Flux Calculation for Vertical Cylinder with Top Cut Off

In summary, the author attempted to solve a problem by drawing a graph and calculating the surface integral. However, he ran into a problem when defining the function Phi. He found that the surface is not a disk of radius 1 and that the only plane where he could project the surface is the xy-plane. He then calculated the surface integral and found that it is negative.
  • #1
DryRun
Gold Member
838
4
Homework Statement
http://s1.ipicture.ru/uploads/20120119/gW4x6Mxi.jpg

The attempt at a solution
I drew the graph. It's a vertical cylinder, with its axis along the z-axis, with the top cut off by the plane z=3 but the cylinder goes down to infinity. The radius of the cylinder is 1.

[tex]\phi(x,y,z)=x^2+y^2-1[/tex]
[tex]∇\vec{\phi}=2x\vec{i}+2y\vec{j}[/tex]
Since the direction of unit normal vector is downward, [itex]-∇\vec{\phi}=-2x\vec{i}-2y\vec{j}[/itex]
[tex]\hat{n}=\frac{-∇\vec{\phi}}{|-∇\vec{\phi}|}[/tex]
[tex]\hat{n}=\frac{-x-y}{\sqrt{x^2+y^2}}[/tex]
[tex]Flux=\int\int\frac{-x^2-y^2}{\sqrt{x^2+y^2}}\,.d \sigma=-\sqrt{x+y}\,.d \sigma[/tex]
Projecting the surface S onto the yz-plane (i cannot project onto the xy-plane as the cylinder starts from z=0 and goes down to infinity):
[tex]x=\sqrt{1-y^2}[/tex]
[tex]x_y=-\frac{y}{\sqrt{1-y^2}}[/tex]
[tex]x_z=0[/tex]
[tex]d\sigma=\frac{1}{\sqrt{1-y^2}}\,.dzdy[/tex]
[tex]S.A.=2\int\int \frac{1}{\sqrt{1-y^2}}\,.dzdy[/tex]
The limits are -∞≤z≤3 and -1≤y≤1 but i think i messed up somewhere above, as these limits look suspicious.

When i visualize this thin slice of the surface at the plane z=3, it seems to me that if i project the thin slice onto the yz-plane, the slice would become invisible, as its thickness is zero. Am i right? But since we're dealing with flux here, then this slice does not have a thickness of zero. Correct?
 
Last edited:
Physics news on Phys.org
  • #2
can you explain a bit?

isn't the surface a disk radius one centred on the z axis and in the z=3 plane, giving [itex] \hat{n}=(0,0,-1)^T[/itex]
 
  • #3
lanedance said:
can you explain a bit?

isn't the surface a disk radius one centred on the z axis and in the z=1 plane, giving [itex] \hat{n}=(0,0,-1)^T[/itex]
No, it's in the plane z=3 and i don't understand how you got [itex] \hat{n}[/itex] directly?
From my notes, i must always follow this method:
[tex]\hat{n}=\frac{∇\vec{\phi}}{|∇\vec{\phi}|}[/tex]
 
  • #4
yeah was editing as you replied
 
  • #5
i'm not sure what you mean about a cylinder, isn't it just the disk as described?
 
  • #6
then only the z component of the field will contribute to the integral
 
  • #7
OK, i think i understand. For flat surfaces, the outward unit normal is +1 is the direction is along the positive side of the axis, otherwise it's negative.
So, for example, for plane x=2, [itex]\hat{n}=(1,0,0)^T[/itex]
And, for plane y=-3, [itex]\hat{n}=(0,-1,0)^T[/itex]
Are these correct?

So, for flat surfaces, there is no need to use this formula:
[tex]\hat{n}=\frac{∇\vec{\phi}}{|∇\vec{\phi}|}[/tex]
Correct?

But if i do use the formula, then i don't get -1. Instead i get this:
[tex]\hat{n}=\frac{-x-y}{\sqrt{x^2+y^2}}[/tex]
Using [itex]x^2+y^2=1[/itex], i can simply:
[tex]\hat{n}=\frac{-x-y}{\sqrt{x^2+y^2}}=-x-y[/tex]
But i don't know how it becomes -1.
 
Last edited:
  • #8
well the normal can be in the positive or negative direction depending on the surface orientation you choose, and that gets important when you start looking at stokes theorem, but yeah they are valid normals

the problem is the function phi you defined, when phi=0, it gives the unit cylinder, NOT the plane z=3.

and though the method is a good one, it's not really required as the surface normal is constant and easily to calculate

if you really want to use it the method, you should start with
phi=z-3
 
  • #9
just to clarify, you might want to re-read the question, the surface is z=3, the boundary is given by x^2+y^2=1 (the unit circle)
 
  • #10
Yes, i did a mistake with [itex]\phi[/itex].
[tex]\phi(x,y,z)=z-3[/tex][tex]∇\vec{\phi}=\vec{k}[/tex][tex]-∇\vec{\phi}=-\vec{k}[/tex][tex]\hat{n}=-\vec{k}[/tex][tex]Flux=\int\int \vec{F}.\hat{n}\,.d \sigma=\int\int-z\,.d \sigma[/tex]
Now, I'm not sure, but i think the only plane where i can project the surface z=3 is the xy-plane. Am i right?
So, projecting on the xy-plane:
[tex]z_x=0[/tex][tex]z_y=0[/tex][tex]\sigma=\sqrt{0+0+1}\,.dxdy[/tex][tex]S.A.=\int\int(-z)\,.dxdy[/tex][tex]S.A.=\int\int(-3)\,.dxdy[/tex]
Transforming to polar coordinates:[tex]S.A.=\int\int(-3)\,.dxdy=\int^{2\pi}_0\int^1_0(-3)\,.rdrd\theta=-3\pi[/tex]
The answer is wrong.
 
Last edited:
  • #11
why is it wrong?
 
  • #12
Thanks for answering, lanedance. I've been waiting for confirmation the whole day.

Well, the answer in my notes is: [itex]\frac{3\pi}{2}[/itex]
Furthermore, the answer in my notes is positive, whereas the answer i got is negative.
 
  • #14
I checked out the example in your link and there is this last part that i can't understand from the website:

At this point we can acknowledge that D is a disk of radius 1 and this double integral is nothing more than the double integral that will give the area of the region D so there is no reason to compute the integral. Here is the value of the surface integral.
eq0065M.gif


My question is: How can he directly know that the answer is [itex]\pi[/itex] without evaluating the integral?
At this point, I'm having a hard time to fully understand, as I'm still learning the fundamentals in this chapter, and my notes are apparently wrong.
 
  • #15
So the full surface vector integral is
[tex]
\int\int \vec{F}\cdot\vec{dS}
[/tex]

the surface differential is equivalent to an infintesimal area on the surface in the direction of the normal
[tex]
\vec{dS}
=\hat{n}dA
[/tex]

Thus the integral becomes
[tex]
\int\int \vec{F}\cdot\hat{n}dA
[/tex]

In our case the dot product is equal to -z, and z=3 on the surface
[tex]
\vec{F}\cdot\hat{n}=-z=-3
[/tex]

So the integral becomes
[tex]
\int\int \vec{F}\cdot\hat{n}dA
=-3\int\int dA
[/tex]

Now what is the area of our surface, a circle of radius 1? you could integrate, as you did with the change to polar coords, or just use the area [itex]\pi r^2=p\i[/itex]
[tex]
\int\int \vec{F}\cdot\hat{n}dA
=\int\int dA
[/tex]
 
  • #16
I see it now.
[tex]\int\int \vec{F}\cdot\hat{n}dA=-3 \int\int dA[/tex]
But since [itex]\int\int dA={\pi}{r}^2[/itex] where r=1
[tex]\int\int \vec{F}\cdot\hat{n}dA=-3 \int\int dA=-3\pi[/tex]
Thanks for the help, lanedance.
 

Related to Flux Calculation for Vertical Cylinder with Top Cut Off

1. What is flux?

Flux is a term used in physics and mathematics to describe the flow or movement of a quantity through a surface or boundary. It can refer to the flow of particles, energy, or other physical quantities.

2. How is flux measured?

Flux is typically measured in units of per unit time, such as particles per second or joules per second. This allows for the comparison of flux across different surfaces or boundaries.

3. What is the importance of evaluating flux across a surface?

Evaluating flux across a surface allows us to understand the rate at which a quantity is moving through that surface. This is crucial in many scientific fields, such as fluid dynamics, electromagnetism, and thermodynamics.

4. How is flux related to the concept of a vector field?

Flux is closely related to the concept of a vector field, which describes the direction and magnitude of a quantity at every point in space. Flux is the amount of that quantity passing through a given surface, and can be calculated by integrating the vector field over that surface.

5. What factors can affect the magnitude of flux across a surface?

The magnitude of flux across a surface can be affected by factors such as the surface area, the velocity or concentration of the quantity, and the angle at which the quantity is passing through the surface. The presence of obstacles or barriers can also affect the flux across a surface.

Similar threads

  • Calculus and Beyond Homework Help
Replies
3
Views
690
  • Calculus and Beyond Homework Help
Replies
6
Views
1K
  • Calculus and Beyond Homework Help
Replies
6
Views
1K
  • Calculus and Beyond Homework Help
Replies
20
Views
622
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
6
Views
868
  • Calculus and Beyond Homework Help
Replies
14
Views
828
  • Calculus and Beyond Homework Help
Replies
8
Views
2K
  • Calculus and Beyond Homework Help
Replies
8
Views
958
  • Calculus and Beyond Homework Help
Replies
1
Views
661
Back
Top