What is the integral of a vector field with the divergence theorem?

In summary: It's defined as 4 \pi \delta x \delta y \delta z but then I don't know how to apply Stokes' (which I guess to use because of the d^3 x and V in the integral. Could I split it into a triple integral and \delta x dx at a time?I know that I have to apply the div theorem somewhere, but this e^{-r} is confusing and what does it mean if the lower limit V is infinity?I haven't seen the integral of \frac{1}{e^r} before but I'm kinda guessing \int \frac{1}{e^r} \, dr
  • #1
SpY]
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Homework Statement


Evaluate the integral
[tex]\int\limits_{V=\infty} e^{-r} \left[ \nabla \cdot \frac {\widehat{r}} {r^2} \right] , d^3 x[/tex]

Homework Equations


Divergence theorem:
[tex]\int\limits_{V} \left ( \nabla \cdot A \right ) \, d^3 x
= \oint\limits_{S} A \cdot \, da}
[/tex]

The Attempt at a Solution


I know that I have to apply the div theorem somewhere, but this [tex]e^{-r}[/tex] is confusing and what does it mean if the lower limit V is infinity?
I haven't seen the integral of [tex]\frac{1}{e^r} [/tex] before but I'm kinda guessing
[tex] \int \frac{1}{e^r} \, dr
= \frac{1}{e^r} \int \frac{1}{u} \frac{du}{e^r}
= ln(e^r)
= r
[/tex]
where I used a substitution [tex]u=e^r[/tex] and [tex]du= e^r dr[/tex]
 
Last edited:
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  • #2
What is the divergence of vec(r)/r^2?

ehild
 
  • #3
It's defined as
[tex]4 \pi \delta x \delta y \delta z [/tex]

but then I don't know how to apply Stokes' (which I guess to use because of the [tex]d^3 x [/tex] and V in the integral. Could I split it into a triple integral and [tex]\delta x dx [/tex] at a time?
 
  • #4
I know that I have to apply the div theorem somewhere, but this [tex]e^{-r}[/tex] is confusing and what does it mean if the lower limit V is infinity?
I haven't seen the integral of [tex]\frac{1}{e^r} [/tex] before but I'm kinda guessing
[tex] \int \frac{1}{e^r} \, dr
= \frac{1}{e^r} \int \frac{1}{u} \frac{du}{e^r}
= ln(e^r)
= r
[/tex]
where I used a substitution [tex]u=e^r[/tex] and [tex]du= e^r dr[/tex]

This is wrong.

[tex]\frac{1}{e^r} = e^{-r}[/tex]

[tex]\int e^{-r} dr[/tex]

[tex]u=-r, du=-dr, -du=dr[/tex]

[tex]\int -e^u du = -e^u = -e^{-r} = \frac{-1}{e^r}[/tex]

The integral of e^-r isn't r, as that would imply that e^-r is a constant number.
 
  • #5
I agree, my above reasoning was useless

Ok so I can integrate the [tex]e^-r[/tex] but I don't think that really matters when there's a delta in the integral... my main problem is how to solve a third order delta integral, probably using the Divergence theorem because of the third order and volume. So we have

[tex]4 pi \int_{V=\infty} e^{-r} {\delta}^3 x z d^3 x[/tex]
 

What is the divergence theorem?

The divergence theorem is a fundamental theorem in vector calculus that relates the surface integral of a vector field to the volume integral of the divergence of that same field.

What is the significance of the divergence theorem?

The divergence theorem allows us to simplify calculations involving vector fields by relating the surface and volume integrals. It also has many important applications in physics and engineering.

How is the divergence theorem used in practice?

In practice, the divergence theorem is used to solve various types of problems involving vector fields, such as calculating fluid flow rates, electric flux, and heat transfer.

What is the formula for the divergence theorem?

The formula for the divergence theorem is ∫∫∫V(∇∙F)dV = ∫∫S(F∙n)dS, where ∇∙F represents the divergence of the vector field F, V is the volume bounded by the surface S, and n is the unit normal vector to the surface at each point.

What are some common misconceptions about the divergence theorem?

Some common misconceptions about the divergence theorem include confusing it with the gradient theorem (which relates line integrals to surface integrals) and not understanding the difference between a vector field and its divergence.

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