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Spherical Integral with abs value in limits

  1. Mar 21, 2017 #1
    1. The problem statement, all variables and given/known data
    This has been driving me crazy I can't for the life of me figure out how to convert the limits of this integral into spherical coordinates because there is an absolute value in the limits and i'm absolutely clueless as to what to do with with.


    2. Relevant equations
    $$\int_{\frac {-3\sqrt{2}} {2}}^\frac {3\sqrt{2}} {2} \int_{y=|x|}^\sqrt{(9-x^2)} \int_{z=\sqrt {x^2+y^2}}^\sqrt{(9-x^2-y^2)} z^2 \, dz \, dy \, dx $$



    3. The attempt at a solution
    Limits so far have been:
    $$\sqrt {x^2+y^2} ≤ z ≤ \sqrt{(9-x^2-y^2)}$$ - this is a cone with a round cap
    $$|x| ≤ y ≤ \sqrt{(9-x^2)}$$
    $$\frac {-3\sqrt{2}} {2} ≤ x ≤ \frac {3\sqrt{2}} {2}$$

    I know from the conversion formulas that:
    z22cos2φ

    and the limits of ρ are:
    0≤ ρ ≤3

    I am just stuck on what to do with with the |x| as I am absolutely clueless on how to convert that ro spherical coordinates or obtain θ from it as it's a piece wise function with two sides.
     
    Last edited: Mar 21, 2017
  2. jcsd
  3. Mar 21, 2017 #2
    Okay, take this idea with a good deal of salt because I'm just learning about this stuff as well, but I think from looking at the limits that the shape you are integrating is totally symmetrical about the yz plane, which is to say that you should just be able to treat the limit as x, then integrate x from 0 to 3/2*sqrt(2) and double the result. All the x's in the entire problem are squared, so I don't think it will interfere with any other limits.
     
  4. Mar 21, 2017 #3

    LCKurtz

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    Do you have a picture of the volume? Do you see that ##\phi## measures from the ##z## axis to the side of the cone? What angle is that? Finally, draw a picture of just ##y=|x|## in the ##xy## plane. Remember that ##\theta## is the same in polar and spherical coordinates. What values of ##\theta## would sweep through the region above ##y = |x|## in the ##xy## plane?
    If you aren't required to use spherical coordinates, cylindrical is also a good way to go on this problem. Actually, it's a good exercise to do it both ways.
     
  5. Mar 21, 2017 #4
    I do not have a picture of the volume but I tried to graph the it and I'm having trouble visualizing the integral because of the odd shape
     
  6. Mar 22, 2017 #5

    haruspex

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    So break the outer integral into two ranges.
     
  7. Mar 22, 2017 #6

    LCKurtz

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    So what about my questions about ##\phi## and ##\theta##? Answer them and you will have the answers to your original post.
     
  8. Mar 23, 2017 #7
    it feels too simple but is it possible that the bounds would be:
    cos-1(-√2/2) ≤ ##\theta## ≤ cos-1(√2/2
    and phi is bounded by
    0 ≤ ##\phi##≤ cos-1(##\phi##=1/√2)
     
  9. Mar 23, 2017 #8

    LCKurtz

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    It is simple because spherical coordinates are the perfect choice for this problem. But you needn't give complicated expressions for the angles. Just state the angles. After all, you should know what angle ##\arccos(\frac 1 {\sqrt 2})## is. Ditto for the ##\theta## limits. When you write down the actual angles you will see your use of ##\arccos## has caused your ##\theta## inequalities to be problematical.
     
  10. Mar 26, 2017 #9
    Ah ok I see so the actual bounds would be:

    Π/4 ≤ θ ≤ /4
    0 ≤ Φ ≤ Π/4

    Then the fact that |x| is a discontinous function doesn't actually matter in spherical coordinates at all?
     
  11. Mar 26, 2017 #10

    LCKurtz

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    ##|x|## is not discontinuous. It's first derivative is. But, no, that doesn't matter. Your limits are now correct, assuming you have ##0\le \rho\le 3##.
     
    Last edited: Mar 26, 2017
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