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How do integrals actually work?

  1. May 23, 2016 #1
    1. The problem statement, all variables and given/known data
    Calculate the electrical energy required to assemble a spherical volume of radius R and charge Q, homogeneous density ρ
    the answer is (3/5)Q/R

    the textbook says you have to build the volume integral one layer of sphere at a time, I'll get back to that later. I like investigating so I decided to try a different way, what I thought made sense was to do a super complicated "double-volume" integral where I take every charge element in the sphere -
    r2111
    and multiply by every other element in the sphere
    r2222
    which turns out to be impossibly complicated. which makes me wonder what would happen if I were to write a computer program to calculate an approximation...

    And then I notice the theorem the author gives out without proof, with pretext of a later proof to come. this is:
    U = 1/(8π)∫E2dv
    so I decide to give it a shot and it turns out pretty well yet sadly, incorrect:
    E(r) = 4πρ(4/3)πR3
    which turns out to be:
    E(r) = Qr/(R^3)
    and I'm suppose to integrate that term, which is not very promising considering I have Q^2 in the integrated term which will not cancel out to Q.

    in the actual answer they take the charge of a spherical volume of radius r multiply it by the next layer and devide by r:
    q*dq/r = ((4/3)πr3ρ * 4πr2drρ)/r
    that term is integrated with the integrand dr, and they get the right answer.

    which is great, and I understand the math perfect, I also understand the concept of multiplying the charge of the spherical volume with radius r, with the added layer, as in two charges being multiplied when calculating U. What I don't understand is why the distances work out in that calculations when the denominator is just r. The way I see it, if you want to find the electrical energy stored in a spherical volume and an additional layer, you will have to divide them into volume elements and then do the whole criss cross multiplication thing between each possible pair, that's just like multiplying the sums:
    (sum of all elements in the spherical volume)*(sum of all elements in layer) = sum of multiplications of every pair of elements from both groups.
    except it's not because for the distances, every multiplied pair of elements is divided by it's unique distance. Is there some kind of theorem unmentioned, about if the geometry of this multiplication is indeed spherical than all the different denominators can be exchanged via the radius of the sphere alone?

    That's one thing, what troubles me even more is that every time I want to do something that has to do with calculus, and was not explicitly and exclusively prescribed by my teacher, I run into insane equations making no sense. Why didn't my first attempt work at all, is there some kind of secret technique your suppose to use with calculus, that they only tell you in second degree math? did I not try hard enough? maybe if I would actually go through all the super impossible math I would actually find out everything cancels out and it works just fine. Whats going on?? And why didn't my second attempt work? is this formula only applicable for very specific situations?

    You know, when Newton invented calculus in his Mathematica Principia he wasn't all that satisfied with it at first. The publication was actually delayed by 20 years, partially because he had trouble accepting this advanced volume integral he had to calculate for the proof about the moon falling, so instead he took his sweet time rendering this genius geometrical proof without any calculus. after three years in high school going through this material once again, I figured maybe I'm looking in the wrong place. would reading Newton's proof for calculus help me understand anything? maybe I'm suppose to start reading about the philosophy of mathematics? I know they talk allot about concepts of numbers and stuff, but maybe there's a section about calculus.
     
  2. jcsd
  3. May 23, 2016 #2

    Jonathan Scott

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    Gold Member

    I expect you're supposed to know about the Shell Theorem, which says that the (gravitational or electrostatic) potential due to a uniform spherical shell is the same as that of a point source at the center with the same strength. If you're careful and patient, you should be able to check it by integrating it yourself, but it's a bit messy.
     
  4. May 23, 2016 #3
    I am familiar with the theorem, what about it?
     
  5. May 23, 2016 #4

    Jonathan Scott

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    Gold Member

    The shell theorem is why the distances work out in that way. If the shells so far are uniform and spherical, their total effect is the same as if a single source equal to the total sum of their source strengths is at the center, which is distance r from the particles of the next new shell at radius r.
     
  6. May 23, 2016 #5
    I did the integral and it came out perfect, and it also works out that if you do the integral than you have actually multiplied every infinitesimal layer by every other infinitesimal layer precisely once. I totally get it now, thanks allot. I still don't understand though why my attempts didn't work out.
     
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