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Volume of region

  1. Jul 25, 2014 #1
    1. The problem statement, all variables and given/known data
    find volume of the region bounded by [tex]x^2+2y^2=2;z=0;x+y+2z=2[/tex]





    3. The attempt at a solution

    I figure "slicing" in the z=0 direction would be the easiest

    the first issue I am having is the upper bound of z, it definitely seems to be 2 but but it's not making sense at the moment how to get that


    when I slice with the z=0 plane I get an ellipse

    y is on the interval [-1,1] and then getting x in terms of y
    [tex]-\sqrt{2-2y^2} \le x \le \sqrt{2-2y^2}[/tex]


    so from that I get the setup

    [tex]\int_{0}^{2}\int_{-1}^{1}\int_{-\sqrt{2-2y^2}}^{\sqrt{2-2y^2}}dxdydz[/tex]

    I feel that is correct but the part mostly bothering be is the upper bound of z.

    using x+y+2z=2 I can see the intercepts for x,y,z are 2,2,1. this is where I am getting stuck
     
  2. jcsd
  3. Jul 25, 2014 #2

    HallsofIvy

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    x+ y+ 2z= 2, which is equivalent to z= 1- x/2- y/2, is the upper bound. Since that involves both x and y, that should be the first integral, not the last.
     
  4. Jul 25, 2014 #3
    [tex]\int_{-1}^{1}\int_{-\sqrt{2-2y^2}}^{\sqrt{2-2y^2}}\int_{0}^{1- x/2- y/2}dzdxdy[/tex]

    more like that then?
     
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