Integrate divergence of a vector over an area

[JMoody]

Hello, I'm hoping somebody can give me some insight on how to solve this problem. This was a solid mechanics exam question and I wasn't able to finish it because I'm rather weak in math.

1. Homework Statement

Homework Equations

Recall divergence theorem for part ii. ∫div(V)dA = ∫V⋅ndS where n is normal to the surface.

The Attempt at a Solution

Part i.
Not an issue, I can solve it easily.

Part ii.
I can apply divergence theorem no problem.
For the n vector field I get:
n₁ = -e₂
n₃ = -e₁
n₂ = .707(e₁ + e₂)

The problem is that I've been out of school for quite a while, and I can't remember how to successfully integrate over the surface (perimeter) now. I realize I need to break it up into parts for each line of the surface, but it's been years since I've taken calculus.

 

[RUber]

For c1, you will have $\int_0^1 V\cdot n_1 \, dx_1$.
For c3, you will have $\int_0^1 V\cdot n_2 \, dx_2$.
For c2, it gets a little tougher.
$\int_{c2} V\cdot n_2 dc2$

 

[JMoody]

For c2, should I do a double integral (x1 and x2, both from 0 to 1) of V⋅n2?

 

[RUber]

It might be easier to directly compute the Area integral
$\int_0^1 \int_0^{1-x_1} div(V) dx_2dx_1$

 

[RUber]

For c2, should I do a double integral (x1 and x2, both from 0 to 1) of V⋅n2?

I don't think that is right, you would be taking the integral over an area. you want the integral over the line.

 

[JMoody]

Thanks RUber, I think you've provided the help I needed to finish it. It was required to perform divergence theorem for the exam though, I'll give it a shot with both methods and see how the solutions compare.

 

[RUber]

It might be easier to directly compute the Area integral
$\int_0^1 \int_0^{1-x_1} div(V) dx_2dx_1$

Nevermind this...that doesn't make any sense.
---
Try integrating over x1, and define x2 in terms of x1 along the line. That should do the trick.

 

[JMoody]

I solved it, I get 1/2 using both methods. Thanks again for the help RUber, it helped set me on the write path.

 

[RUber]

When I worked it out, I still had the $\frac 1 {\sqrt{2}}$ in the answer.

 

[JMoody]

I'm not sure where the square root comes from, but I believe the accepted solution was 1/2 when we went over the exam (he just didn't take the time to go over something trivial like integration since most of my peers are coming straight out of undergrad >.<). I calculated it both by using divergence theorem to integrate over the line (more complicated) and by integrating the divergence over the area directly.

 

[RUber]

Aha, I forgot to multiply by the $\sqrt{2}$ in the dS term.
Good work.

 

[BvU]

Why (in post #7) "Nevermind this...that doesn't make any sense." ?
I thought I had div V = 3x₂ and that would be a simple integral, so I'm grossly overlooking sonething ?

 

[RUber]

Thanks BvU, I was clearly overthinking things and convinced myself I was wrong.