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Help me integrate this volume please!

  1. Feb 26, 2008 #1
    1. The problem statement, all variables and given/known data

    Volume enclosed by: z= x^2 + y^2 and z= 2y.
    This should look like 1/2 of a bowl only on the +y side of the x y z plane I believe. I want to know if integrating in different order or cord system will affect the result. I am having alot of problems attempting to find the answer to this part of the equation:

    V= Int(0->2) [4/3 (2y-y^2)^(3/2) dy] = Pi/2 ???

    My teacher says he found the volume to be Pi/2. Which i think might be incorrect.

    I need to show the process of integration of dydzdx and dxdydz and compare the results. Or change to spherical coordinates.

    2. The attempt at a solution

    I am having an horrendous time finding the correct bounds for this i believe is my problem. Along with the fact i do not understand how to use my accessible integration tables to correctly solve the last part.

    Starting with:

    V= Int[0->2]dy Int[y^2 -> 2y]dz Int [-sqrt(z-y^2) -> sqrt(z-y^2)]

    I move down correctly to,

    V= Int(0->2) [4/3 (2y-y^2)^(3/2) dy]

    V= Pi/2 ????

    Can some one please explain to me the correct integration table substitution of this part? Every table i look at has things in the form of:

    Int[sqrt(2au-u^2)du , Int[sqrt(u^2-a^2) or Int[sqrt(a^2-u^2)

    How do you convert my problem which seems to have the form of Int[{(sqrt(2au-u^2))^3}du] ?? I've tried of substitution and i feel uncertain of the result. If someone could explain how that integral equals Pi/2 i should be able to change to spherical cords or integrate in a different order.

    Thanks in advance!!
     
    Last edited: Feb 26, 2008
  2. jcsd
  3. Feb 26, 2008 #2
    I would solve this using Mathematica on my computer, but for some reason it rolled back my registration and i cant access it. I would like to have some work done for this by tomorrow, and i know that my professor does not like when we "cheat" using a computer to do integrals anyways hahaha.
     
  4. Feb 26, 2008 #3
    Your professor's answer is correct.

    To start you off, for your integral, [itex]z[/itex] is going from [itex]x^2+y^2[/itex] to [itex]2y[/itex]. Now you're left with the image of a circle in the x-y plane. What should your limits of integration be there?

    You could try a different coordinate system as well - since the projection in the x-y plane is a circle, what do you think would be best?
     
  5. Feb 26, 2008 #4
    Thanks for fast response pingpong.

    Ok, so how did you figure his answer was correct? Were you able to use an integral table? cause i would really like to know that process.... i have no clue on how to do
    V= Int(0->2) [4/3 (2y-y^2)^(3/2) dy]= Pi/2 . with out a computer's help that is.

    I'm still confused on the limits of x and y also... Wont the projection onto the x/y plane be just 1/2 of a circle? Since the values of -y wont be included in the volume?

    I would like to try to use spherical coordinates but I am just so uncertain of how to set it up, it seems like every example i've done is much simpler... All 3 calc books i have dont give me any good examples, or i'm not quick enough to understand them.
     
  6. Feb 26, 2008 #5
    Just getting frustrated on this problem now... i feel that i've lost understanding on how to set up the limits let alone understand how the integral is going to work.

    Any help would be amazing
     
  7. Feb 28, 2008 #6
    So it's been a couple days so far, and i've still not been able to change the cords of this problem to spherical or cylindrical and reproduce my answer of pi/2.....

    Can some one look at my bounds for when i tried to change to Cylindrical Cords?

    Original problem restated----Volume bounded by z= x^2 + y^2 and z=2y.

    Cylindrical Cords attempt.
    Bounds of Z, theta and r.
    r^2<z<2rsin(theta) -pi < theta < pi 0<r<2

    V= (Triple integral with bounds above) r^2 cos(theta) dzdrd(theta)
    Which i evaluate to an answer not equal to pi/2.
     
    Last edited: Feb 28, 2008
  8. Feb 28, 2008 #7

    HallsofIvy

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    Where did the "r^2 cos(theta)" come from? That looks like spherical coordinates. The differential for cylindrical coordinates is r dz dr d(theta).
     
  9. Feb 28, 2008 #8
    Thats a good point halls of ivy.... some how i manage to combine the spherical and change it to cos instead of sin. Let me attempt it with that change!
     
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