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Integral in physics

  1. Nov 9, 2005 #1
    Can you help me know more about the integral in physics??? Sometimes, I see some problems use this to solve. How can I figure out??? thank you.
  2. jcsd
  3. Nov 10, 2005 #2
    You usually use the integral when you want to do a sum of infinitisemally (Blah, my spelling sucks.. :P) small parts of something... or a sum of infinitesmally small parts of something multiplied by something else...
    For example... work..
    W = Fs (Where s is the distance in the direction of the force... leaving out the vector dot product)
    If you were to, instead of take s as a whole distance, break s up into differential elements of infinitesmally small value ds, the work done over this infinitesmally small distance would be:
    dW = F ds
    And that uses the concept of integrals in the sense that you want the TOTAL work done... i.e. the sum of all those differential elements:
    W = dW1 + dW2 + dW3..... + dWn
    Where n is where the segments finally end at the final distance.
    And since the integral is essentially an infinite amount of sums of small bits of... stuff.... you can use its concept to find the total sum.
    It's especially useful for the work done by a spring; since the force is non-constant overall, which would mean ks^2 wouldn't work as a formula...
    So, you use the concept of the integral...
    At any one point, the difference in work from the point just before it would be:
    dW = F ds
    You know that the spring force is F = ks.... so: (where k is a constant)
    dW = ks ds (Where ds is an infinitesmally small difference in distance)
    Since it would be long and boring to do this by hand, we use integration:
    W = integral (ks)ds = 0.5*ks^2
    And that's where that formula comes from...

    I probably made a mistake somewhere in my vague, useless, inaccurate attempt at helping you out, so I don't mind if someone corrects me...

    By the way, I think this should be in another forum. :P
  4. Nov 10, 2005 #3
    I'm working with the equation B=((mu sub naught)*i)/(2*pi*R). I know that R is changing, but I don't know what dB would look like. I thought that R in the original equation would be dR in the new one, but it doesn't make sense to me that dR would be to the negative one power. Do you have any ideas?
  5. Nov 10, 2005 #4
    well as in equation
    dB=((mu sub naught)*i)/(2*pi*dR) u can c dRis one of the denominator, to make it a neumerator, multiply and divide the equation with R^-1, there u have your dR with negative one in power. this is just done for the sake of simplicity coz it is sometimes easier to solve an intergral with variable in neumerator rather that otherwise.
  6. Nov 11, 2005 #5


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    I assume you're working with the equation for an infinitely long wire. The equation you have there was derived from a more general equation so if r varies you will not have to do any sort of integral. That equation is the complete picture in the limit that the wire is infinitely long. So in this case B varies as 1/r with a bunch of constants in front of it.
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