Surface integral

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Homework Statement


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.


Homework 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 [itex]\cdot[/itex] dS, where a and dS are both vectors. Is that right?


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?

Please help :)
 

Answers and Replies

  • #2
Dick
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Homework Statement


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.


Homework 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 [itex]\cdot[/itex] dS, where a and dS are both vectors. Is that right?


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?

Please help :)
Yes, you have figured out how to set the problem up. Now keep going.
 
Last edited:
  • #3
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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
Dick
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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?
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
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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
HallsofIvy
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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
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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
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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
Dick
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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?
No, that's not what you get. What value are you integrating over the square?
 
  • #10
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Surfaceintegral(xz2dxdy) = ∫∫xz2dxdy

setting z=0 gives xy and setting x=y=1 gives 1..
 
  • #11
Office_Shredder
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setting z=0 gives xy and setting x=y=1 gives 1..
Setting z=0 certainly does not give xy! What is 0*x*y equal to?
 
  • #12
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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?
 
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  • #13
HallsofIvy
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And you do not have an "integration constant" because this is a definite integral.
 

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