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Math Help - problem in region bounded by a curve and a straight line

  1. #16
    MHF Contributor Unknown008's Avatar
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    \dfrac{\pi}{16} \left[49(7) - \frac{14(7)^2}{2} + \frac{7^3}{3}\right] - \left[49(3) - \frac{14(3)^2}{2} + \frac{3^3}{3} \right] - \pi\left[4(4) - \frac{4^2}{2}- 4(3) - \frac{3^2}{2} \right]

    This line should be:

    \dfrac{\pi}{16} \left[49(7) - \frac{14(7)^2}{2} + \frac{7^3}{3}\right] - \left[49(3) - \frac{14(3)^2}{2} + \frac{3^3}{3} \right] - \pi\left[(4(4) - \frac{4^2}{2})- (4(3) - \frac{3^2}{2}) \right]

    \dfrac{\pi}{16} \left[49(7) - \frac{14(7)^2}{2} + \frac{7^3}{3}\right] - \left[49(3) - \frac{14(3)^2}{2} + \frac{3^3}{3} \right] - \pi\left[4(4) - \frac{4^2}{2}- 4(3) + \frac{3^2}{2} \right]

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  2. #17
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    i get:

    \dfrac{\pi}{16} \left[49(7) - \frac{14(7)^2}{2} + \frac{7^3}{3}\right] - \left[49(3) - \frac{14(3)^2}{2} + \frac{3^3}{3} \right] - \pi\left[4(4) - \frac{4^2}{2}- 4(3) + \frac{3^2}{2} \right]

    = \frac{\pi}{16}\left[\frac{64}{3}\right] -\pi\left[\frac{1}{2}\right]

    = \pi\frac{4}{3} - \pi\left[\frac{1}{2}\right]

    = \frac{5}{6}\pi unit^3

    right?
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  3. #18
    MHF Contributor Unknown008's Avatar
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    \pi\left[4(4) - \frac{4^2}{2}- 4(3) + \frac{3^2}{2} \right] = \pi (16 - 8 - 12 + 4.5) = 0.5\pi

    Giving final answer as \frac56 \pi\ units^3
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  4. #19
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    okey sir..thanks very much..i really appreciate with you..thanks!!
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