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Double Integrals in Polar Coordinates

  1. Dec 4, 2013 #1
    1. The problem statement, all variables and given/known data

    Use polar coordinates to find the volume of the given solid.

    Enclosed by the hyperboloid -x2 - y2 + z2 = 1 and the plane z = 2

    2. Relevant equations

    r2 = x2 + y2, x = rcosθ, y = rsinθ

    ∫∫f(x,y)dA = ∫∫f(rcosθ,rsinθ)rdrdθ

    3. The attempt at a solution

    -x2 - y2 + 4 = 1 → x2 + y2 = 3

    0 ≤ r ≤ √3
    0 ≤ θ ≤ 2∏

    f(x,y) = √(1 + x2 + y2)
    f(rcosθ,rsinθ) = √(1 + r2)

    V = ∫∫r√(1 + r2)drdθ

    u = 1 + r2
    du = 2rdr

    V = ∫∫1/2√ududθ
    V = ∫1/3(u3/2)dθ
    V = 1/3∫43/2 - 1dθ
    V = 7/3∫dθ
    V = (7/3)θ = 14∏/3
     
  2. jcsd
  3. Dec 4, 2013 #2
    Oops, I just realized the integral should be ∫∫(2 - √(1 + r^2))rdrdθ, not ∫∫r√(1 + r^2)drdθ. Sorry about that.
     
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