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math

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find the volume of the solid by rotating y=1-x^2, y=0 around the xaxis

  • math - ,

    using discs, of thickness dx

    v = ∫[-1,1] πr^2 dx
    where r = y = 1-x^2
    So,

    v = π∫[-1,1] (1-x^2)^2 dx = 16/15 π

    Using shells, of thickness dy, we have to account for the two branches of the parabola, so it's easier just to use symmetry and get

    v = 2∫[0,1] 2πrh dy
    where r = y and h = x = √(1-y)
    v = 4π∫[0,1] y√(1-y) dy = 16/15 π

  • math - ,

    pi y^2 dx from -1 to +1
    same as 2 pi y^2 dx from 0 to 1

    2 pi (1-2 x^2 + x^4) dx from 0 to 1

    2 pi (1-0) - 4 pi (1^3/3-0) + 2 pi (1^5/5-0)

    2 pi - 4 pi/3 + 2 pi/5

    (pi/15) ( 30 - 20 + 6)

    (16/15) pi

    Check my arithmetic !

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