Problem with the Divergence Theorem

In summary, the conversation is about a student seeking help with a homework question involving triple integrals. They have already solved part a) but are struggling with part b) and are looking for clarification on the correct approach. They also mention a possible mistake in part 3a) and seek confirmation before contacting their lecturer. Another person in the conversation points out a mistake in the original expression. The conversation ends with the student expressing gratitude for the help.
  • #1
Claire84
219
0
I was wondering if someone could give me a hand here with 2b) on the following link.
http://www.am.qub.ac.uk/users/j.mccann/teaching/ama102/2003/assignments/assign_8.pdf
For part a) I got it to be equal to 3x^2+3y^2+3z^2+2y-2xy, and I'm hoping that's right!

However, for part b) I can't seem to get the answer they're after at all. We haven't covered any examples of this in our lectures yet and we won't have any lectures before the homework has to be handed in due to May Day, so any help would be much appreciated. Just to make sure I started it off right, could you just check if this is correct or not (sorry, I'm no good with LaTex!)-

triple integral of (3r^2sin^2(theta)cos^2(f)+ 3r^2sin^3(theta)sin^2(f) + 3r^2cos^2(theta) + 2rsin(theta)sin(f) - 2r^2sin^2(theta)cos(f)sin)f))r^2sin(theta)drdfdtheta

where the integral with respect to r is within the limits 0 and a, with respect to f is 0 to 2pi and with respect to theta is 0 to pi (where f is the asimuthal angle or whatever it's called). I'd really appreciate if you could just ehck I've done that bit okay, because it'd be a bit pointless me running through it if the mistake was in the first line.

Btw, the answer I get at the end has an 8 on the numerator instead of a 12, so the answer isn't completely far out so I'm hoping it's just a wee mistake somewhere. Thanks! :smile:


Btw, sorry if this is posted in the wrong forum!
 
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  • #2
The divergence should be 3r^(2). I'll look a bit further into this..
 
  • #3
Ok:
If you go back to your vector field, you'll see that from both the j'th and k'th component, you will gain a 2xy term, but with opposite signs.
In your original expression, you've ended up with 2y-2xy instead
 
  • #4
Ah so shold it look like this then-

3x^2 + 3y^2 + 3z^2 ?

Gah I can be so stupid with even the simplest things!

Thanks for helping btw! :smile:
 
  • #5
That's what I got, too (3r^(2)).
 
  • #6
Btw, for part 3a), that's a mistake, right? I mean I keep getting xi - yj + (-y^2 - x)k and I asked one of the phd students about it and they got the same but I just want to check before I email the lecturer about it...
 
  • #7
Agreed, he'll be writhing in shame..
 
  • #8
Hope that wasn't sarcasm there. :tongue:
 

1. What is the Divergence Theorem and why is it important?

The Divergence Theorem is a mathematical principle that relates the flow of a vector field through a closed surface to the behavior of the vector field inside the surface. It is important because it allows us to calculate the flux (flow) of a vector field through a closed surface without having to integrate over the entire surface. This makes calculations much simpler and more efficient.

2. What is the problem with the Divergence Theorem?

The problem with the Divergence Theorem is that it only applies to certain types of vector fields and closed surfaces. If these conditions are not met, the theorem cannot be used and alternative methods of calculation must be used.

3. What types of vector fields and closed surfaces does the Divergence Theorem apply to?

The Divergence Theorem applies to vector fields that are continuous and have continuous first-order partial derivatives. It also applies to closed surfaces that are smooth, meaning they have no sharp corners or edges.

4. How do you know when you can use the Divergence Theorem?

In order to use the Divergence Theorem, you must first check if the vector field and closed surface meet the requirements mentioned in question 3. If they do, then you can use the theorem to calculate the flux. If not, you will need to use another method of calculation.

5. Can the Divergence Theorem be applied in three dimensions?

Yes, the Divergence Theorem can be applied in three dimensions. In fact, it is most commonly used in three dimensions to calculate the flux of a vector field through a closed surface. It can also be extended to higher dimensions, but this is beyond the scope of most applications.

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