- #1
namu
- 33
- 0
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
[itex]
\vec{ F}=x\hat i+y\hat j-z\hat k
[/itex]
out of the closed cone
[itex]
x^2+y^2=z^2 \,\,\,\,\,\,\,0\leq z\leq1
[/itex]
Let us first do this the easy way using the divergence theorm.
[itex]
\int\int \vec F \cdot d \textbf{S}=\int \int \int \nabla \cdot \vec F dV
[/itex]
[itex]
\nabla \cdot \vec F=1
[/itex]
[itex]
\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}
[/itex]
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.
[itex]
\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
[/itex]
where S1 is the cone and S2 is the lid of the cone, namely the unit disk z=1.
[itex]
f=x^2+y^2-z^2=0
[/itex]
[itex]
\nabla f=(2x,2y,-2z)
[/itex]
[itex]
\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}
[/itex]
[itex]
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
[/itex]
[itex]
\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
[/itex]
Over the disk [itex]\hat n=\hat k[/itex] and z=1
[itex]
\int\int \vec F \cdot d \textbf{S}_2=-\int_0^{2\pi}d\theta\int_0^1 r dr =-\pi
[/itex]
Hence,
[itex]
2\pi-\pi=\pi
[/itex]
So this is the wrong answer. I don't see where I went wrong. Can someone please help? Thank you.
I want to compute the flux integral for
[itex]
\vec{ F}=x\hat i+y\hat j-z\hat k
[/itex]
out of the closed cone
[itex]
x^2+y^2=z^2 \,\,\,\,\,\,\,0\leq z\leq1
[/itex]
Let us first do this the easy way using the divergence theorm.
[itex]
\int\int \vec F \cdot d \textbf{S}=\int \int \int \nabla \cdot \vec F dV
[/itex]
[itex]
\nabla \cdot \vec F=1
[/itex]
[itex]
\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}
[/itex]
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.
[itex]
\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
[/itex]
where S1 is the cone and S2 is the lid of the cone, namely the unit disk z=1.
[itex]
f=x^2+y^2-z^2=0
[/itex]
[itex]
\nabla f=(2x,2y,-2z)
[/itex]
[itex]
\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}
[/itex]
[itex]
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
[/itex]
[itex]
\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
[/itex]
Over the disk [itex]\hat n=\hat k[/itex] and z=1
[itex]
\int\int \vec F \cdot d \textbf{S}_2=-\int_0^{2\pi}d\theta\int_0^1 r dr =-\pi
[/itex]
Hence,
[itex]
2\pi-\pi=\pi
[/itex]
So this is the wrong answer. I don't see where I went wrong. Can someone please help? Thank you.