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Homework Help: Evaluate volume of paraboloid

  1. May 5, 2009 #1
    1. The problem statement, all variables and given/known data

    evaluate volume of paraboloid z = x2 + y2 between the planes z=0 and z=1

    3. The attempt at a solution

    i figured we would need to rearrange so that F(x,y,z) = x2 + y2 - z

    then do a triple integral dxdydz of the function F. the limits for the first integral dz would be z=1 and z=0. and i dont know what the other limits would be (y1,y2 and x1, x2?)

    but this first integral gave an answer of -1/2. this would mean that the volume would end up being zero which i dont think is right.

    then i thought that maybe i should say the function is z(x,y) = x2 + y2 and integrating dxdy.

    the dy limits would then be
    [when z=1] y2 = 1 - x2
    y = sqrt[1-x2]

    [when z=0] y2 = -x2
    y = sqrt[-x2] = xi i=complex number

    the dx limits would then be
    [when z=1] x2 = 1 - y2
    x = sqrt[1-y2]

    [when z=0] x2 = -y2
    x = sqrt[-y2] = yi i=complex number

    but this seems like a dead end
    any suggestions would be helpful

    ok just tried another thing: the lower limits for dx and dz being zero taking out the complex numbers.

    this gives: after dy=> 2y dx = 2sqrt(1-x2)dx = x*sqrt(1-x2) + sin-1x

    i then put in the limits 1 an 0 and got pi/2

    still dont know if this is right though
    Last edited: May 5, 2009
  2. jcsd
  3. May 5, 2009 #2


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    Homework Helper

    The volume is pi/2, but I can't say I understand how you arrived at that answer. To check if your expression is correct you could take the planes z=0 and z=2, which should give you a volume 2pi.
    Last edited: May 5, 2009
  4. May 5, 2009 #3


    Staff: Mentor

    Not true. When z = 0, x2 + y2 = 0, which means that both x and y are 0.
  5. May 5, 2009 #4
    ~ ok well i tried it and got pi so im not sure what i have done wrong...

    ~ i have basically done a double integral dxdy (ie dy was done first then dx of that)...

    ~ this gave 2y which i then needed to sub in the limits.
    when z=0, y=0 so this is the lower limit
    when z=2, y= sqrt(2-x2)

    ~ this gave me 2sqrt(2-x2), this integrates to give:
    2*[sin-1(x/sqrt(2)) + (x/2)(sqrt(2-x2))]

    ~ the limits for this are
    when z=0, x=0 so this is the lower limit
    when z=2, y= sqrt(2)
  6. May 5, 2009 #5


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    I find this pretty hard to explain without drawing pictures, but here goes.

    [itex]f(x,y)=x^2+y^2[/itex]. This is a paraboloid as stated and we can take [itex]x^2+y^2 \leq z\leq 1[/itex]. If we slice this paraboloid in small slices parallel to the x-y plane we get a lot of circles. These can be described as [itex]\sqrt{1-x^2}[/itex], thus [itex]-\sqrt{1-x^2} \leq y \leq \sqrt{1-x^2}[/itex]. All that remains now is the range for x, which is easy, take y=0 then [itex]-1 \leq x \leq 1[/itex].

    This gives the integral:

    [tex]\int_{-1}^1 \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} \int_{x^2+y^2}^1 dz dy dx[/tex]

    There is a simpler way, but I have to go now I'll be back in a few hours.
  7. May 5, 2009 #6
    ok then just to check if you use the planes between z=0 and z=4 is the volume 8pi?
  8. May 5, 2009 #7


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    Yep that is correct you can use the formula [itex]\frac{1}{2} \pi r h[/itex] with r the radius of the greatest circle (top or bottom of the parabola depending on its orientation) and h its height.

    While the method in my previous post will work for general cases sometimes looking at symmetry and clever thinking saves you a lot of computing time. For example in this case where the problem is completely rotational symmetric around the z-axis. You can obtain this paraboloid by simply rotating [itex]y=\sqrt{z}[/itex] around the z-axis. now cut the paraboloid in slices, whose shapes are circles with radii [itex]\sqrt{z}[/itex]. The area of one slice is therefore [itex]\pi z[/itex] adding all the circles together between the interval z=0 and z=1 yields the volume of the paraboloid.

    [tex]\int_0^1 \pi z dz=\frac{\pi}{2}[/tex]

    Lots of text, but a very easy calculation.
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