Answer
$\dfrac{abc}{24}(a+4)$
Work Step by Step
The Divergence Theorem states that $\iint_S \overrightarrow{F}\cdot d\overrightarrow{S}=\iiint_Ediv \overrightarrow{F}dV $
Here, $S$ is a closed surface and $E$ is the region inside that surface.
$div F=\dfrac{\partial a}{\partial x}+\dfrac{\partial b}{\partial y}+\dfrac{\partial c}{\partial z}$
This implies that $div F=\dfrac{\partial (z)}{\partial x}+\dfrac{\partial (y)}{\partial y}+\dfrac{\partial (zx)}{\partial z}$
$div F=0+1+x=1+x$
Now, we have
$\iiint_E (1+x)dV=\int_{0}^{a}\int_0^{b(1-x/a)} \int_0^{c(1-x/a-y/b)} (1+x) dzdydx$
$=\int_{0}^{a}\int_0^{b(1-x/a)} (1+x)(z)_0^{c(1-x/a-y/b)}dydx$
We need to use $u$- Substitution by plugging in $u=1-\dfrac{x}{a} ; du=\dfrac{-dx}{a}; dx=-adu$ and the limit becomes: when $x=0$ then $u=1$ ; when $x=a$, then $u=0$
Now, we have
$=(ac) \int_{0}^{1} (1+a-au)[\dfrac{bu^2}{2}]du$
$=\dfrac{abc}{2} \int_{0}^{1} (u^2+au^2-au^3) du$
$=\dfrac{abc}{2}[\dfrac{u^3}{3}+\dfrac{au^3}{3}-\dfrac{au^4}{4}]_0^1 du$
$=\dfrac{abc}{2} \times \dfrac{1}{3}+\dfrac{a(1)^3}{3}-\dfrac{a(1)^4}{4}$
$=\dfrac{abc}{24}(a+4)$
