Answer
Using Fubini's Theorem we show that the solid has volume: $V = AL$.
Work Step by Step
Referring to Figure 18, let the region on the $xy$-plane be ${\cal R} = \left[ {0,L} \right] \times \left[ {0,d} \right]$. So, the volume $V$ is given by
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal R} f\left( {x,y} \right){\rm{d}}y{\rm{d}}x$
Assuming that $f\left( {x,y} \right)$ is continuous, thus by Fubini's Theorem the double integral can be written as an iterated integral:
$V = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal R} f\left( {x,y} \right){\rm{d}}y{\rm{d}}x = \mathop \smallint \limits_{x = 0}^L \left( {\mathop \smallint \limits_{y = 0}^d f\left( {x,y} \right){\rm{d}}y} \right){\rm{d}}x$
But $\mathop \smallint \limits_{y = 0}^d f\left( {x,y} \right){\rm{d}}y$ is the area $A$, that is, the area of the front face of the solid.
So,
$V = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal R} f\left( {x,y} \right){\rm{d}}y{\rm{d}}x = \mathop \smallint \limits_{x = 0}^L \left( {\mathop \smallint \limits_{y = 0}^d f\left( {x,y} \right){\rm{d}}y} \right){\rm{d}}x$
$ = A\mathop \smallint \limits_{x = 0}^L {\rm{d}}x$
$ = AL$
Hence, the volume $V = AL$.
