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What is the volume of the solid bounded by $x+y+z=1$, $x=0$, $y=0$, and $z=0$? I had problems defining the limits of integration.

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You need $x+y+z<1$ so you have $$ \int_0^1\left(\int_0^{1-x}\left(\int_{0}^{1-x-y}dz\right) dy\right)dx$$

You need $z< 1-x-y$ and $y<1-x$, otherwise you go outside the simplex.

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    You mean for the dx and dy to be on the outside, right? – 2017-02-12
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    I mean to indicate that $x$ is the first integral, $y$ is the second and $z$ is the third. I like putting $dx$ right after the integral sign, before the expression cause it's easier to indicate what belongs to what. Will change to what you're used to though. – 2017-02-12
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    Got it, thanks! – 2017-02-12