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Area of a circle and integral

  1. Jun 13, 2012 #1
    [tex] \int_{-\sqrt{r^2-x^2}}^{\sqrt{r^2-x^2}} \sqrt{r^2-x^2-y^2}dy [/tex]

    I can calculate the above integral [part of a double integral] by the conventional way [somewhat long], however my book says that this integral equals to [tex]\frac {\pi}{2}(r^2-x^2)[/tex] because the integral is actually the area of half a circle. I have difficulty to understand how this integral has something to do with half of a circle. [The author of my book thinks this is fact is trivial, however for me it is not =(]

    And if the above integral is the area of half a cicle than what about:
    [tex] \int_{-\sqrt{r^2-x^2}}^{\sqrt{r^2-x^2}} (r^2-x^2-y^2)dy [/tex]
    I'll appreciate any help.

    [EDIT: I think now I understand why the first integral is the area of half a cirlce, now thinking what I can tell about the second integral...]
    Last edited: Jun 13, 2012
  2. jcsd
  3. Jun 13, 2012 #2
    Well, if you look at the bounds, y = sqrt(r^2 - x^2)

    You can rearrange to get x^2 + y^2 = r^2
  4. Jun 13, 2012 #3
    Since you are integrating with respect to y, [itex]r^2 - x^2 = t^2[/itex] is a constant. So the integral becomes,

    [tex]\int_{-t}^{t} \sqrt{t^2-y^2}dy[/tex]

    Now, the function you are integrating can be taken as,

    [tex]X = \sqrt{t^2 - y^2}[/tex]

    And this is the equation of the semi-circle.

    Or, fully in circle form(the square root gives only positive solutions implying half circle),

    [tex]X^2 + y^2 = t^2[/tex]

    Where t is the radius. So, the integral you are trying to find is simply the area of a half circle of radius t.
  5. Jun 13, 2012 #4
    Thanks for the answers however I already understood the [tex]\int_{-\sqrt{r^2-x^2}}^{\sqrt{r^2-x^2}} \sqrt{r^2-x^2-y^2}dy[/tex] case.

    Looking now at [tex]\int_{-\sqrt{r^2-x^2}}^{\sqrt{r^2-x^2}} (r^2-x^2-y^2)dy[/tex]

    Can I compute this integral in similar way? I seems like this time it is an ellipse.
  6. Jun 13, 2012 #5
    Why do you think it is an ellipse?

    Using the same idea, you would have,

    [tex]X + y^2 = t^2[/tex]

    This doesn't resemble any of the standard shapes, I believe. ie circle, ellipse, square etc.
  7. Jun 13, 2012 #6
    So I guess my idea is wrong, I'm trying to calculate the following integral:

    [tex]\iint\limits_{x^2+y^2\leq R^2}(R^2-x^2-y^2)dxdy[/tex]

    Any ideas?
  8. Jun 13, 2012 #7
    I wonder what the idea was....

    How did you try this one?
  9. Jun 13, 2012 #8
    I tried a lot of things the last one was:
    https://dl.dropbox.com/u/27412797/hard_integral.jpeg [Broken]
    But calculating this integral is a lot of work, I was given a hint that I can use the symmetry of the area which is a circle, but still unable to find the proper way to use this symmetry.
    Last edited by a moderator: May 6, 2017
  10. Jun 13, 2012 #9


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    estro, this is such a huge scan! You could have used Microsoft Paint or any other image editor to cut the white parts and downsize it for viewing convenience. :smile:
  11. Jun 13, 2012 #10
    Sorry for that.
    Just cropped the scan, please refresh =)
  12. Jun 13, 2012 #11


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    Have you done polar coordinates? It's much simpler to evaluate this double integral if you first convert to polar coordinates.
    [tex]\iint\limits_{x^2+y^2\leq R^2}(R^2-x^2-y^2)dxdy=\int^{\theta=2\pi}_{\theta=0} \int^{r=R}_{r=0} (R^2-r^2).rdrd\theta[/tex]
    Note that ##x^2+y^2=r^2## where ##r## is the radius of the circle.
    ##x^2+y^2\leq R^2## describes a disc of radius R.
  13. Jun 13, 2012 #12
    I'm not allowed to use polar coordinates on this question. The instructor told us symmetry [of the area of integration] should be enough to solve it. =(
  14. Jun 13, 2012 #13


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    $$\iint\limits_{x^2+y^2\leq R^2}(R^2-x^2-y^2)dxdy$$
    First, try to describe the region of integration:
    For y fixed, x varies from ##x =-\sqrt{R^2-y^2} ## to ##x =\sqrt{R^2-y^2} ##
    y varies from ##y =-R## to ##y=R##
    Now that you have the limits, you should be able to evaluate the double integral. :smile:
  15. Jun 13, 2012 #14
    Hmm, the only way I can currently think of using symmetry is by splitting the term as [itex]\sqrt{R^2-x^2-y^2} \cdot \sqrt{R^2-x^2-y^2}[/itex]. Its not a very less tedious method, though(but it is interesting!)
  16. Jun 13, 2012 #15
    This is exactly what I did [see the attached scan], I'm looking for more elegant solution... =)

  17. Jun 13, 2012 #16
    Yes, I tried this one as well bu maybe I missed something let me recheck.
  18. Jun 13, 2012 #17
    Actually I have pretty nice solution, but I'm not sure my instructor meant it to be solved this way. I will post it in a hour. [I think you will like it]
  19. Jun 13, 2012 #18
    Nice, I'll ponder over it a bit more till then, getting somewhere. :smile:
  20. Jun 13, 2012 #19
    In the following proof I pretty much used only simple tools and intuition. I'm still looking for another way to solve it let me know if you have other ideas.

    The drawing is only 1/4 of the actual 3D body. [My drawing skills are as bad as my math=)]
    https://dl.dropbox.com/u/27412797/short_proof_2.JPG [Broken]
    Last edited by a moderator: May 6, 2017
  21. Jun 13, 2012 #20
    Nice and elegant :approve:
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