Can You Solve the Divergence Theorem for a Cylindrical Vector Field?

In summary: The integral over the side is \int_{\theta= 0}^{2\pi}\int_{r= 0}^3 \vec{F}\cdot (\vec{k} drd\theta)+ \vec{F}\cdot (-\vec{k}drd\theta)= 2\int_{\theta= 0}^{2\pi}\int_{r=0}^3 z^2 drd\theta= 2\int_{\theta= 0}^{2\pi} 9z^2 d\theta= 18\pi z^2.In summary, the divergence theorem was used to compute the flux integral for the vector field F = <
  • #1
Baumer8993
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Homework Statement


Verify the divergence theorem by computing both integrals for the vector field
F = <x^3, y^3, z^2> over a cylindrical region define by x^2+y^2 ≤ 9.


Homework Equations


Divergence Theorem, and Flux Integrals.


The Attempt at a Solution


I did the divergence theorem, and got 279 pi for my answer. I did the integral in cylindrical, and regular with the same answer so that I know is correct. I know I need to break this up into three different surfaces. I can do the top, and the bottom easily since N = <0, 0, 1> for the top, and N = < 0, 0, -1> for the bottom, then I just plug in the z value for it.

The side of the cylinder is giving me trouble. I have looked up ways to do it online, but they make no sense, or are disorganized. Where do I start, and then where do I go from there?
 
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  • #2
You haven't completely stated the problem. You say it is over the cylidrical region [itex]x^2+ y^2\le 9[/itex] but don't say what the ends are. What values of z?

You want to integrate over the surface of the cyinder. Any point on the circle [itex]x^2+ y^2= 9[/itex] can be written [itex](3cos(\theta), 3sin(\theta))[/tiex] with [itex]\theta[/itex] going from 0 to [itex]2\pi[/itex]. They cylinder has axis along the z-axis so we can take any point on the cylinder as given by [itex]\vec{r}(\theta, z)= 3cos(\theta)\vec{i}+ 3sin(\theta)\vec{j}+ z\vec{k}[/itex].

The [itex]\vec{r}_\theta= -3sin(\theta)\vec{i}+ 3cos(\theta)\vec{j}[itex] and [itex]\vec{r}_z= \vec{k}[/itex] and the cross product of those vectors gives the vector differential of surface area, [itex](3cos(\theta)\vec{i}+ 3 sin(\theta)\vec{j})d\theta dz[/itex].

The two ends will have vector differential [itex]\vec{k} drd\theta[/itex] and [itex]-\vec{k}drd\theta[/itex] with r going from 0 to 3 and [itex]\theta[/itex] going from 0 to [itex]2\pi[/itex]
 

FAQ: Can You Solve the Divergence Theorem for a Cylindrical Vector Field?

What is the Divergence Theorem?

The Divergence Theorem is a mathematical theorem that relates the flux (flow) of a vector field through a closed surface to the divergence of the vector field within the enclosed volume.

How is the Divergence Theorem useful?

The Divergence Theorem is useful in many areas of science, including fluid dynamics, electromagnetism, and engineering. It allows for easier calculation of flux through complex surfaces and can also be used to solve differential equations.

What is the formula for the Divergence Theorem?

The formula for the Divergence Theorem is: SF ⋅ dA = ∫V&nabla ⋅ F dV, where F is the vector field, S is the closed surface, dA is the surface element, and V is the enclosed volume.

What is the relationship between the Divergence Theorem and Gauss's Law?

Gauss's Law is a special case of the Divergence Theorem, where the vector field is the electric field and the closed surface is a Gaussian surface. The Divergence Theorem allows for the calculation of the electric flux through any closed surface, while Gauss's Law specifically relates the flux to the enclosed charge.

How can the Divergence Theorem be proved?

The Divergence Theorem can be proved using the fundamental theorem of calculus and the concept of Green's theorem. It can also be proved using the concept of divergence itself, which is defined as the limit of a flux divided by the enclosed volume as the volume approaches zero.

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