March 29, 2017

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Let V be the volume of the three-dimensional structure bounded by the region 0≤z≤1−x^2−y^2. If V=a/bπ, where a and b are positive coprime integers, what is a+b?

  • calculus (point me in the right direction please?) - ,

    One easy way to do this is to consider the figure to be a solid of revolution formed by revolving the parabola z = 1-x^2 around the z-axis.

    v = ∫[0,1] πr^2 dz
    where r = x
    v = π∫[0,1] (1-z) dz
    = π/2

    The other, more direct way, of course, is to do a volume integral:

    v = 4∫[0,1]∫[0,√(1-x^2)]∫[0,1-x^2-y^2] dz dy dx
    = 4∫[0,1]∫[0,√(1-x^2)] 1-x^2-y^2 dy dx
    = 4∫[0,1] (1-x^2)y - 1/3 y^3 [0,√(1-x^2)] dx
    = 4∫[0,1] (1-x^2)^(3/2) - 1/3 (1-x^2)^(3/2) dx
    = 8/3∫[0,1] (1-x^2)^(3/2) dx
    = 8/3 * 1/8 (x√(1-x^2) (5-2x^2) + 3arcsin(x)) [0,1]
    = π/2

  • calculus (point me in the right direction please?) - ,

    thank you steve!

  • calculus (point me in the right direction please?) - ,

    Another easy way is to use cylindrical coordinates, like polar coordinates with a z-axis.

    Just as dA = r dr dθ is the area element in polar coordinates,

    dV = r dz dr dθ in cylindrical coordinates.

    So, now we have z = 1-r^2
    v = ∫[0,1]∫[0,2π]∫[0,1-r^2] r dz dr dθ
    = ∫[0,1]∫[0,2π] r(1-r^2) dr dθ
    = ∫[0,1] 2πr(1-r^2) dr
    = 2π(1/2 r^2 - 1/4 r^4)
    = π/2

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