How to find flux through certain sides of a surface with divergence theorem.

In summary, to calculate the flux through only some of the sides of a composite surface, you must parameterize each individual surface and then use the formula for a flux integral to calculate the flux through each surface separately.
  • #1
SiggyYo
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Homework Statement


Given a vector field [itex]\textbf{F}[/itex] and a composite (with this I mean cuboids, cylinders, etc. and not spheres for example) surface [itex]S[/itex], how do I calculate the flux through only some of the sides of [itex]S[/itex]? I am interested in a general way to do this, but right now I am struggling with the following:

Let [itex]\textbf{F}=\frac{xz}{x^2+y^2} \textbf{i}+\frac{yz}{x^2+y^2}\textbf{j}+ze^{x^2+y^2}\sqrt{x^2+y^2}\textbf{k},[/itex] defined on [itex]\mathbb{R}^3[/itex]

and [itex]E\subset\mathbb{R}^3[/itex] be the solid body given by
[itex]\sqrt{x^2+y^2}\leq z\leq2,\ x^2+y^2\geq 1.[/itex] (This is a sort of hollow stumped cone)

a) A part of [itex]E[/itex]'s edge is the cylinder part [itex]S_0[/itex] given by
[itex]x^2+y^2=1, 1\leq z\leq 2[/itex]
Show that the flux of [itex]F[/itex] inward through [itex]S_0[/itex] is [itex]-3\pi.[/itex]

Homework Equations


The divergence theorem.

The Attempt at a Solution


[itex]div{\textbf{F}}=e^{x^2+y^2}\sqrt{x^2+y^2}-2z\frac{x^2-y^2}{x^2+y^2},[/itex]
and with [itex]x=\cos{\theta}[/itex] and [itex]y=\sin{\theta}[/itex] we get
[itex]div{\textbf{F}}=e-2z(1-\sin^2{\theta}).[/itex] Our triple integral then becomes

[itex]\int_1^2\int_0^1\int_0^{2\pi}re-2rz(1-\sin^2{\theta})\,d\theta\,dr\,dz=2e\pi^2-\frac{3}{2}\pi[/itex].

And that is how far I've gotten. Assuming all of this is correct, my problem now is that this is the flux through the entire cylinder, along with its top and bottom. How to I calculate the flux through these discs, so that I can subtract them from the total flux?

Thanks in advance.

SiggyYo
 
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  • #2
SiggyYo said:

Homework Statement


Given a vector field [itex]\textbf{F}[/itex] and a composite (with this I mean cuboids, cylinders, etc. and not spheres for example) surface [itex]S[/itex], how do I calculate the flux through only some of the sides of [itex]S[/itex]? I am interested in a general way to do this, but right now I am struggling with the following:

Let [itex]\textbf{F}=\frac{xz}{x^2+y^2} \textbf{i}+\frac{yz}{x^2+y^2}\textbf{j}+ze^{x^2+y^2}\sqrt{x^2+y^2}\textbf{k},[/itex] defined on [itex]\mathbb{R}^3[/itex]

and [itex]E\subset\mathbb{R}^3[/itex] be the solid body given by
[itex]\sqrt{x^2+y^2}\leq z\leq2,\ x^2+y^2\geq 1.[/itex] (This is a sort of hollow stumped cone)

a) A part of [itex]E[/itex]'s edge is the cylinder part [itex]S_0[/itex] given by
[itex]x^2+y^2=1, 1\leq z\leq 2[/itex]
Show that the flux of [itex]F[/itex] inward through [itex]S_0[/itex] is [itex]-3\pi.[/itex]

Homework Equations


The divergence theorem.


The Attempt at a Solution


[itex]div{\textbf{F}}=e^{x^2+y^2}\sqrt{x^2+y^2}-2z\frac{x^2-y^2}{x^2+y^2},[/itex]
and with [itex]x=\cos{\theta}[/itex] and [itex]y=\sin{\theta}[/itex] we get
[itex]div{\textbf{F}}=e-2z(1-\sin^2{\theta}).[/itex] Our triple integral then becomes

[itex]\int_1^2\int_0^1\int_0^{2\pi}re-2rz(1-\sin^2{\theta})\,d\theta\,dr\,dz=2e\pi^2-\frac{3}{2}\pi[/itex].

And that is how far I've gotten. Assuming all of this is correct, my problem now is that this is the flux through the entire cylinder, along with its top and bottom. How to I calculate the flux through these discs, so that I can subtract them from the total flux?

Thanks in advance.

SiggyYo

I didn't check all your work, but the limits on your last integral for r aren't correct. r doesn't go from 0 to 1; it goes from 1 to 2.

The way you calculate flux through a surface S is to parameterize it:
##\vec R(u,v) = \langle x(u,v),y(u,v),z(u,v)\rangle##. Then you use the formula for a flux integral$$
\iint_S \vec F\cdot d\vec S = \pm\iint_{(u,v)}\vec F(u,v)\cdot \vec R_u\times
\vec R_v\, dudv$$ where the ##\pm## sign is chosen so that the direction of ##\pm\vec R_u\times\vec R_v## agrees with the orientation. In your problem, for the cylinder part you might use the parameterization$$
\vec R(\theta,z)=\langle \cos\theta,\sin\theta,z\rangle$$to get started. Then you have two more surfaces to work similarly with their appropriate parameterizations.
 
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Related to How to find flux through certain sides of a surface with divergence theorem.

1. What is the divergence theorem?

The divergence theorem, also known as Gauss's theorem, is a mathematical statement that relates the flux through a closed surface to the divergence of a vector field inside the surface. It is often used in physics and engineering to calculate the flow of a vector field through a surface.

2. How do you find the flux through a specific side of a surface using the divergence theorem?

To find the flux through a specific side of a surface using the divergence theorem, you first need to calculate the divergence of the vector field at each point on the surface. Then, you can integrate the divergence over the surface to find the total flux through that side of the surface. This can be done using either a surface integral or a triple integral, depending on the complexity of the surface and vector field.

3. What is the significance of finding the flux through certain sides of a surface?

The flux through certain sides of a surface can provide valuable information about the behavior of a vector field. For example, in fluid dynamics, the flux through a surface can tell us how much fluid is flowing through a specific area. In electromagnetism, the flux through a surface can indicate the strength of an electric or magnetic field in that region.

4. Can the divergence theorem be applied to any vector field and surface?

Yes, the divergence theorem can be applied to any vector field and any closed surface. However, it is important to note that the surface must be closed, meaning it forms a complete boundary around a specific region in space. Additionally, the vector field must be continuous and differentiable over the surface.

5. Are there any limitations to using the divergence theorem to find flux through surfaces?

One limitation of the divergence theorem is that it only applies to closed surfaces. This means that it cannot be used to find the flux through open surfaces, such as the surface of a sphere. Additionally, the vector field must be defined and continuous over the entire surface for the theorem to be applicable.

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