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Calc volume help

  1. Oct 21, 2003 #1
    Calc help plz!!!

    Plz explain to me what the problem below is talking about? I don't really understand what it's saying...I can't picture the graph of the function...
    "The base of a solid is a quadrant of a circle of radius a. Each cross section perpendicular to one edge of the base is a semicircle whose diameter lies in the base. Find the volume."
     
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  3. Oct 22, 2003 #2

    jamesrc

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    Sounds like a funny looking shape; kind of like a loaf of bread or something. Anyway, the base is just a quarter of a circle, so you have x^2+y^2 = a^2

    Then you've got semicircles in the z-axis (I'm setting these semicircles' diameters parallel to the x-axis.)

    To find the volume, I think you just integrate the area of the semicircles along the y-axis:

    V = integral( A(r)dy) (limits from 0 to a, x-sectional area A is a function of the radius)
    A = (pi r^2)/2 (for each semicircle)
    r = x/2 (x is a function of y, this should be clear if you drew it the way I said I did)
    and then use the first equation...

    I end up with V = (pi a^3)/4, though I'm prone to screwing up algebra
     
  4. Oct 22, 2003 #3

    HallsofIvy

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    More half a croissant than a loaf of bread but I like that description!
    Here's a way to construct it (following what jamesrc said):

    Get a thin sheet of styrofoam. Draw a quarter circle on a sheet of paper (with radius a multiple of the thickness of your styrofoam). Mark off points on one radius (say, the horizontal one) of the quarter circle equal to the thickness of the styrofoam. Draw vertical lines at each mark across the quarter circle and measure their length.
    Cut out semicircles of styrofoam having those distances as diameter.
    Glue the semicircles on the quarter circle- in the correct places, of course.
    Take this to class and impress your teacher!

    This model is an approximation to the figure. The thinner the styrofoam is in comparison to the radius of your quarter circle (in other words the more multiples of the thickness you make your radius) the "smoother" it will be. Actually, what you have is a model for the "Riemann sums" that define the integral.

    Since your quarter circle has radius a, it is part of the graph of
    (again, as jamesrc said) x2+ y2= a2. Taking our quarter circle to be the one in the first quadrant, we have y= √(a2- x2) and that will be the "diameter" of the semicircle constructed at x. The radius of that semicircle is then (1/2)√(a2- x2) of course and, since the area of a semicircle is (1/2)πr2, the area of each such semicircle is (1/2)π(1/2)√(a2- x2))= (1/8)π(a2- x2.
    Taking the thickness of your "styrofoam" to be Δ x, the volume of each semicircular piece is (1/8)π(a2- x2)Δx and the volume of all of them put together is
    Σ(1/8)π(a2- x2); Delta; x. That would be the volume of your model- an approximation to the volume of the original figure.

    Now the usual limit "magic" gives the exact volume of the original figure and turns the Riemann sum into an integral as (1/8)π∫(a2- x2)dx where the integral is from x= 0 to x= a. That's an easy integral.

    (I get (1/12)πa3. I wonder if jamesrc didn't "screw up" the integral rather than the algebra!)
     
  5. Oct 22, 2003 #4

    jamesrc

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    Good grief. I screwed up the algrebra.
    My second to last step was
    V = (pi/8)*(a^3 - (a^3/3))
    and I managed to get pi*a^3/4

    pi*a^3/12 is more like it.

    Next time I try to help I promise I won't work the answer out on a napkin.
     
  6. Oct 22, 2003 #5

    HallsofIvy

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    Absolutely, napkins get wrinkled and that messes up your algebra!

    (Plus the fact that I keep spilling coffee on them!)
     
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