Integrating a 3 - form in R3

We can integrate a 3 form using this result.
Example: Find
$\int_S \omega^3$
where
$\omega^3 =dx_1 \wedge dx_2 \wedge dx_3$
and
$S$
is the image of the cube
$K=\{(u_1,u_2,u_3):0 \leq u_1 ,\: u_2 , \: u_3 \leq 1 \}$
under the function
$\mathbf{f} (u_1 , u_2 , u_3) =(u_1 , u_2^2 , u_3^3)$
.
\begin{aligned} \\ &= \left| \begin{array}{ccc} \frac{\partial f_1}{\partial u_1} & \frac{\partial f_1}{\partial u_2} & \frac{\partial f_1}{\partial u_3} \\ \frac{\partial f_2}{\partial u_1} & \frac{\partial f_2}{\partial u_2} & \frac{\partial f_2}{\partial u_3} \\ \frac{\partial f_3}{\partial u_1} & \frac{\partial f_3}{\partial u_2} & \frac{\partial f_3}{\partial u_3} \end{array} \right| \\ &= \left| \begin{array}{ccc} 1 & 0 &0 \\ 0 & u_2 & 0 \\ 0 & 0 & 3u_3^2 \end{array} \right| \\ &=6u_2u_3^2\end{aligned}

Then
\begin{aligned} \int_S \omega^3 &= \int_K 6u_2 u_3^2 du_1 du_2 du_3 \\ &= \int^1_0 \int^1_0 \int^1_0 6u_2 u_3^2 du_1 du_2 du_3 \\ &= \int^1_0 \int^1_0 6u_2 u_3^2 du_2 du_3 \\ &= \int^1_0 [3u^2_2 u_3^2]^{1}_{u_2=0} du_3 \\ &= \int^1_0 3u_3^2 du_3 \\ &= [u^3_3]^1_0 =1 \end{aligned}