1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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?

    Screenshot2012-05-25at40757AM.png


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

    sharks

    User Avatar
    Gold Member

    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

    sharks

    User Avatar
    Gold Member

    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

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    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].
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook