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Volume of a tetrahedron by Triple Integral

  1. May 9, 2014 #1
    1. The problem statement, all variables and given/known data
    By using triple integral, find the volume of the tetrahedron bounded by the coordinate planes and the plane 2x+3y+2z=6.


    2. Relevant equations

    Volume= ∫vdv=∫∫∫dxdydz

    3. The attempt at a solution

    find intercepts of the plane on the axes,
    x-intercept=3
    y-intercept=2
    z-intercept=3


    then i don't know how to get limits of integration in the formula
     
  2. jcsd
  3. May 9, 2014 #2

    LCKurtz

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    Draw a picture of the plane in the first octant by joining those 3 points. Then use that picture for the limits.
     
  4. May 10, 2014 #3
    Yes i get that. If I were finding the area of in 2D, i would draw lines parallel to x or y axis and find the curves between which they lie. These would be the limits of inner integral. then i find the lowest and highest value of the outer integral and that becomes the limits for it... but in 3D i would have to draw planes parallel to say xy plane...but don't know which curves they lie between.
     

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  5. May 10, 2014 #4

    LCKurtz

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    It's the same idea in 3D. If you are going to do the inner integral in the ##z## direction first, you go from ##z## on the bottom surface (the xy plane) to ##z## on the top surface (the plane). Once you have done that you will have an xy integral and you can look at the triangle in the xy plane for the limits, just as you would do in 2D.

    If you wanted to go in the x direction first for some reason, you would go from x on the back surface to x on the front surface then look in the yz plane for the dydz limits.
     
  6. May 10, 2014 #5
    got it thanx.
     
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