1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    Cyosis

    User Avatar
    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

    Mark44

    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

    Cyosis

    User Avatar
    Homework Helper

    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

    Cyosis

    User Avatar
    Homework Helper

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Evaluate volume of paraboloid
Loading...