Help me with Surface intergral!

  • Thread starter Learner123
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  • #1
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Homework Statement



Evaluate [tex]\int[/tex][tex]\int[/tex] (x^2 +4y + z)dS where S is the portion of the plane 3y +2z = 6 with 0<x<3

Homework Equations


choosing x, y as variables
dS = [tex]\sqrt{}(partial dz/dx)^2 + (partial dz/dy)^2 +1[/tex]

The Attempt at a Solution


If i choose x, y as variables or x, z as variables, then when i choose limit of intergral, x will go from 0 to 3 whereas y or z will go from -infinite to +infinite, and solving the intergral results of infinite result (seem to be not right to me)
if choosing y,z as variables, then i can't "translate" f = x^2 + 4y + z into parameterized f since x does not depend on y and z in the plane 3y +2z = 6 (in the common problems, the plane involves x, so i just forming x = g(y,z) and then sub x into f which is in the intergral)
Any help would highly appriciated.
 

Answers and Replies

  • #2
HallsofIvy
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Are you sure you have read the problem completely? If the only condition is that "0< x< 3", then, yes, the surface is of infinite extent and the integral does not exist. Is it possible that you are asked to integrate over the part of the plane in the first octant with 0< x< 3?
 
  • #3
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the problem is given exactly above, nothin say about first obtant (i also thought about 1st obtant, too when i read the word "portion of", but it seems not that) so i guess the intergral is infinite is right, maybe the instructor gave the wrong problem. I wasn't sure so i asked, thank you for pointing out.
 

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