# Volume of region

I need to find the volume of the region determined by:

First octant, $z+y=1$ and $z+x=1$.

My response is the result of the integrals:

$$\int_0^1 \int_0^x \int_0^{1-y} dzdydx+\int_0^1\int_0^x \int_0^{1-x}dzdydx$$

Is this correct? Thank you for your help.

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No, it is not correct. It should rather be $$\int_0^1 \int_0^x \int_0^{1-\min(x,y)} \mathrm{d}z \mathrm{d} y \mathrm{d}x$$ See the picture:

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thanks. How do I get the $\min(x,y)$? Why my response is not correct? – Hiperion Apr 13 '12 at 16:04
@Hiperion Sorry my previous comment made no sense. The reason your result is not correct, is that you want to integrate over region where $0<x+z<1$ and $0<y+z<1$ both hold. The first integral integrates over a region, where $0<x+z<1$ may not hold. – Sasha Apr 13 '12 at 16:21
Since you need both $x+z<1$ and $y+z<1$, it means that $z<1-x \land z<1-y$, that is $z < \max(1-x, 1-y) = 1- \min(x,y)$ – Sasha Apr 13 '12 at 16:24