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If $C_1=\{(x,y,z) such\,\, that \,\,0<x=y=z<1\}$ and $C_2=\{(x,y,z) such\,\, that \,\, 0<x<y<z<1\}$ find the volume of each..For the second one the bounareis of the triple integral are $x=0$ to $y$ and $y$ from $0=z$ and $z=0$ to $1$ what about the boundaries of $C_1$?

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I believe you should write your integral like this: $$I_{C1} = \int_0^1\int_0^1\int_0^1f(x, y, z)dx.dy.dz$$ And for C2: $$I_{C2} = \int_y^1\int_x^z\int_0^yf(x, y, z)dx.dy.dz$$

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  • $\begingroup$ I dont think so.. because this represent a straight line in 3D so the volume will be zero... I think it is different from $0<x,y,z<1$ $\endgroup$
    – Klaus82
    Commented Oct 27, 2016 at 0:52
  • $\begingroup$ I see what you mean. But I interpreted C1 and C2 as mere boundaries of a triple integral of a function that was not given to us. $\endgroup$ Commented Oct 27, 2016 at 1:09
  • $\begingroup$ Sorry, I probably misunderstood your question. $\endgroup$ Commented Oct 27, 2016 at 1:10

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