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Homework Help: Integrals as the area under a curve

  1. Jun 7, 2012 #1
    I have always seen the integral as the area under a curve. So for instance, if you integrated over the upper arc of a circle you would get ½[itex]\pi[/itex]R2.
    But then, I learnt to do integrals in spherical coordinates and something confuses me. If you do the integral from 0 to 2[itex]\pi[/itex], you don't get the area of a circle - you get the length of the circumference. Why is that? Certainly you are integrating over the arc of a circle? I can see, that you would need to integrate from 0 to R to actually get the formula for the area of a circle, but then you do a double integral, whereas in cartesian coordinates you would only integrate over y = ±√(R2-x2). Isn't there some sort of mismatch here?

    Edit: Yes, indeed when thinking of it, I am surprised that you do surface integrals as double integrals, when a single integral already yields and area?
  2. jcsd
  3. Jun 7, 2012 #2
    The integral of what?
  4. Jun 8, 2012 #3
    Well just ∫rdθ from 0:2[itex]\pi[/itex]
  5. Jun 8, 2012 #4
    The reason you don't get an area can be shown on dimensional grounds. When you integrate [itex]\int y(x) \; dx[/itex], you get an area because [itex]y,x[/itex] both have the same dimensions (usually length).

    When you integrate [itex]\int r \; d\theta[/itex], you don't get an area because while [itex]r[/itex] has dimensions of length, [itex]\theta[/itex] does not. The problem is that when you write an integral, it doesn't know what the underlying geometry is. For all it knows, [itex]r,\theta[/itex] belong to a rectangular coordinate system. To get the right result, you have to consider the geometry in question and put in something physically sensible.

    Alternatively, you could say that integerals really don't spit out areas unless they're double integrals, and really what you've done when you integrate is

    [itex]\int_a^b \int_0^{y(x)} 1 \; dy \; dx[/itex]

    Setting up the polar coordinates integral the same way should get the correct result without having to stop and think about it in a fuzzy way.
  6. Jun 8, 2012 #5


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    Certainly integrals can be used to find "area under a curve" but that is only an application- it is not what the integral is. You are very much over simplifying to think of the integral as being automatically "area".

    As far as [itex]\int_0^{2\pi} r d\theta[/itex] is concerned, it is, of course, [itex]2\pi r[/itex], the circumference of the circle. That's because, if we were to divide the entire circle, of radius r, into very thin sectors, with base angle [itex]d\theta[/itex], the lenth of the arc would be [itex]r d\theta[/itex]. I don't understand why you would think it would be the area.

    If you divide a circle, of radius r, into very thin sectors, with base angle [itex]d\theta[/itex], we can approximate each sector by as triangle with base [itex]rd\theta[/itex] , as before, and height r. Such a triangle has area "1/2 base times height" or [itex](1/2)(r d\theta)(r)= (1/2)r^2 d\theta[/itex]. We can approximate the area of the entire circle by [itex]\sum (1/2)r^2 d\theta= (1/2)r^2 \sum(d\theta)[/itex] since "1/2" and "r" are constants. Of course, [itex]\sum d\theta= 2\pi[/itex] so the are is [itex](1/2)r^2(2\pi)= \pi r^2[/itex].
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