How to evaluate a surface integral with three points?

  • #1

Homework Statement


Let G=x^2i+xyj+zk


And let S be the surface with points connecting (0,0,0) , (1,1,0) and (2,2,2)

Find ∬GdS. (over S)

Homework Equations




The Attempt at a Solution



I parametrised the surface and found 0=2x-2y. I’m not sure if this is correct. And I’m also uncertain about how to proceed.
 

Answers and Replies

  • #2
Ray Vickson
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Homework Statement


Let G=x^2i+xyj+zk


And let S be the surface with points connecting (0,0,0) , (1,1,0) and (2,2,2)

Find ∬GdS. (over S)

Homework Equations




The Attempt at a Solution



I parametrised the surface and found 0=2x-2y. I’m not sure if this is correct. And I’m also uncertain about how to proceed.
Check to see if the three given points satisfy your proposed surface equation.
 
  • #3
Okay thanks. So I can conclude that that is the correct surface equation.

Therefore:

r(y,z)= (y,y,z)

G(r(x,z))=(y^2, y^2, z)

Then the integral will be:

∫∫(y^2, y^2, z)(1+2y+1)dydz

Is my reasoning correct? If yes, how do I determine the terminals?

If no, what am I doing incorrectly?
 
  • #4
LCKurtz
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Homework Statement


Let G=x^2i+xyj+zk


And let S be the surface with points connecting (0,0,0) , (1,1,0) and (2,2,2)

Find ∬GdS. (over S)
OK, so you are using i,j,k notation for vectors and G is a vector. What do you mean by "the" surface connecting these points. There are infinitely many such surfaces. And when you write ##\iint GdS## what does that mean if G is a vector? Is dS a vector? Is there a missing dot product there?

Homework Equations




The Attempt at a Solution



I parametrised the surface and found 0=2x-2y. I’m not sure if this is correct. And I’m also uncertain about how to proceed.

Were you given the surface is a plane? If so, why not tell us? So you are using ##y=x##.

Okay thanks. So I can conclude that that is the correct surface equation.

Therefore:

r(y,z)= (y,y,z)
Is that a vector? What happened to the ijk notation for your vectors?

G(r(x,z))=(y^2, y^2, z)

Then the integral will be:

∫∫(y^2, y^2, z)(1+2y+1)dydz

Is my reasoning correct? If yes, how do I determine the terminals?

If no, what am I doing incorrectly?
In that integral you apparently have a vector multiplied by a scalar so you are expecting a vector answer? And where did the 1+2y+1 come from. It is all very confusing. Perhaps I am misunderstanding the problem.
 
Last edited:
  • #5
Ray Vickson
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Okay thanks. So I can conclude that that is the correct surface equation.

Therefore:

r(y,z)= (y,y,z)

G(r(x,z))=(y^2, y^2, z)

Then the integral will be:

∫∫(y^2, y^2, z)(1+2y+1)dydz

Is my reasoning correct? If yes, how do I determine the terminals?

If no, what am I doing incorrectly?
Are you integrating over the triangle with vertices (0,0,0), (1,1,0) and (2,2,1)? Your problem statement did not say that explicitly, but my default assumption is that you want that triangle.
 

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