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Integration question again

  1. Feb 23, 2013 #1
    1. The problem statement, all variables and given/known data
    A goldfish bowl is a glass sphere of inside diameter 20cm. Calculate the volume of water it contains when the maximum depth is 18cm.





    The attempt at a solution

    I don't really have an idea of how to attempt this, all I did so far was a draw a little sketch of the bowl and put in the dimensions

    Hmm.. should I just find the volume of the sphere, sketch the cross-section of it, and then try to use the principles of solids of revolution to find the volume?
     
  2. jcsd
  3. Feb 23, 2013 #2

    Curious3141

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    Consider the vertical cross-section of the bowl. Let the centre of the bowl be the origin (0,0). y-coordinates range from -10 to +8.

    Now consider the horizontal circular cross section of a disc of water taken at a certain y-coordinate. Find its radius via Pythagoras theorem, and hence its area.

    Hence figure out the volume of an infinitesimally small cylinder having that cross-section and a vertical height dy.

    Now do the integration, imposing the correct bounds for y.
     
  4. Feb 23, 2013 #3
    Umm wouldn't the radius be 10cm, because the diameter of the bowl is 20cm?

    Cause maybe I drew my diagram badly I'm not seeing how I can use Pythagoras ' theorem.
    :S
     
  5. Feb 24, 2013 #4

    Curious3141

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    Right through the centre (at y = 0), yes the radius of the cross-section would be 10cm.

    But what about other y values. Hint: think of a right triangle, the hypotenuse being the constant radius of the bowl (10cm), the vertical height being y and the horizontal base being the radius of the cross-section.
     
  6. Feb 24, 2013 #5
    Is the radius 6cm?
     
  7. Feb 25, 2013 #6

    Curious3141

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    Huh? The radius of the horizontal cross-section varies continuously depending on the level you're taking it at.

    Did you make a proper sketch?
     
  8. Feb 25, 2013 #7
    Hm... Maybe I didn't on my diagram 10 is the hypotenuse and 8 is the perpendicular height.
     
  9. Feb 26, 2013 #8

    Curious3141

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    10cm is always the hypotenuse.

    y varies from -10 (bottom) to +8 (top of water level). Do you understand this?

    8cm is only the height when you're considering the area of the top surface of the water (y = 8). At that point, the radius is ##\sqrt{10^2 - 8^2} = 6cm##. Agree?

    At the bottom of the bowl (y = -10), the radius is zero, because the bottom is just a point, not a circle. Agree?

    The radius of the cross-section right through the level of y = 0 (center of the sphere) is 10cm (simply the radius of the sphere). Agree?

    Now you can take the cross section of water at *any* water level between the bottom and the top, not just those "special" levels. Your job is to find an expression, in terms of y, for the radius of this cross-section. Can you do this?

    Remember, what you get will be in terms of y - it'll have a y in the expression, not just a number.
     
    Last edited: Feb 26, 2013
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