Calculus (Solid of Revolution)

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The region R is bounded by the x-axis, x = 2, and y = x^2. What is the the volume of the solid formed by revolving R about the line x = 2?

  • Calculus (Solid of Revolution) -

    y=x^2 and x=2 intersect at (2,4)
    the radius of the rotated region = 2 - x
    = 2 - √y


    V = π∫ (2-√y)^2 dy from 0 to 4
    = π∫ (4 - 4√y + y) dy from 0 to 4
    = π[ 4y - (8/3)y^(3/2) + (1/2)y^2 ] from 0 to 4
    = π (16 - (8/3)(8) + 8 - 0)
    = 8π/3

    check my arithmetic and thinking.

  • Calculus (Solid of Revolution) -

    y=x^2 intersects x=2 at (2,4)

    using shells,
    v = ∫2πrh dx [0,2]
    where
    r = 2-x
    h = y

    v = 2π∫(2-x)(x^2)dx [0,2]
    = 2π∫2x^2 - x^3 dx [0,2]
    = 2π(2/3 x^3 - 1/4 x^4)[0,2]
    = 2π(16/3 - 4)
    = 2π(4/3)
    = 8π/3

    using discs,

    v = ∫πr^2 dy [0,4]
    = π∫(2-√y)^2 dy [0,4]
    = π∫(4 - 4√y + y)dy [0,4]
    = π(4x - 8/3y√y + 1/2 y^2)[0,4]
    = π(16 - 64/3 + 8)
    = 8π/3

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