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Math Help - Calc 1 ellipse volume

  1. #1
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    Calc 1 ellipse volume

    Find the volume of the solid generated by revolving about the x-axis the region bounded by the x-axis and the upper half of the ellipse

    x^2/a^2 + y^2/b^2 = 1

    (and thus find the volume of a prolate spheroid)

    Here a and b are positive constants with a>b

    Volume of the solid of revolution?

    Okay, so this one is tripping me up a bit... here's what I've done:

    I tried to solve for y and got --
    y = square root(b^2(1 - x^2/a^2)) or y = b*sqrt(1 - x^2/a^2)

    I thought I would need to square the function and ultimately multiply by pi:

    pi int_a^b (b^2(1-x^2/a^2) dx and I moved the b^2 to the beginning with pi and integrated to get:
    b^2* pi [ x - x^3/3a^2]

    so my final answer, keeping the b and a...
    b^2*pi[(b-(b^3)/(3a^2))-(a-(a^3)/(3a^2))]

    unfortunately this is not the correct answer. any help?? thanks!
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  2. #2
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    Here we use the formula V = \pi \int y^2 dx

    \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1.

    \frac{y^2}{b^2} = 1 - \frac{x^2}{a^2}

    y^2 = b^2 \left(1 - \frac{x^2}{a^2}\right)
    y^2 = \frac{b^2}{a^2} (a^2 - x^2)

    substitute into V formula. Set the limits to a and -a (x intersections which enclose the upper area).

    V = \pi \int \limits_{-a}^a \frac{b^2}{a^2} (a^2 - x^2) dx

    V = \pi \frac{b^2}{a^2} \int \limits_{-a}^a (a^2 - x^2) dx

    V = \pi \frac{b^2}{a^2} \left[a^2 x - \frac{x^3}{3} \right]_{-a}^a

    V = \pi \frac{b^2}{a^2} \left[ \left(a^2 *a - \frac{a^3}{3} \right) - \left(a^2 *-a - \frac{(-a)^3}{3} \right) \right]

    V = \pi \frac{b^2}{a^2} \left( \frac{2a^3}{3} + \frac{2a^3}{3} \right)

    V = \pi \frac{b^2}{a^2} * \frac{4a^3}{3}

    V = \frac{4 \pi a^3 b^2}{3a^2}

    V = \frac{4}{3} \pi ab^2
    Last edited by nzmathman; December 1st 2008 at 11:21 PM.
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