• Support PF! Buy your school textbooks, materials and every day products Here!

Divergence theorem with inequality

  • #1

Homework Statement



F(x,y,z)=4x i - 2y^2 j +z^2 k

S is the cylinder x^2+y^2<=4, The plane 0<=z<=6-x-y

Find the flux of F

Homework Equations



The Attempt at a Solution



What is the difference after if I change the equation to inequality?

For example :
x^2+y^2<=4, z=0

x^2+y^2<=4 , z=6-x-y

x^2+y^2=4, z=6-x-y

thanks!
 
Last edited:

Answers and Replies

  • #2
Math_QED
Science Advisor
Homework Helper
2019 Award
1,601
643

Homework Statement



F(x,y,z)=4x i - 2y^2 j +z^2 k

S is the cylinder x^2+y^2=4, 0<=z<=3

Homework Equations



The Attempt at a Solution


[/B]
I find that the answer is 84 pi

What is the difference after if I change the equation to inequality?

For example :
x^2+y^2<=4, z=3

x^2+y^2<=4 , z=0

X^2+y^2=4, z=3

thanks!
Make a sketch and everything will become clear.
 
  • #3
LCKurtz
Science Advisor
Homework Helper
Insights Author
Gold Member
9,544
756

Homework Statement



F(x,y,z)=4x i - 2y^2 j +z^2 k

S is the cylinder x^2+y^2=4, 0<=z<=3

Homework Equations



The Attempt at a Solution


[/B]
I find that the answer is 84 pi
Answer to what??

What is the difference after if I change the equation to inequality?

For example :
x^2+y^2<=4, z=3

x^2+y^2<=4 , z=0

X^2+y^2=4, z=3

thanks!
One inequality gives the surface of the cylinder and the other gives the solid cylinder.
 
  • #4
Sorry, I made some mistake previously and I modified the question.
If I want to find the flux of the bottom face, top face, side surface of cylinder and the tangent surface of the cylinder,
How can I express it in inequality?
 
  • #5
LCKurtz
Science Advisor
Homework Helper
Insights Author
Gold Member
9,544
756
Sorry, I made some mistake previously and I modified the question.
If I want to find the flux of the bottom face, top face, side surface of cylinder and the tangent surface of the cylinder,
How can I express it in inequality?
Bottom: ##\vec R(x,y) = \langle x,y,0\rangle,~0\le x^2+y^2\le 4##
Top: ##\vec R(x,y) = \langle x,y,3\rangle,~0\le x^2+y^2\le 4##
Side: ##\vec R(\theta,z) = \langle 2\cos\theta,2\sin\theta,z\rangle,~0\le \theta\le 2\pi,~0\le z\le3##
I have no idea what you mean by the "tangent surface".
You could, and likely should, use polar coordinates for the first two instead of ##x## and ##y## parameters.
 

Related Threads on Divergence theorem with inequality

Replies
3
Views
1K
Replies
1
Views
2K
  • Last Post
Replies
1
Views
3K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
9
Views
2K
  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
2
Views
5K
  • Last Post
Replies
5
Views
1K
Top