am making a silly mistake & its just one of those days that I can't eyeball it. Here is the problem.

I want to compute the flux integral for

$\vec{ F}=x\hat i+y\hat j-z\hat k$

out of the closed cone

$x^2+y^2=z^2 \,\,\,\,\,\,\,0\leq z\leq1$

Let us first do this the easy way using the divergence theorm.

$\int\int \vec F \cdot d \textbf{S}=\int \int \int \nabla \cdot \vec F dV$

$\nabla \cdot \vec F=1$

$\int \int \int \nabla \cdot \vec F dV=\int_0^{2\pi} d\theta \int_0^1 r dr \int_r^1 dz=\frac{\pi}{3}$

This agrees with the formula for the volume of a cone of radius one and height one. Now, let us apply the flux integral directly.

First let's look at the cone.

$\int\int \vec F \cdot d \textbf{S}=\int\int \vec F \cdot d \textbf{S}_1+\int\int \vec F \cdot d \textbf{S}_2$

where S₁ is the cone and S₂ is the lid of the cone, namely the unit disk z=1.

$f=x^2+y^2-z^2=0$

$\nabla f=(2x,2y,-2z)$

$\hat n=\frac{\nabla f}{|\nabla f|}=\frac{(2x,2y,-2z)}{2 \sqrt{x^2+y^2+z^2}}=\frac{(2x,2y,-2z)}{2 \sqrt{2}z}$

$d\textbf{S}_1=\hat n \sqrt{\frac{\partial f}{\partial x}^2+\frac{\partial f}{\partial y}^2+\frac{\partial f}{\partial z}^2} dx dy=\hat n 2 \sqrt{x^2+y^2+z^2} dx dy= \hat n 2 \sqrt{2}z dx dy =(2x,2y,-2z)dx dy$

$\int\int \vec F \cdot d \textbf{S}_1=\int\int \vec F \cdot (2x,2y,-2z)dx dy=\int\int 2x^2+2y^2+2z^2 dx dy =\int\int 4x^2+4y^2 dx dy=\int_0^{2\pi}d\theta \int_0^1 r dr 4 r^2=2\pi$

Over the disk $\hat n=\hat k$ and z=1

$\int\int \vec F \cdot d \textbf{S}_2=-\int_0^{2\pi}d\theta\int_0^1 r dr =-\pi$

Hence,

$2\pi-\pi=\pi$

So this is the wrong answer. I don't see where I went wrong. Can someone please help? Thank you.

 

[abiyo]

You applied the divergence theorem correctly. The flux over the circle is also correct.
The mistake you made is when you computed the flux over S1: the cone. To define
the normal for z=f(x,y) first write it in a format as follows:
$$\sqrt{x^{2}+y^{2}}-z=0$$ Then the normal $N$ will be
$$N=<\frac{x}{z},\frac{y}{z},-1>$$ From here you can do the work and show
that the flux you get will be $$\frac{4\pi}{3}$$ This will confirm the divergence
theorem. Let me know if this makes sense or if you have any further questions.

 

[namu]

You applied the divergence theorem correctly. The flux over the circle is also correct.
The mistake you made is when you computed the flux over S1: the cone. To define
the normal for z=f(x,y) first write it in a format as follows:
$$\sqrt{x^{2}+y^{2}}-z=0$$ Then the normal $N$ will be
$$N=<\frac{x}{z},\frac{y}{z},-1>$$ From here you can do the work and show
that the flux you get will be $$\frac{4\pi}{3}$$ This will confirm the divergence
theorem. Let me know if this makes sense or if you have any further questions.

I worked it out and I got the correct answer. Thank you!