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Math Help - Papus theorem

  1. #1
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    Question Papus theorem

    Question 1

    Use the Theorem of Pappus to find the volume of the solid generated when the region bounded by the x-axis and the semicircle y= sqrt(a^2 -x^2) is revolved about

    a) the line y=-a
    b) the line y=x-a


    Question 2
    Use the Theorem of Pappus to find the centroid of the triangular region with vertices (0,0), (a,0), and (0,b) , where a> 0 and b>0 .
    Hint Revolve the region about the x-axis to obtain y bar and about the y-axis to obtain x bar.

    I don't quite understand the Pappus theorem. Please do help you know how to solve them. Thank you very much.
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  2. #2
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    #1:

    \overline{x}=0 from the symmetry of the region, \frac{{\pi}a^{2}}{2} is the area of the semicircle, 2{\pi}\overline{y} is the distnace traveled by the centroid to generate the sphere so \frac{4}{3}{\pi}a^{3}=(\frac{{\pi}a^{2}}{2})(2{\pi  }\overline{y}), \;\ \overline{y}=\frac{4a}{3{\pi}}

    Now for part a, V=\left[\frac{1}{2}{\pi}a^{2}\right]\left[2{\pi}\left(a+\frac{4a}{3{\pi}}\right)\right]=\frac{1}{3}{\pi}(3\pi+4)a^{3}

    See if you can tackle part b.


    #2:

    Revolve the region about the x-axis to get \overline{y} and about the y-axis to get \overline{x}.

    The region generates a cone of volume \frac{1}{3}{\pi}ab^{2}

    when revolved about the x-axis, the area of the region is

    \frac{1}{2}ab, so \frac{1}{3}{\pi}ab^{2} =\left(\frac{1}{2}ab\right)(2{\pi}\overline{y}) , \;\ \overline{y}=\frac{b}{3}.


    A cone of volume \frac{1}{3}a^{2}b is generated when the

    region is revolved about the y-axis so \frac{1}{3}{\pi}a^{2}b =\left(\frac{1}{2}ab\right)(2{\pi}\overline{x}), \;\ \overline{x}=\frac{a}{3}.

    The centroid is.........?
    Last edited by galactus; September 3rd 2007 at 08:16 AM.
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  3. #3
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    Hi galatus,

    Thank you very much for your reply.

    I don't fully understand this question.

    I know the Pappus theorem is
    V = area R * distance traveled by the centroid.

    You have showed me


    So the second part is the distance traveled by the centroid. Why equals 2pi ( a + (4a)/(3pi))? UM... just don't get it!
    Last edited by kittycat; September 5th 2007 at 03:12 PM.
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  4. #4
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    Hello again, Kittycat. As before, you are giving up too soon. There is nothing here that should be a surprise to you.
    1. Your axis of revolution is y = -a.
    2. Your Centroid is at (0,\frac{4a}{3\pi})
    3. The distance from the Centroid to the Axis of Rotation is the Radius of the circle travelled. \left(\frac{4a}{3\pi} - (-a)\right)
    4. 2\pi\;r is the formula for the circumference of a circle, given its radius.
    P.S. Pep talk - pep talk - pep talk. More confidence. There is more stuff in your brain than you think. Reach in a bring it out.
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  5. #5
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    Question

    hi TKHunny,

    I am not that great - don't have so much good stuff in my brain as you thought.

    I think you are great - with brilliant intellects!

    What about for part b - the line y=x-a?
    Can you explaining to me how to get the distance from the centroid to the line y=x-a?

    Thank you very much.
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  6. #6
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    Quote Originally Posted by kittycat View Post
    I am not that great - don't have so much good stuff in my brain as you thought.
    Hogwash!

    Can you explaining to me how to get the distance from the centroid to the line y=x-a?
    Several ways. One of the most useful formulas ever, the Distance from a point (a,b) to a line cx + dy + e = 0 is:

    \frac{|c*a + d*b + e|}{\sqrt{c^{2}+d^{2}}}

    It is a beautiful thing.
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