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Homework Help: Multiple integrals in polar form

  1. May 25, 2012 #1
    1. The problem statement, all variables and given/known data

    do you see how the integral of r is .5?


    I don't get how that follows?
  2. jcsd
  3. May 25, 2012 #2


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    First, label your limits:
    [tex]\int^{\theta=\pi}_{\theta=0}\int^{r=1}_{r=0} rdrd\theta[/tex]Then, integrate the inner-most integral and from there, integrate outwards, one integral at a time. So, in this case, integrate w.r.t.r first and then w.r.t.θ.
  4. May 25, 2012 #3
    still confused
  5. May 25, 2012 #4


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    This is the first step in solving any integral. You should understand which variable relates to the limits, then make the following modification.:
    [tex]\int^{\pi}_{0}\int^{1}_{0} rdrd\theta=\int^{\theta=\pi}_{\theta=0}\int^{r=1}_{r=0} rdrd\theta[/tex]
    Next, break the integrals down, starting with the inner-most integral:
    [tex]\int^{r=1}_{r=0} r\,.dr=answer[/tex]
    [tex]\int^{\theta=\pi}_{\theta=0} answer\,.d\theta=final\;answer[/tex]
  6. May 25, 2012 #5
    The book says the answer to this

    [tex]\int^{r=1}_{r=0} r\,.dr[/tex]

    is .5, I don't get that.
  7. May 25, 2012 #6
    Ok, I got it.

    the integral of r is r2/2, hence 1/2
  8. May 25, 2012 #7


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    The way I would do this is to note that the integral gives the area between the x-axis and the curve [itex]y= \sqrt{1- x^2}[/itex] from x= -1, to 1. [itex]y= \sqrt{1- x^2} is the part of [itex]y^2= 1- x^2[/itex] or [itex]x^2+ y^2= 1[/itex]. That is, it is the area of a semi-circle of radius 1 and so is [itex]\pi/2[/itex].

    Of course, [itex]\int_0^\pi d\theta= \pi[/itex] and [itex]\int_0^1 r dr= 1/2[/itex], as you say, so the double integral is [itex]\pi/2[/itex].
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