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Why use integral?

  1. Jul 2, 2013 #1
    Hi, im 15 years old and I became fasinated about these sumbjects of integral and differentiation.
    My question is, as far as I know integral is used to calculate area under a curve but I dont know why in many physics equasions its been used where there is not even a graph and a curve to find the area!!

    The question doesn't even ask to find any area but still I can see Einstein used lots of integrals in general relativity.

    I mean where would you use integral or differentiation even though no graph and no curve is given and no area is needed to be found.

    Note: I might not be even right in my question, if so please point it out.
    Thanks.
     
  2. jcsd
  3. Jul 2, 2013 #2
    Hi, im 15 years old and I became fasinated about these sumbjects of integral and differentiation.
    My question is, as far as I know integral is used to calculate area under a curve but I dont know why in many physics equasions its been used where there is not even a graph and a curve to find the area!!

    The question doesn't even ask to find any area but still I can see Einstein used lots of integrals in general relativity.

    I mean where would you use integral or differentiation even though no graph and no curve is given and no area is needed to be found.

    Note: I might not be even right in my question, if so please point it out.
    Thanks.
     
  4. Jul 2, 2013 #3

    mathman

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    Many physics models are in terms of derivatives, which describe local behavior of some system. To get a global solution requires some sort of integration.
     
  5. Jul 2, 2013 #4

    WannabeNewton

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    Integrals actually show up very rarely in general relativity (GR) compared to other theories like quantum mechanics and electromagnetism. In GR your sight will be drowned out by a sea of derivatives everywhere :)
     
  6. Jul 2, 2013 #5
    The short answer is that "area under the curve" is only one application/one way of thinking about the integral. I could go ahead and give you the mathematical definition of an integral which would probably be a bit confusing, but eventually it might make sense, but instead try an experiment.

    If an object is moving at a certain velocity for a certain amount of time, then it moves a distance equal to that velocity times the amount of time. Draw a graph of ##v(t)=1## and draw a graph of ##x(t)## that corresponds to that. Do the same thing with another simple velocity function like ##v(t)=t##.

    This is probably the simplest case of an integral in physics. In accordance to what mathman posted, note that ##v(t)=\frac{dx}{dt}##. I'm purposefully leaving a lot out, so if the point is unclear, let me know.
     
  7. Jul 2, 2013 #6
    Basically a lot of things in physics have to do with rates of change and the relationship between different variables.

    For example on a speed-time graph (y axis: speed, x axis: time), the gradient of the slope is acceleration, and the area under the curve is distance.

    Finding the area under the curve has given us new information about the "curve" that we were investigating. As you get further on into physics and mathematics, integration becomes a key tool in finding out useful information about the equations that you use. Graphs and "curves" don't need to be used since people who write papers on general relativity don't need geometric interpretations to guide them anymore.

    And high-5 to a fellow 15 year old :smile:
     
  8. Jul 2, 2013 #7
    The pictorial representation of a derivative as a tangent slope and an integral as the area under a curve are mostly instructional or didactic tools that are presented for conceptual purposes. Certainly, you can find the area under a curve using an integral, but it's application extends far beyond this.

    Specifically, physicists model real world phenomena largely through the construction of differential equations (DE). These equations are designed to identify how different aspects of the system change under various circumstances and how much they change. Once these parameters have been established and incorporated into full equations or coupled sets of equations, the equation(s) is "solved" through integrating or "anti-differentiating" the equation in order to get rid of the derivative terms. Once this is achieved, all that is needed is for a set of initial conditions to be entered into the solved equation in order to witness how a system evolves in time under its native dynamics. Of course, most DE's, such as most solutions to Einstein's GR equation, do not have exact or analytical solutions so they must be estimated using a technique called numerical integration.

    In any case, hope that helped. If you continue your studies in calculus, you'll gain a better understanding of the flexibility of the use of the integral probably sooner than later. You're right at the beginning stage now.
     
  9. Jul 2, 2013 #8

    WannabeNewton

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  10. Jul 2, 2013 #9

    atyy

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    In physics, the integral is always the area under a curve - or the volume enclosed. Are there any exceptions to this?

    Philosophical question: would you allow it to start from an integral?
     
  11. Jul 2, 2013 #10

    WannabeNewton

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    I'll allow anything for a price :biggrin: I'm not really a fan of integrals but I'd allow it just to fit in with all the cool people :smile:
     
  12. Jul 2, 2013 #11

    atyy

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    So you really like "differential geometry", but you're not a fan of "integral geometry" (I just made up the word :redface:) and forms?
     
  13. Jul 2, 2013 #12

    WannabeNewton

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    Well "differential forms" has the word differential in it so I love the subject. But death to "integral geometry"!
     
  14. Jul 2, 2013 #13

    atyy

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    Well argued!

    OK, maybe I was too hasty with saying all integrals in physics are areas under curves.

    The most common integral in physics is something like the "work done", which is the integral of force along a path. More generally, this is the signed definite integral, which perhaps is more a weighted sum of oriented lengths, areas or volumes. Is it still legitimate to think of those as "area under a curve"?
     
  15. Jul 2, 2013 #14

    WannabeNewton

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    I don't know honestly. I never think of integrals as area under a curve in physics as I find it rarely if ever helps me solve problems. I think of them in the way I learned to think of them from reading "An Introduction to Mechanics" by Kleppner and Kolenkow and doing the problems in the text, which is as a sum of infinitesimal elements associated with whatever quantity we are interested in, as I described in the link in post #6:

    "As far as physics goes I just view them as sums of infinitesimal elements...For example if I have charge distributed over a region I would start with an infinitesimal element of the region on which there is some charge and add up all such elements throughout the region to get the total charge. This ends up just being the integral of the charge density over that region. Geometrically it's a way of dealing with quantities associated with continuous distributions as opposed to discrete distributions... "

    Another example would be a man pulling a block with a rope that has non-negligible mass, with the problem being to figure out the tension in the rope as a function of position. One could look at the free body diagrams of infinitesimal rope elements and then integrate along the entire rope, i.e. add up the infinitesimal elements, to find the total tension across the rope.
     
  16. Jul 2, 2013 #15
    Thanks, I found that quiet useful.
     
  17. Jul 2, 2013 #16
    Thanks everyone, but I didn't really get the last dicussion anyways.
     
  18. Jul 2, 2013 #17
    Sometimes people go off on a tangent. Read WannabeNewton's link and DiracPool's post. See what makes sense and what doesn't there.
     
  19. Jul 3, 2013 #18

    atyy

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    Naively, I think of integration as something like ∫f(x,y)dxdy. Then dxdy is the area of a small parallelogram, and f(x,y) is the height, so f(x,y)dxdy is "height * area" which is "volume" under the surface f(x,y), so it seems to me the same concept as "area under a curve".

    In the case of the charge density, let's say surface charge density over an area, the integral has the same form, so it can also be thought of as "area under a curve".

    Spivak (Calculus on Manifolds, p100) says if ω is a k-form on [0,1]k, then ω = f dx1 ∧ ... ∧ dxk, for a unique function f. Then he defines ∫ω = ∫f , and expresses it a few lines later as ∫f(x1,...,xk) dx1...dxk, which seems similar to the "area under the curve" idea.

    Earlier in the book, when discussing Fubini's theorem (p57), he says the volume under the graph of f is something like ∫f (I'm paraphrasing).

    Is that a definition that isn't general? I guess the two things that are different are that in some cases, like a sphere, I can't cover the whole area I'm integrating over with one set of coordinates. Secondly, the area is oriented, so integrating along a line in one direction gives the negative of the integrating in the other direction. But neither of those seem markedly different from the idea of "area under a curve".

    Edit: There seems to be something similar at the top of p9 of http://www.math.ucsd.edu/~aminor/2013winter20E/forms_TerenceTao.pdf , which discusses the relationship between integration of forms and the Lebesgue or Riemann integral.

    ???
     
    Last edited: Jul 3, 2013
  20. Jul 3, 2013 #19

    HallsofIvy

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    Exceptions? I have to say that is simply not true! In physics, the integral is distance traveled, given a speed function. Or, in physics, the integral is speed attained given an acceleration function. Or, in physics, the integral is the mass of an object given a density function. Or, in physics, the integral is the work done on an object moving under a given force function.

    I don't know what "it" refers to here. Would I allow what to "start form an integral"? And what do you mean by "start from an integral"?
     
  21. Jul 3, 2013 #20

    atyy

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    Let's take the distance travelled example. The distance d for constant speed v travelled for time t is d=vt, which is the area under the speed-time graph. That's what I was thinking of when I said integration is the area under a curve.

    By "it" I meant general relativity. I was referring to the Hilbert action, which is defined using integration. If one extremizes the Hilbert action, the Einstein field equations result. So in this sense, the Hilbert action is a "starting point" for general relativity. It's especially useful if one wants to implement the equivalence principle, because if minimally coupled matter actions are added to the Hilbert action, we get the correct Einstein field equations, equations of motion for the matter, with the form of the stress-energy tensor such that the principle of equivalence is respected.
     
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