# Surface integral

1. Feb 18, 2012

### aaaa202

1. The problem statement, all variables and given/known data
Calculate the surface integral of the vector field a=xy i + (x+1) j + xz^2 k over a square in the xy-plane with length 1 and whose unit normal points in the positive direction of the z axis.

2. Relevant equations
This is the problem. There are many different types of surface integrals. Though since it's a vector field I should assume that they're asking for an integral of the type I = ∫ a $\cdot$ dS, where a and dS are both vectors. Is that right?

3. The attempt at a solution
I'm really not sure, as my book's examples on these are quite poor. I think what I want to find is a general expression for a small vector area in the square. Is that just given by dydx times the unit normal to the xy-plane?

2. Feb 18, 2012

### Dick

Yes, you have figured out how to set the problem up. Now keep going.

Last edited: Feb 18, 2012
3. Feb 18, 2012

### aaaa202

Okay great :)

But can you perhaps explain geometrically what this surface integral will yield?

And what justifies viewing an area element af something times the normal vector to it?

4. Feb 18, 2012

### Dick

It yields what's called the 'flux' of the vector field through the surface. It's useful for example in electrostatics. See http://en.wikipedia.org/wiki/Flux But I wouldn't worry too much about what it means. Just try and compute it. It's REALLY easy in this case.

5. Feb 18, 2012

### aaaa202

hmm okay, I'm still not sure if I'm doing the write thing.
The surface vector element is dS = dxdy k
So the integral over the surface becomes:

Surfaceintegral(xz2dxdy) = ∫∫xz2dxdy
But I don't get a numer just something dependent on z, so either my approach is wrong or I fail at elimination z.

6. Feb 18, 2012

### HallsofIvy

Staff Emeritus
Do you realize that this is a really "trivial" problem? The integral is in the xy-plane. What is the z coordinate of every point in the xy-plane?

(That's also why they can say " a square in the xy-plane with length 1" without having to specify which square!)

7. Feb 18, 2012

### aaaa202

hmm okay I thought about that, but not knowing the geometric interpretation I wasn't quite sure if I could just put z=0....

And as I'm not in general sure WHY my approach is even correct, I would not say it's trivial for me (maybe for you). Why is it that you put an define a vector area element in the direction as the normal unit?

8. Feb 18, 2012

### aaaa202

But then with your comment I'd just get the area of the square as the value of my integral.. Shouldn't I get something else?

9. Feb 18, 2012

### Dick

No, that's not what you get. What value are you integrating over the square?

10. Feb 19, 2012

### aaaa202

Surfaceintegral(xz2dxdy) = ∫∫xz2dxdy

setting z=0 gives xy and setting x=y=1 gives 1..

11. Feb 19, 2012

### Office_Shredder

Staff Emeritus
Setting z=0 certainly does not give xy! What is 0*x*y equal to?

12. Feb 19, 2012

### aaaa202

zero.. ahh yes, I was too fast. But then you get an integration constant and then something with either x or y depending on which integration you did first.

Edit: Nah wait, integral of zero must always be zero (never thought of that). So everything is zero. Yes okay. But I still don't really understand what is happening. Why do you "vectorize" an area and what does the dot product represent?

Last edited: Feb 19, 2012
13. Feb 19, 2012

### HallsofIvy

Staff Emeritus
And you do not have an "integration constant" because this is a definite integral.