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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.

1 Answers 1

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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:

enter image description here

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    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)2012-04-13