Calculate the vector field over surface S using Gassian Law

[skrat]

Homework Statement

Integrate vector field $\vec{F}=(x+y,y+z,z+x)$ over surface $S$, where $S$ is defined as cylinder $x^2+y^2=1$ (without the bottom or top) for $z\in \left [ 0,h \right ]$ where $h>0$

Homework Equations

The Attempt at a Solution

Since cylinder is not closed (bottom and top are not included) I firstly have to take this into account, therefore Gaussian law is:

$\int \int _{S}+\int \int _{O}=\int \int \int _{Cylinder}\nabla\vec{F}dV$

so

$\int \int \int _{Cylinder}3dV=3\pi h$

Now, to calculate the vector filed over surface S, we need to firstly calculate the vector field over top and bottom (surface O).

Polar coordinates $x=r\cos\phi$ and $y=r\sin\phi$ for $r\in \left [ 0,1 \right ]$ and $\phi \in \left [ 0,2\pi \right ]$.

Than $\vec{F}(\phi , r)=(r\cos\phi + r\sin\phi , r\sin\phi + h,h+r\cos\phi )$ and $r_{phi}\times r_r=(0,0,-r)$

Than for bottom where $z=0$: (Direction of $r_{phi}\times r_r$ is ok)

$\int_{0}^{2\pi }\int_{0}^{1}-r^2cos\phi drd\phi =0$

but for $z=1$: (Direction of $r_{phi}\times r_r$ sign needs to be corrected)

$\int_{0}^{2\pi }\int_{0}^{1}(rh+r^2cos\phi ) drd\phi =\pi h$

So $\int \int _{S}=2\pi h$

Or is my idea of surface orientation completely messed up?

 

[LCKurtz]

Homework Statement

Integrate vector field $\vec{F}=(x+y,y+z,z+x)$ over surface $S$, where $S$ is defined as cylinder $x^2+y^2=1$ (without the bottom or top) for $z\in \left [ 0,h \right ]$ where $h>0$

Homework Equations

The Attempt at a Solution

Since cylinder is not closed (bottom and top are not included) I firstly have to take this into account, therefore Gaussian law is:

$\int \int _{S}+\int \int _{O}=\int \int \int _{Cylinder}\nabla\vec{F}dV$

so

$\int \int \int _{Cylinder}3dV=3\pi h$

Now, to calculate the vector filed over surface S, we need to firstly calculate the vector field over top and bottom (surface O).

Polar coordinates $x=r\cos\phi$ and $y=r\sin\phi$ for $r\in \left [ 0,1 \right ]$ and $\phi \in \left [ 0,2\pi \right ]$.

Than $\vec{F}(\phi , r)=(r\cos\phi + r\sin\phi , r\sin\phi + h,h+r\cos\phi )$ and $r_{phi}\times r_r=(0,0,-r)$

Than for bottom where $z=0$: (Direction of $r_{phi}\times r_r$ is ok)

$\int_{0}^{2\pi }\int_{0}^{1}-r^2cos\phi drd\phi =0$

but for $z=1$: (Direction of $r_{phi}\times r_r$ sign needs to be corrected)

$\int_{0}^{2\pi }\int_{0}^{1}(rh+r^2cos\phi ) drd\phi =\pi h$

So $\int \int _{S}=2\pi h$

Or is my idea of surface orientation completely messed up?

I didn't check all your arithmetic, but that looks correct about the orientation. The difference between this problem and the one in the other thread is that the divergence theorem specifies the outward pointing normal so you know how the surface is oriented. That's why you know how to check the sign, unlike in the other thread.

 

[skrat]

LCKurtz, thank you is not enough for all your help in both threads... However, THANK YOU!
I will probably have more problems like this by the end of this week, but you helped a lot!

Cheers