# Surface Flux Integral

• Apr 30th 2010, 04:11 AM
caliber
Surface Flux Integral
Hi People

New here, so please be kind!

I am doing some work for my maths module, I have got so far but I am not sure if it is correct, and if not, why not. It is important to me to getting it correct because this will also be on my exam so any help you can give me will be very much appreciated.

The question:

Region V of a cylinder is denoted by $\displaystyle x^2+y^2\leq 1, 0 \leq z \leq 2$ and S is the surface of this cylinder.

Vector field $\displaystyle F(x,y,z) = xi + yi + z(x^2+y^2)k$

Now, I have been instructed earlier in the question to convert F into cylindrical co-ordinates, so...

$\displaystyle F(\rho,\varphi,z) = \rho \widehat{\rho} + z\rho^{2}\widehat{z}$

I am then instructed to find the surface integral. I do this, but am unable to do it in cylindrical co-ordinates. Is this correct?

I have a feeling it is not, because at the end of it I am left with an answer simply of 8pi

Once again, thanks for any help and if you need any more workings and such I can provide them.
• Apr 30th 2010, 07:24 AM
Jester
Quote:

Originally Posted by caliber
Hi People

New here, so please be kind!

I am doing some work for my maths module, I have got so far but I am not sure if it is correct, and if not, why not. It is important to me to getting it correct because this will also be on my exam so any help you can give me will be very much appreciated.

The question:

Region V of a cylinder is denoted by $\displaystyle x^2+y^2\leq 1, 0 \leq z \leq 2$ and S is the surface of this cylinder.

Vector field $\displaystyle F(x,y,z) = xi + yi + z(x^2+y^2)k$

Now, I have been instructed earlier in the question to convert F into cylindrical co-ordinates, so...

$\displaystyle F(\rho,\varphi,z) = \rho \widehat{\rho} + z\rho^{2}\widehat{z}$

I am then instructed to find the surface integral. I do this, but am unable to do it in cylindrical co-ordinates. Is this correct?

I have a feeling it is not, because at the end of it I am left with an answer simply of 8pi

Once again, thanks for any help and if you need any more workings and such I can provide them.

First off, there are 3 surfaces - top, bottom and side. Top and bottom are straight forward leading to $\displaystyle \pi$ (top) and $\displaystyle 0$ (bottom). The side is a little more complicated. Parametrize the surface

$\displaystyle {\bf r} = < \cos \theta, \sin \theta, z>,\;\;\; 0 \le \theta \le 2 \pi,\;\;\; 0 \le z \le 2.$

So the flux is given by

$\displaystyle \iint_D {\bf F} \cdot {\bf r}_{\theta} \times {\bf r}_z dA = \int_0^2 \int_0^{2 \pi} dz d \theta\ = 4 \pi\;\; \text{since}\; {\bf F} \cdot {\bf r}_{\theta} \times {\bf r}_z = 1$

Total flux is therefore $\displaystyle 4\pi + \pi = 5 \pi$.

To check, use the divergence theorem.
• May 3rd 2010, 04:03 AM
caliber
Quote:

Originally Posted by Danny
First off, there are 3 surfaces - top, bottom and side. Top and bottom are straight forward leading to $\displaystyle \pi$ (top) and $\displaystyle 0$ (bottom). The side is a little more complicated. Parametrize the surface

$\displaystyle {\bf r} = < \cos \theta, \sin \theta, z>,\;\;\; 0 \le \theta \le 2 \pi,\;\;\; 0 \le z \le 2.$

So the flux is given by

$\displaystyle \iint_D {\bf F} \cdot {\bf r}_{\theta} \times {\bf r}_z dA = \int_0^2 \int_0^{2 \pi} dz d \theta\ = 4 \pi\;\; \text{since}\; {\bf F} \cdot {\bf r}_{\theta} \times {\bf r}_z = 1$

Total flux is therefore $\displaystyle 4\pi + \pi = 5 \pi$.

To check, use the divergence theorem.

Hi

Thanks for that. How would I use divergence theorem for this? I am aware of divergence theorem, but not sure how to use it for cylindrical co-ordinates.

Thanks
• May 3rd 2010, 05:58 AM
Jester
Quote:

Originally Posted by caliber
Hi

Thanks for that. How would I use divergence theorem for this? I am aware of divergence theorem, but not sure how to use it for cylindrical co-ordinates.

Thanks

I think it's better to use the Div thm. in cartesian and then switch the triple integral. So, if

$\displaystyle {\bf F } = <x,y,(x^2+y^2)z> \; \text{so}\; {\bf \nabla \cdot F} = 1 + 1 + x^2+y^2$

so $\displaystyle \iiint\limits_V ( 2 + x^2 + y^2) dV = \int_0^{2 \pi} \int_0^1 \int_0^2 (2 + r^2) r dz\, dr\, d\, \theta$.
• May 3rd 2010, 08:03 AM
caliber
Thanks, However on working through this i get:

$\displaystyle \int_0^{2 \pi} \int_0^1 \int_0^2 (2 + r^2) r dz\, dr\, d\, \theta = \int_0^{2 \pi} \int_0^1 \int_0^2 (2r + r^3) dz\, dr\, d\, \theta$

$\displaystyle = \int_0^{2 \pi} \int_0^1 2(2r + r^2) dr\, d\, \theta$

$\displaystyle = \int_0^{2 \pi} -(8/3) d\, \theta$

$\displaystyle = -16\pi/3$

which is incorrect?
• May 3rd 2010, 09:17 AM
Jester
Quote:

Originally Posted by caliber
Thanks, However on working through this i get:

$\displaystyle \int_0^{2 \pi} \int_0^1 \int_0^2 (2 + r^2) r dz\, dr\, d\, \theta = \int_0^{2 \pi} \int_0^1 \int_0^2 (2r + r^3) dz\, dr\, d\, \theta$

$\displaystyle = \int_0^{2 \pi} \int_0^1 2(2r + r^2) dr\, d\, \theta$

$\displaystyle = \int_0^{2 \pi} -(8/3) d\, \theta$

$\displaystyle = -16\pi/3$

which is incorrect?

First let me say that you're integrating $\displaystyle 2r + r^3$ over a positive range i.e., $\displaystyle [0,2 \pi] \times [0,1] \times [0,2]$ so the answer must be non-negative. Check your integration.
• May 3rd 2010, 09:46 AM
caliber
Quote:

Originally Posted by Danny
First let me say that you're integrating $\displaystyle 2r + r^3$ over a positive range i.e., $\displaystyle [0,2 \pi] \times [0,1] \times [0,2]$ so the answer must be non-negative. Check your integration.

apologies, please disregard the negative sign.