# Divergence theorem with inequality

## 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

## 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:

member 587159

## 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

## 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.

LCKurtz
Homework Helper
Gold Member

## 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

## 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!

One inequality gives the surface of the cylinder and the other gives the solid cylinder.

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?

LCKurtz
Homework Helper
Gold Member
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.