Surface integrals (without real integration)

[mewmew]

Given F= $$(ix+jy) Ln(x^2+y^2)$$

and given S, which is a cylinder of radius r, and height h(in the z axis) evaluate $$\int\int_s F.n \,ds$$. It says that you shouldn't need to do any work if you think about it enough. I figured I could find the area of the main part to be $$2 \pi r h$$ then multiply that by $$Ln(r^2)=2Ln(r)$$ to get the answer but I am off by a factor of r in my answer. I don't think the caps to the cylinder contribute to this as the normal is orthogonal to F.

One more question, what exactly does a surface integral return? I feel stupid but I can't seem to find out exactly what the physical meaning of the result of a surface integral is. Thanks for the help.

 

[topsquark]

Given F= $$(ix+jy) Ln(x^2+y^2)$$

and given S, which is a cylinder of radius r, and height h(in the z axis) evaluate $$\int\int_s F.n \,ds$$. It says that you shouldn't need to do any work if you think about it enough. I figured I could find the area of the main part to be $$2 \pi r h$$ then multiply that by $$Ln(r^2)=2Ln(r)$$ to get the answer but I am off by a factor of r in my answer. I don't think the caps to the cylinder contribute to this as the normal is orthogonal to F.

One more question, what exactly does a surface integral return? I feel stupid but I can't seem to find out exactly what the physical meaning of the result of a surface integral is. Thanks for the help.

A surface integral is the integral of a function over a surface, as opposed to a line, etc. Unit-wise you are gaining an extra meter^2. As far as a physical meaning, it depends on what you are integrating.

What's happening in this integration would be easier to understand if you switch everything to polar coordinates.

-Dan

 

[mewmew]

I think I'm just confusing myself more If I integrate that vector function around a cylinder than it will be 0. I can imagine that we would have constant vectors of magnitude $$Ln[r^2]$$ all in the radial direction around the cylinder for any z value. I guess I still don't have a good enough understanding of what a surface integral really is to be able to find an easy way to express this. I think its time to re-read some div,grad,curl.

 

[siddharth]

As topsquark said, switch over to polar.

In fact, I suggest you work this out the hard way and then look back.