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Suppose we have a unit square $[0,1] \times [0,1]$ and some function $f(x,y)$. Say you want to find the volume below the the plane $y=x$. Would it be

$$\int_{0}^{1} \int_{y}^{1} f(x,y) \ dx \ dy$$?

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In $3$D (i.e. $xyz$), $y=x$ is not a line. It's a plane. – user2468 Mar 21 '12 at 0:29
And it's a vertical plane, so it's not clear what's meant by "below". – Gerry Myerson Mar 21 '12 at 0:36
James, a plane will divide the 3D space into two sides. It is not hard to visualize/draw in 3D. Which side do you think about? – user2468 Mar 21 '12 at 0:38
@J.D.: All I mean is the region for which $y<x$ in $[0,1] \times [0,1]$. I am correct right? – James Mar 21 '12 at 0:42
Actually, I was guessing what you wanted in the previous comment. If $f(x,y)\ge 0$ and if $R$ is the region in the $x$-$y$ plane for which $(x,y)\in [0,1]\times[0,1]$ and $y<x$, then the volume of the solid bounded above by the graph of $f$ and below by $R$ is the integral in your post. – David Mitra Mar 21 '12 at 1:00

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