Is there a relationship between the Laplacian of a bivariate function and an iterated integral describing its "under-the-curve" volume within a certain domain and range? Just curious... If there is no such relationship, as is probably the case, could someone tell me what the purpose of the Laplacian, in multivariate functions, is exactly? Or, more generally, what the purpose of divergence is in vector fields? Correct me, please, if I'm misusing the terminology.
Hmmm... I think I've myself found the answer to my first question, which was badly phrased, but I would still like the second answered. Can anyone explain to me what divergence is, exactly? Mathematically, it's fairly simple... but visually ... errr Please Register or Log in to view the hidden image!
Isn't divergence where the area of a function goes to infinity? As it is impossible to visualise infinity, I can see your problem Please Register or Log in to view the hidden image!
The divergence of a vector field is basically how much "stuff" is leaving a region. Do you know the Divergence (or Gauss') Theorem yet? That might help your understanding. John-This is a different use of the word divergence.
Oh sorry yeh, he said . I missed that in the first post Please Register or Log in to view the hidden image!
I had thought divergence was basically a measure of how "different" the vector field became over space. I guess that if I combine that idea with a defined space-region with defined volume and surface, it is "how much stuff is leaving". Thanks, shmoe.
OK. I need major aid understanding multiple integrals. First off, there's Fubini's theorem. I undertand it, 'cept I don't know if it applies to both integrals over area and integrals over volume. In other words, does it apply to all multiple integrals? Next, an integral-over-area of a function is a double integral used on a bivariate function, and will result in a volume, correct? Just as an integral over the x-axis is single, used on a real function, and results in an area? Thirdly, does an integral over volume, used on a trivariate function, result in a '4-dimensional' content?
Yes it does, provided you meet the requirements of the theorem. Integrating a continuous function over a closed, bounded region will do it. There are weaker conditions, but I don't know a precise statement off-hand. Yes, yes, and yes, remembering that negative function values count as negative area/volume/whatever.
Excellent! Now I need to solidify my understanding of Gauss' Theorem. The notation ∫ dV V is a volume integral, and equivalent to ∫∫∫ dz dy dx, yes? If so, would ∫ dA A or ∫ dS S be the correct notation for an integral over area, or ∫∫ dy dx?
The dV is fine for integrating over a 3-d set. The dA is also fine for a 2-d set. The dS is not though, as ds is usually used for line integrals (the "ds" standing for the distance you move down the line). These are pretty standard in texts I can think of. As for your limits of integration, you can use whatever you like here. To make things confusing, if you're integrating over a solid 3-d shape (like the unit ball), you'll sometimes see ∫ dV V to mean the integral over the entire ball and ∫ dA S to mean the integral over the Surface of the ball.
