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Surface integrals and flux

  • #1

Homework Statement


calculate the upward flux of f(x,y,z) = <yz,2x+y,y^2+z>

Let S be the portion of the cylinder z=4-y^2 lying in the first octant to the right of the plane y=4.
a parametrization into the u v plane is:r(u,v)=(u,v,4-v^2)
region is a rectangle in the uv plane with bounds, (0,0) , (0,2) and (4,0)

Homework Equations


[tex]\int[/tex][tex]\int[/tex] F [tex]\bullet[/tex](ru [tex]\times[/tex]rv) dA



The Attempt at a Solution



ru x rv = 0i + 2vj + 1k

so then i have [tex]\int[/tex] <4v-v^3, 2u+v, v^2 + 4 - v^2 > x <0 + 2v + 1 >

[tex]\int[/tex] (2v^2 + 4uv + 4)dudv
4[tex]\int[/tex] (v^2/2 + uv + 1) dudv


i know i can't go to polar so i integrate

is this the correct integral?

0 to 2, outside integral, 4* 0 to 4 inside integral ( v^2/2 + uv + 1 ) dudv ?
 

Answers and Replies

  • #3
Defennder
Homework Helper
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5
so then i have [tex]\int[/tex] <4v-v^3, 2u+v, v^2 + 4 - v^2 > x <0 + 2v + 1 >
You're not writing it out properly. It's supposed to be (0 2v 1)^T There isn't any "+".

[tex]\int[/tex] (2v^2 + 4uv + 4)dudv
4[tex]\int[/tex] (v^2/2 + uv + 1) dudv
How did you get v^2 ? And I still don't understand your notation. Why is there 2 separate integrals instead of a single double integral?
 
  • #4
You're not writing it out properly. It's supposed to be (0 2v 1)^T There isn't any "+".

How did you get v^2 ? And I still don't understand your notation. Why is there 2 separate integrals instead of a single double integral?
I'm sorry there's not. That's just the next step to solving the integral.

then i take the dot product of <4v-v^3, 2u+v, v^2 + 4 - v^2 > dot <0, 2v,1 >

I get < 0, 4uv + 2v^2, v^2 -v^2 + 4 > which simplifies into 2v^2 + 4uv + 4

so i integrate between...

[tex]\int[/tex] 0 to 2 and [tex]\int[/tex] 0 to 4

integrand of : 2v^2 + 4uv + 4 (dudv)
 
  • #5
Defennder
Homework Helper
2,591
5
Well ok, I can get that. So what's the problem now?
 
  • #6
HallsofIvy
Science Advisor
Homework Helper
41,833
955
I'm sorry there's not. That's just the next step to solving the integral.

then i take the dot product of <4v-v^3, 2u+v, v^2 + 4 - v^2 > dot <0, 2v,1 >

I get < 0, 4uv + 2v^2, v^2 -v^2 + 4 > which simplifies into 2v^2 + 4uv + 4

so i integrate between...

[tex]\int[/tex] 0 to 2 and [tex]\int[/tex] 0 to 4

integrand of : 2v^2 + 4uv + 4 (dudv)
So it is
[tex]\int_{v= 0}^2\int_{u= 0}^4 (2v^2+ 4uv+ 4)dudv[/tex]?

Looks pretty straightforward to me. What is the problem? That can, of course, be separated into
[tex]2\int_0^4 du\int_0^2 v^2dv+ 4\int_0^4 udu\int_0^2 vdv+ 4\int_0^4du\int_0^2dv[/tex]
 
  • #7
Defennder
Homework Helper
2,591
5
For some reason the "u=0" and "v=0" isn't showing up properly in Latex.
 

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