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Computing Flux Integral

  • Thread starter andrec
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  • #1
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Hi, so this is my first time on this forum and I've just gotten very frustrated with trying to understand this one problem. I got part A by using divergence theorem but I don't know how go about computing part B.

Suppose F⃗ (x,y,z)=⟨x,y,5z⟩. Let W be the solid bounded by the paraboloid z=x2+y2 and the plane z=4. Let S be the closed boundary of W oriented outward.

(a) Use the divergence theorem to find the flux of F⃗ through S.
s F⃗ ⋅dA⃗ = 56pi

(b) Find the flux of F⃗ out the bottom of S (the truncated paraboloid) and the top of S (the disk).
Flux out the bottom =
Flux out the top =


Thanks for your help.

BTW, side question. What exactly is the difference in the Surface and Flux integral.
 

Answers and Replies

  • #2
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The flux integral is over a closed surface, so you can invoke the Divergence Theorem. For part b, you need to consider open surfaces (on which the divergence theorem no longer applies). You're going to have to figure out the normal vectors to the surfaces and actually do these 2d integrals by hand, so to speak.
 
  • #3
SammyS
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Hi, so this is my first time on this forum and I've just gotten very frustrated with trying to understand this one problem. I got part A by using divergence theorem but I don't know how go about computing part B.

Suppose F⃗ (x,y,z)=⟨x,y,5z⟩. Let W be the solid bounded by the paraboloid z=x2+y2 and the plane z=4. Let S be the closed boundary of W oriented outward.

(a) Use the divergence theorem to find the flux of F⃗ through S.
s F⃗ ⋅dA⃗ = 56pi

(b) Find the flux of F⃗ out the bottom of S (the truncated paraboloid) and the top of S (the disk).
Flux out the bottom =
Flux out the top =


Thanks for your help.

BTW, side question. What exactly is the difference in the Surface and Flux integral.
Hello andrec. Welcome to PF !

For (b), it's much easier to calculate the flux through the top surface, z=4. Use that and the result for part (a) to get the flux through bottom surface, z=x2+y2 .
 

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