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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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    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, then the volume of the solid bounded above by the graph of $f$ and below by $R$ is the integral in your post.2012-03-21

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