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Volume of bounded by 2 surfaces

  1. Sep 19, 2009 #1
    1. The problem statement, all variables and given/known data

    I need to find the volume of the body bounded by the following surfaces:
    z = x2 + y2
    z = 1 - x2 - y2

    2. Relevant equations

    Volume of a body between z=o and the upper surface:
    [tex]\iint_{D} z(x,y) dA[/tex]

    3. The attempt at a solution

    Ok, this is something I need to do with a double integral, not a triple integral.
    So those are 2 symmetric paraboloids, creating in between a body shaped like a lens.
    Since I need to find the volume of a body from z=o to the surface, I thought about this.
    The 2 surfaces are symmetrical, so if I take their intersection (which is a circle radius 2-1/2 at z=1/2) and shift it down to z=0, I can calculate the volume of the upper half of the solid and then multiply by 2.
    I'll need it in polar coordinates of course, so:
    [tex]x=r\cos{\theta}, y=r\sin{\theta}[/tex]
    and the function after shifting it is:
    z = 1/2 - x2 - y2 = 1/2 - r2(sin2θ + cos2θ) = 1/2 - r2
    So the integral is:
    [tex]
    \int_{0}^{2\pi} \int_{0}^{\frac{1}{\sqrt{2}}} (\frac{1}{2} - r^2) r dr d\theta = \int_{0}^{2\pi} \int_{0}^{\frac{1}{\sqrt{2}}} (\frac{r}{2} - r^3) dr d\theta = [/tex]

    [tex]\frac{1}{4} \int_{0}^{2\pi} (r^2-r^4)_{0}^{\frac{1}{\sqrt{2}}} d\theta = \frac{1}{8} \int_{0}^{2\pi} d\theta = \frac{\pi}{4}[/tex]

    And since it was the upper half, I need to multiply by 2 to get the entire volume, it being [itex]\frac{\pi}{2}[/itex].

    Was I correct?
     
  2. jcsd
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