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Math Help - ∫∫∫ xy dV Triple Integration. Help with bounds!

  1. #1
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    ∫∫∫ xy dV Triple Integration. Help with bounds!

    ∫∫∫ xy dV, where the solid is a tetrahedron with vertices (0,0,0), (1,0,0),(0,2,0), and (0,0,3)

    I tried this with 0<x<1, 0<y<-2x+2, and 0<z<3 with no luck =\ Are my bounds wrong in this case?

    What would be really helpful is if someone could explain how I am supposed to figure out the bounds. I don't even know how to figure them out which means I can't do any of the problems assigned
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by crabchef View Post
    ∫∫∫ xy dV, where the solid is a tetrahedron with vertices (0,0,0), (1,0,0),(0,2,0), and (0,0,3)

    I tried this with 0<x<1, 0<y<-2x+2, and 0<z<3 with no luck =\ Are my bounds wrong in this case?

    What would be really helpful is if someone could explain how I am supposed to figure out the bounds. I don't even know how to figure them out which means I can't do any of the problems assigned
    yes, your limits for z are wrong. see here for the method of finding bounds. post #2 gives a shortcut for special cases (this one). posts #7 and #9 give the longer, more conventional way of tackling it
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    thanks for the link! this is going to take me a while to sort out ^^

    ah nevermind, z should equal 3 - 3x - 3y/2
    Last edited by crabchef; July 13th 2008 at 10:23 PM.
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    Quote Originally Posted by crabchef View Post
    ∫∫∫ xy dV, where the solid is a tetrahedron with vertices (0,0,0), (1,0,0),(0,2,0), and (0,0,3)

    I tried this with 0<x<1, 0<y<-2x+2, and 0<z<3 with no luck =\ Are my bounds wrong in this case?

    What would be really helpful is if someone could explain how I am supposed to figure out the bounds. I don't even know how to figure them out which means I can't do any of the problems assigned
    The upper surface of the tetrahedron is a plane with points (1,0,0), (0,2,0), and (0,0,3) which has equation x+ \frac{y}{2}+ \frac{z}{3}= 1. That can be rewritten \frac{z}{3}= 1- x- \frac{y}{2} and then z= 3- 3x- \frac{3y}{2}. That is the upper bound for your tetrahedron.
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