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Vector Calculus: Surfave Integrals

  • Thread starter MaxManus
  • Start date
  • #1
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Vector Calculus: Surface Integrals

Homework Statement


Find the surface integral of u[/B dot n over S where S is part of the surface z = x + y^2 with z < 0 and x > -1, u is the vector field u = (2y,x -1,0) and n has a negative z component


Homework Equations





The Attempt at a Solution


How do you "find" the surface. I have just started ont the subject and I have no idea how to see what is the surface and the region of integration.

Solution according to the book
The surface is written parametically as (x,y,x+y^2)
two vectors parallel to the surface are (1,0,1) and (0,1,2y)
Their cross product = (-1,-2y,1)
ndS = (-1,-2y,1)dxdy
changing the direction of n
ndS = (1,2y,-1)dxdy
u dot ndS = xdxdy
region of integration x+y^2 < 0, x > -1, so doing the x integration first, -1<x<-y^2 and
-1 <y <1
 
Last edited:

Answers and Replies

  • #2
lanedance
Homework Helper
3,304
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hey MaxManus

to visualise the surface, consider the curves given by y = 0 and x = 0, ie the surface slices by the xz & yz planes, this should be a good starting point
 
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