Answer
Using geometry as in Example 2, we evaluate:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal R}^{} \left( {15 - 3x} \right){\rm{d}}A = 112.5$
Please see the sketch attached.
Work Step by Step
Similar to Example 2, we use geometry to evaluate the double integral $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal R}^{} \left( {15 - 3x} \right){\rm{d}}A$.
Referring to the attached figure, we see that the double integral is equal to the volume $V$ of the solid wedge underneath the graph of $z = 15 - 3x$.
Now, the triangular face of the wedge has area $A = \frac{1}{2}\cdot5\cdot15 = \frac{{75}}{2}$.
The volume of the wedge is equal to the area $A$ times the length $3$. Therefore,
$V = \frac{{75}}{2}\cdot3 = \frac{{225}}{2} = 112.5$
Hence, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal R}^{} \left( {15 - 3x} \right){\rm{d}}A = 112.5$.
