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September 6th, 2017, 10:58 PM   #1
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parameterization in Surface integration

I want to surface integrate over a surface of the plane $\displaystyle S : 2x+3y+6z =12$ which lie in the 1st octant. Should I use parameterization, if I should, how?
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September 7th, 2017, 03:52 AM   #2
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A surface, being two dimensional, requires two parameters. You are given 2x+3y+6z=12 which you can write as x= 6- (3/2)y- 3z. Take y and z as parameters- or, if you prefer, s and t, writing x= 6- (3/2)s- 3t, y= s, z= t.

Obviously, you can do that with any of the variables: we can just as well write y= 4- (2/3)x- 2z and take x= s, y= 4- (2/3)s- 2t, z= t or we can write z= 2- (1/3)x- (1/2)y and take x= s, y= t, z= 2-(1/3)s- (1/2)t.

Or, if you don't like fractions, take x= 3s, y= 2t, z= 2- s- t.
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September 7th, 2017, 05:40 AM   #3
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Originally Posted by Country Boy View Post

Or, if you don't like fractions, take x= 3s, y= 2t, z= 2- s- t.
Poor fractions
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September 7th, 2017, 08:11 AM   #4
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Yeah, I'm so mean to fractions- but they started it!
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September 7th, 2017, 08:31 AM   #5
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I think you're being irrational.
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September 7th, 2017, 04:52 PM   #6
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Get real!
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September 11th, 2017, 05:10 AM   #7
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Suppose i need to integrate a function
F = 36z - 36 + 18y
over the surface, how should i place the limits?
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September 12th, 2017, 04:56 AM   #8
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I thought I had already responded to this! In my first post I suggested the parameterization x= 3s, y= 2t, z= 2- s- t. Since we can write s and t in terms of x and y separately, we want to project the surface down to the xy-plane.

The plane 2x+3y+6z=12 crosses the z= 0 plane where 2x+ 3y= 12 and that cuts the x-axis where 2x= 12 or x= 6. To integrate over that, we can take x going from 0 to 6 and, for each x, y going from 0 to y= (12- 2x)/3= 4- (2/3)x.

Since x= 3s and y= 2t, s= x/3 goes from 0 to 2 and, for each s, t goes from 0 to (4- (2/3)x)/2= 2- (1/3)x= 2- (1/3)(3s)= 2- s.

That is, integrate $\displaystyle \int_0^6 \int_0^{2- s} .... dtds$.
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