Quick Divergence Theorem question

In summary, the divergence theorem states that the flux density of a vector field is found by differentiation of its vector field vector.
  • #1
Stef42
10
0

Homework Statement


Use the divergence theorem in three dimensions

[tex]\int\int\int\nabla\bullet V d\tau= \int \int V \bullet n d \sigma[/tex]

to evaluate the flux of the vector field

V= (3x-2y)i + x4zj + (1-2z)k

through the hemisphere bounded by the spherical surface x2+y2+z2=a2 (for z>0) and the x-y plane

Hint: The direct evauation of the flux may not be the easiest way to proceed

Homework Equations





The Attempt at a Solution


I found it pretty simple which means I probably messed up (and I'm not sure what the hint is talking about
ok so the divergence is
[tex]\nabla \bullet V = 3-2=1 [/tex]

and the integral over the volume of the hemisphere (using spherical polar coordinates) is
[tex] \int_{0}^a r^2 d \tau \int_{0}^{2\pi} d\phi \int_{0}^{\pi/2} sin\theta d\theta = \frac{2\pi a^3}{3} [/tex]

So am I doing it completely wrong? I don't know the answer but if anyone could look through it and spot anything I would really appreciate it
 
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  • #2
Hi Stef42! :smile:

(have a del: ∇ and a theta: θ and a phi: φ and a sigma: σ and use \cdot instead of \bullet :wink:)
Stef42 said:
Use the divergence theorem in three dimensions

[tex]\int\int\int\nabla\bullet V d\tau= \int \int V \bullet n d \sigma[/tex]

to evaluate the flux of the vector field

Hint: The direct evauation of the flux may not be the easiest way to proceed

I found it pretty simple which means I probably messed up (and I'm not sure what the hint is talking about

Looks fine to me …

I think the hint just means don't use V.n, which is the flux, use ∇.V … though that's a bit unnecessary since they've already told you to :rolleyes:
 
  • #3
thanks for the speedy response tiny-tim :-)
Ok so just to clear up:
∇.V= divergence of vector field V
V.n= flux
So then the sigma integral is flux density ?
So what did I calculate? flux density aswell? wasn't I supposed to find the flux? :confused:
 
  • #4
Stef42 said:
So what did I calculate? flux density aswell? wasn't I supposed to find the flux? :confused:

oooh, I get confused between flux and flux density …

anyway, V.n is definitely fluxy, and ∇.V definitely isn't :wink:
 
  • #5
well, at least I'm not the only one who gets confused, damn vector calculus :frown:
well from my lectures notes:
"The divergence represents the flux density of the vector field and, because of the derivative operation, has an associated Fundamental Theorem of Calculus called the divergence theorem : (quotes equation)"

hmm think I'll go talk to him tomorrow :)
 

1. What is the Quick Divergence Theorem?

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

2. How is the Quick Divergence Theorem different from the standard Divergence Theorem?

The Quick Divergence Theorem is a simplified version of the standard Divergence Theorem, which is used in more complex mathematical calculations. It assumes that the vector field is constant over the surface, making it easier to apply in certain situations.

3. What are the practical applications of the Quick Divergence Theorem?

The Quick Divergence Theorem is commonly used in physics and engineering to calculate the flow of fluids and gases through a given surface. It can also be applied in electromagnetism to calculate electric and magnetic fields.

4. Can the Quick Divergence Theorem be used in 3-dimensional space?

Yes, the Quick Divergence Theorem can be applied in 3-dimensional space, just like the standard Divergence Theorem. It can be extended to higher dimensions as well.

5. How is the Quick Divergence Theorem proven?

The Quick Divergence Theorem can be proven using the fundamental theorem of calculus and the Cauchy integral theorem, which relate the integral of a function over a region to the values of the function on the boundary of the region.

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