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Volume of revolution

  1. Apr 24, 2006 #1
    I have a question about finding the volume of a hemisphere, i've got that bit sorted but the next bit asks,

    if the bowl is partially filled what percentage of the bowl is filled.

    I think i understand the method but i need to find the x value when the bowl is filled 4.5cm of a hemisphere of radius 9cm
  2. jcsd
  3. Apr 24, 2006 #2


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    Have you rotated about the x-axis or y-axis?

  4. Apr 24, 2006 #3


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    You have the equation of the shape (outline) of the vessel, you have the height (plane) it needs to be filled to...

    ie. integrate the outline^2 between 0 and this height and times by pi.

    Of course - you want the height:redface: but from what you write... this is 4.5cm ?!?
  5. Apr 24, 2006 #4
    about the y axis.

    the shape is a circle - x^2 + y^2

    is this what you mean by outline?

    the equation i have been given is, x^2 + y^2 = 81
  6. Apr 24, 2006 #5


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    then [tex]\pi\int_0^{4.5}f(x)^2dx[/tex] should be your answer.

    from my immediate thought...
  7. Apr 24, 2006 #6


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    No, I don't think you have the limits of integration right. Since the water will "fill" from the bottom of the circle up, the intgration should be from x= 9 to x= 4.5 (or, more correctly, -9 to -4.5 so the water doesn't spill out!). Also the problem asked for the percentage that was filled. If the water is 1/2 up the side of the bowl, what percentage is filled?
  8. Apr 24, 2006 #7
    if i use -9 and -4.5 as the limits of integraton, what is the function of x -
    f(x) i use?

    as i have been given the equation for a circle it has confused me a little.
  9. Apr 25, 2006 #8


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    Úsing my above equation with...

    [tex]f(x)=(81-(x-9)^2)^{1/2}[/tex] as the outline should work.

    ie. [tex]\pi\int_0^{4.5}(18x-x^2)dx[/tex]

    Giving a numerical value of the water filling 31% of the hemisphere - which seems about right...
    Last edited: Apr 25, 2006
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