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Polar Coordinates

  1. Aug 28, 2014 #1

    Zondrina

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    1. The problem statement, all variables and given/known data

    My answer seems to differ from the books answer, so I'm wondering where something has gone wrong.

    Find the volume inside the prism bounded by the planes ##y = x##, ##y = 0##, ##x = \frac{a}{\sqrt{2}}##, ##z = 0## and the cone ##az = h\sqrt{x^2 + y^2}##.

    2. Relevant equations

    We want the area under ##z = \frac{h}{a} \sqrt{x^2 + y^2}##

    3. The attempt at a solution

    When i visualized this, I saw a triangle in the x-y plane formed by ##y = x##, ##y = 0##, and ##x = \frac{a}{\sqrt{2}}##, which gets cut by ##z = \frac{h}{a} \sqrt{x^2 + y^2}## in the x-y-z plane.

    So we want:

    $$V_R = \int \int_R \frac{h}{a} \sqrt{x^2 + y^2} dA$$

    Using polar co-ordinates gives:

    ##(1) rsin(\theta) = 0 \quad##
    ##(2) rsin(\theta) = rcos(\theta)##
    ##(3) rcos(\theta) = \frac{a}{\sqrt{2}}##
    ##(4) f(rcos(\theta), rsin(\theta)) = \frac{h}{a} r##

    Looking at ##(2)## implies ##\theta = \frac{\pi}{4}## so that ##0 ≤ \theta ≤ \frac{\pi}{4}##.

    The limits for ##r## have proven a little tricky. The answer is listed as:

    ##\frac{a^2 h}{12} [1 + \frac{3 \sqrt{2}}{2} \log(1 + \sqrt{2})]##

    I have obtained the answer ##\frac{a^2 h \pi}{12}## using ##0 ≤ r ≤ a##, which is obviously incorrect, but indicates my limits for ##\theta## make physical sense, so I believe the limits for ##r## are the issue. I can't seem to visualize them properly right now.
     
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  3. Aug 28, 2014 #2

    jedishrfu

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    ....deleted... My answer was off...
     
  4. Aug 28, 2014 #3

    Zondrina

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    Calculus is the bread and butter of all that is true. So forgive me if I'm firm about using it.

    Where is the cylinder coming from exactly? The lines ##y=0##, ##y=x## and ##x = \frac{a}{\sqrt{2}}## give a triangle with a varying height in the x-y plane (based on ##\frac{a}{\sqrt{2}}##).
     
  5. Aug 28, 2014 #4

    jedishrfu

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    I thought I saw a cylinder with cone defining its upper edge but had forgotten about the x=a/sqrt(2) plane
     
  6. Aug 28, 2014 #5

    vela

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    Why are you even using polar coordinates here? It's more straightforward to describe the area of integration using cartesian coordinates.
     
  7. Aug 28, 2014 #6

    Zondrina

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    Re-arranging the equation for the cone a bit:

    ##az = h \sqrt{x^2 + y^2} \Rightarrow \frac{z^2}{h^2} = \frac{x^2}{a^2} + \frac{y^2}{a^2}##

    I see that it opens up in the z-direction and the top of the cone should be a perfect circle of radius ##a##. I still can't see how to get the radius otherwise.

    The book says it is possible to find the solution with polar co-ordinates, and I would like to find a solution using them.
     
  8. Aug 28, 2014 #7

    LCKurtz

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    You have that triangle you described in the xy plane. Use it to get the polar limits. ##r## goes from ##0## to the ##r## value on the line ##x=\frac a{\sqrt 2}##. What is the equation of that line in polar coordinates?
     
    Last edited: Aug 28, 2014
  9. Aug 28, 2014 #8

    Zondrina

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    That would correspond to equation ##(3)## in my first post I believe.

    $$r \cos(\theta) = \frac{a}{\sqrt{2}}$$

    Though I doubt that ##0 ≤ r ≤ \frac{a}{cos(\theta) \sqrt{2}}##
     
  10. Aug 28, 2014 #9

    LCKurtz

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    Why do you doubt it? If you start at ##r=0## and move through the region in the ##r## direction, isn't that the line you hit? And I would suggest expressing that with ##\sec\theta## when you set up the integral.
     
  11. Aug 28, 2014 #10

    Zondrina

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    I am never going to believe wolfram again.

    I think I understand now that you have clarified that. Thank you.
     
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