Answer
1. First way: in the order $dxdydz$
$\mathop \smallint \limits_{z = 1}^3 \mathop \smallint \limits_{y = 0}^1 \mathop \smallint \limits_{x = 0}^2 \left( {xy + z} \right){\rm{d}}x{\rm{d}}y{\rm{d}}z = 10$
2. Second way: in the order $dydxdz$
$\mathop \smallint \limits_{z = 1}^3 \mathop \smallint \limits_{x = 0}^2 \mathop \smallint \limits_{y = 0}^1 \left( {xy + z} \right){\rm{d}}y{\rm{d}}x{\rm{d}}z = 10$
The two results agree.
Work Step by Step
1. First way: in the order $dxdydz$
$\mathop \smallint \limits_{z = 1}^3 \mathop \smallint \limits_{y = 0}^1 \mathop \smallint \limits_{x = 0}^2 \left( {xy + z} \right){\rm{d}}x{\rm{d}}y{\rm{d}}z$
$ = \mathop \smallint \limits_{z = 1}^3 \mathop \smallint \limits_{y = 0}^1 \left( {\left( {\frac{1}{2}{x^2}y + xz} \right)|_0^2} \right){\rm{d}}y{\rm{d}}z$
$ = \mathop \smallint \limits_{z = 1}^3 \mathop \smallint \limits_{y = 0}^1 \left( {2y + 2z} \right){\rm{d}}y{\rm{d}}z$
$ = \mathop \smallint \limits_{z = 1}^3 \left( {\left( {{y^2} + 2yz} \right)|_0^1} \right){\rm{d}}z$
$ = \mathop \smallint \limits_{z = 1}^3 \left( {1 + 2z} \right){\rm{d}}z$
$ = \left( {z + {z^2}} \right)|_1^3 = 3 + 9 - 1 - 1 = 10$
So, $\mathop \smallint \limits_{z = 1}^3 \mathop \smallint \limits_{y = 0}^1 \mathop \smallint \limits_{x = 0}^2 \left( {xy + z} \right){\rm{d}}x{\rm{d}}y{\rm{d}}z = 10$.
2. Second way: in the order $dydxdz$
$\mathop \smallint \limits_{z = 1}^3 \mathop \smallint \limits_{x = 0}^2 \mathop \smallint \limits_{y = 0}^1 \left( {xy + z} \right){\rm{d}}y{\rm{d}}x{\rm{d}}z$
$ = \mathop \smallint \limits_{z = 1}^3 \mathop \smallint \limits_{x = 0}^2 \left( {\left( {\frac{1}{2}x{y^2} + yz} \right)|_0^1} \right){\rm{d}}x{\rm{d}}z$
$ = \mathop \smallint \limits_{z = 1}^3 \mathop \smallint \limits_{x = 0}^2 \left( {\frac{1}{2}x + z} \right){\rm{d}}x{\rm{d}}z$
$ = \mathop \smallint \limits_{z = 1}^3 \left( {\left( {\frac{1}{4}{x^2} + xz} \right)|_0^2} \right){\rm{d}}z$
$ = \mathop \smallint \limits_{z = 1}^3 \left( {1 + 2z} \right){\rm{d}}z$
$ = \left( {z + {z^2}} \right)|_1^3 = 3 + 9 - 1 - 1 = 10$
So, $\mathop \smallint \limits_{z = 1}^3 \mathop \smallint \limits_{x = 0}^2 \mathop \smallint \limits_{y = 0}^1 \left( {xy + z} \right){\rm{d}}y{\rm{d}}x{\rm{d}}z = 10$.
The two results agree.
