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Voulme of irregular shape

  1. May 21, 2006 #1
    q: (set up the integral that you would use to find the volume of the solid described, evaluate the integral)
    The region below the plane z+y=1 and inside the cylinder x^2+y^2 equal or less than: 1,0 equal and less to z equal and less to 1

    this is what i have been trying:
    tried
    -sqrt1-x^2 <= y <= sqrt1-x^2
    0 <= z <= 1-y
    -1 <= x <= 1

    the books gives the answer: pi-2/3

    oki guys/gals, really can't do it and it's just f--king enoying me soo much.. so if someone could do this or at least give me a sweet description how to do this.. !spending more time on this than my actual revision!
     
  2. jcsd
  3. May 21, 2006 #2

    AKG

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    You should show some work, but it sounds like you're really stuck. Start by drawing a picture. If you look at the part of this region for y < 0, you get half a cylinder. If you want to find the volume of this by integrating along the y-axis, then think of this volume as a bunch of thin sheets side by side of height (in the z-direction) 1, width (in the x-direction) [itex]2\sqrt{1-y^2}[/itex], and thickness (in the y-direction) dy. The other half of this irregular shape is more irregular, because the height changes, but you can apply the same idea, and get an expression for the volume of this shape in terms of integrals. Set up an expression, and try evaluating. If you still get stuck, show your work.
     
    Last edited: May 21, 2006
  4. May 26, 2006 #3

    HallsofIvy

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    Looks to me like a good candidate for polar coordinates. In polar coordinate, r runs from 0 to 1 and [itex]\theta[/itex] from 0 to [itex]2\pi[/itex]. For each r and [itex]\theta[/itex], z runs from 0 to [itex]1- y= 1- rsin(\theta)[/itex]. Of course, the "differential of area" in polar coordinates is [itex]r dr d\theta[/itex].

    The volume is
    [tex]\int_{r= 0}^1 \int_{\theta= 0}^{2\pi} (1- r sin(\theta))rd\theta dr[/tex]
     
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