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Math Help - VOlume and parabolic cylinder

  1. #1
    n22
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    VOlume and parabolic cylinder

    Hi ,
    I would very much appreciate help and guidance withi this .
    It is quite tough.thanks.
    Find the volume of the region bounded by the parabolic cylinder z=1-x^2 and the planes z=0,y=0 and y+z=2

    (also I dont quite fully understand ..but do I always find the projections -why and how many projections do I need to find ?-sorry abt the stupid question-ineed to check)
    Last edited by n22; July 2nd 2013 at 05:50 AM.
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    Re: VOlume and parabolic cylinder

    Start by doing a sketch of your region.
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  3. #3
    n22
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    Re: VOlume and parabolic cylinder

    VOlume and parabolic cylinder-jpg.png

    Please see my sketch and add any neccessary changes ..i did this on paint
    thanks
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    Re: VOlume and parabolic cylinder

    OK, that's a good start. Now since your region opens up along the y-axis, that would be a good first variable to determine bounds for. What bounds can you come up with for y?
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    n22
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    Re: VOlume and parabolic cylinder

    Quote Originally Posted by Prove It View Post
    OK, that's a good start. Now since your region opens up along the y-axis, that would be a good first variable to determine bounds for. What bounds can you come up with for y?
    would the limits for z be 0≤z≤2-y

    and y goes from where to where? hintsplz
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    Re: VOlume and parabolic cylinder

    No, start with y, the bounds will depend on x and z.
    Then work on z, the bounds will depend on x.
    Then your x bounds will only be numbers.
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    n22
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    Re: VOlume and parabolic cylinder

    Quote Originally Posted by Prove It View Post
    No, start with y, the bounds will depend on x and z.
    Then work on z, the bounds will depend on x.
    Then your x bounds will only be numbers.
    I tried to make a projection onto the y-z plane to make it easier for visualization..please checkVOlume and parabolic cylinder-jpg.png
    y limits: 0≤y≤(1+x) got this from intersection of the planes..
    z limits:0≤z≤(1-x)
    x limits:1≤x≤0
    Please check.thanks
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    Re: VOlume and parabolic cylinder

    Quote Originally Posted by n22 View Post
    I tried to make a projection onto the y-z plane to make it easier for visualization..please checkClick image for larger version. 

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    y limits: 0≤y≤(1+x) got this from intersection of the planes..
    No. Picture drawing lines parallel to the y-axis. They will be bounded by the planes y = 0 and y + z = 2. How can you write the second plane as y = something?

    z limits:0≤z≤(1-x)
    Correct.

    x limits:1≤x≤0
    Since when is \displaystyle 1 \leq 0. Assuming you meant \displaystyle 0 \leq x \leq 1 then you would be correct.
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  9. #9
    n22
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    Re: VOlume and parabolic cylinder

    [QUOTE=Prove It;791795]No. Picture drawing lines parallel to the y-axis. They will be bounded by the planes y = 0 and y + z = 2. How can you write the second plane as y = something?

    would it be 0≤y≤2-z
    how did you get x=0?? I seem to be getting x=√-1 which is obviously wrong.
    when y=0,z=2
    when y=2-z,z=0 these were then subbed into x=sqrt(1-z)

    sorry about the stupid questions.
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    Re: VOlume and parabolic cylinder

    Yes \displaystyle 0 \leq y \leq 2 - z is correct.

    I got x = 0 because you're told the region is bounded by the plane x = 0. I was pointing out that while the bounds you got for x were wrong, you wrote them down the wrong way, since \displaystyle 1 \leq x \leq 0 means you are saying \displaystyle 1 \leq 0, which it obviously isn't.

    Anyway, now that you have the correct bounds, set up your triple integral.
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