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Volume of dome -shape

  1. Aug 14, 2011 #1
    Volume of "dome"-shape

    1. The problem statement, all variables and given/known data
    Find the volume between of the shape that is confined between the two cylinders and the xy-plane. Maybe I have understood it wrong, but after drawing them it looks like the intersection is a dome-looking shape. So I'm guessing the volume that I have to find is between the dome and the ground (xy-plane).


    2. Relevant equations
    The two cylinders
    [itex]x^2 = 4 - 4z[/itex]
    [itex]y^2 = 4 - 4z[/itex]



    3. The attempt at a solution
    I'm not sure how to do this. I know that [itex]x = y = +- 2[/itex] when [itex]z = 0[/itex] and [itex]x = y = 0[/itex] when [itex]z = 1[/itex].
    I thought that maybe adding the two equations can give an equation for how z varies with x and y.
    [itex]x^2 + y^2 = 8 - 8z[/itex] and then solve for z and take the integral of z.
    In polar coordinates, the radius varies from 0 to 2. That gives me a volume of 3 pi.
    The volume is supposed to be 8... any ideas?
     
  2. jcsd
  3. Aug 14, 2011 #2
    Re: Volume of "dome"-shape

    The function for the height of the dome at a given (x,y) is going to have to be piece wise because at some points it's the first cylinder and at some points the second cylinder. That's probably why you're having trouble finding the function to integrate. My advice is to split the xy-plane into 4 triangles for which the volume above them is equal. Then you can find the non-piecewise function above a single triangle, integrate that, and multiply by 4.

    Actually, there is even more symmetry to the problem. You can use symmetry to work out that the volume of each triangle based part is 1/8th the volume of a length 4, radius 2 cylinder.
     
  4. Aug 15, 2011 #3
    Re: Volume of "dome"-shape

    Thanks, I understand now that I traced the domain on paper. However, how do I get the function for the height?
    Do I add the two equations, solve for z and use that, or do I use one half of the sum of the two equations and solve for z?

    Also, could you explain the last part? I'm not that good at understanding symmetry...
     
  5. Aug 15, 2011 #4
    Re: Volume of "dome"-shape

    You should find the intersections of the two functions. This is the line where the dome has a "sharp edge".
    To do this you should equal [itex]f_1(x,y)=f_2(x,y)[/itex]
    What you obtain are the two lines [itex]y=x, y=-x[/itex].
    This lines, together with the other boudaries where [itex]f_1(x,y)=f_2(x,y)=0[/itex], will divide the domain on the xy plane into 4 parts.
    You should integrate under one of these areas per time.
    Then you should understand that you can is very simple to obtain the whole solid.

    Someone else suggested you can further divide that xy plane into 8 areas and so simplify the task more.
     
  6. Aug 15, 2011 #5
    Re: Volume of "dome"-shape

    Ohh, thanks! My math skills are a little rusty hehe... was just looking at an example in my book that is almost identical. I totally forgot about the whole projection-on-the-xy-plane.
     
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