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Homework Help: Evaluate the integral by changing to cylindrical coordinates.

  1. Jun 3, 2010 #1
    I wish I knew how to type this out with the proper symbols but here it goes. It says to change the following to cylindrical coordinates and evaluate

    (x^2 + y^2)^(1/2) dz dy dx where -3<=x<=3, 0<=y<=(9-9x^2)^1/2, 0<=z<=9-x^2-y^2


    2. Relevant equations



    3. The attempt at a solution

    I got 162pi/5

    Would anybody mind confirming?

    Cheers!
     
  2. jcsd
  3. Jun 3, 2010 #2

    HallsofIvy

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    That upper limit on the y-integral can't be "[itex]\sqrt{9- 9x^2}[/itex]" since that would give an ellipse with semi-axis on the x-axis 1- x can't go from 0 to 3 inside that!

    I will assume this is actually
    [tex]\int_{x=0}^3\int_{y=0}^{\sqrt{9- x^2}} \int_{z= 0}^{9- x^2- y^2}dzdydx[/tex]

    That, now, is an integration over the eighth of a sphere, in the first quadrant, with center at (0, 0, 0) and radius 3.

    In cylindrical coordinates, r will go from 0 to 3 while [itex]\theta[/itex] goes from 0 to [itex]\pi/2[/itex]. z goes from 0 to [itex]9- x^2- y^2= 9- r^2[/itex] and the differential of volume is [itex]r dr d\theta[/itex].

    To see the LaTex code for formulas in a post, just click on the formula.
     
  4. Jun 3, 2010 #3
    Yes that is the correct integral. IM curious as to why its from 0 to pi/2. If the x value is from -3 to 3, isnt that a half a circle, hence 0 to pi?
     
  5. Jun 4, 2010 #4

    HallsofIvy

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    You are right- I misread the problem. If you look at my first integral, you will see I have x running from 0 to 3 rather than -3 to 3.

    Yes, [itex]\theta[/itex] should go from 0 to [itex]\pi[/itex], not [itex]\pi/2[/itex].
     
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