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Math Help - Triple Integrals

  1. #1
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    Triple Integrals

    I am having quite a bit of trouble with triple integrals and I keep coming to this problem.

    Find the volume between the plane x + y + z = 1 and the xy - plane, for x + y \leq 2, 0 \leq x, and 0 \leq y

    So I started with the bounds which are:
    0 \leq z \leq 1 - x - y
    0 \leq y \leq 2 - x
    0 \leq x \leq 2

    And so the triple integral should look like:

    \int_{0}^{2}\int_{0}^{2-x}\int_{0}^{1-x-y}dzdydx  \rightarrow \int_{0}^{2}\int_{0}^{2-x}z\mid_{0}^{1-x-y}dydx \rightarrow \int_{0}^{2}\int_{0}^{2-x}(1-x-y) dydx  \rightarrow \int_{0}^{2} y - xy - \frac{y^2} {2}\mid_{0}^{2-x}dx \rightarrow \int_{0}^{2} (2-x)-(2x-x^2)- (\frac{4-4x+x^2} {2}) dx

    After simplifying I ended up with:

    \int_{0}^{2} -x +\frac{x^2} {2} dx \rightarrow -\frac{x^2} {2} + \frac{x^3} {6} And After evaluating I ended up with -\frac{2} {3}. I am not sure where I went wrong but the answer is supposed to be 1.

    I just wanted to say thanks for everyone that has been helping me the past few days with this kind of stuff, I appreciate it a lot.
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  2. #2
    Moo
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    Hello,

    Whew ! I finally spotted the mistake !

    You're taking z>0, while, since the plane x+y+z=1 intersects the xy-plane, there are some negative z that satisfy the conditions !

    Actually, we have z=1-x-y>1-2=-1, because x+y<2 (large inequalities, but I'm too lazy to do the latex)

    So try it with the boundaries for z : -1 and 1-x-y

    (I haven't done the calculations yet)
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