<s>[INST] Application of Triple Integrals Any easy way to solve this integration?
$\int_{0}^{3} \int_{0}^{\frac{6-2x}{3}} \int_{0}^{\frac{6-3y-2x}{2}} xyz \ dzdydx$
My work
$\int \int \int_D xyz \ dV = \dfrac{1}{2}\int_{0}^{3} \int_{0}^{\frac{6-2x}{3}}  xy(z^2 )\biggr|_{0}^{\frac{6-3y-2x}{2}} \ dy \ dx$
$\int \int \int_D xyz \ dV = \dfrac{1}{8}\int_{0}^{3} \int_{0}^{\dfrac{6-2x}{3}}  xy(6-3y-2x)^2 \ dy \ dx$
$\int \int \int_D xyz \ dV = \dfrac{1}{8}\int_{0}^{3} x \left [y\dfrac{(6-3y-2x)^3}{3(-3)} -\dfrac{(6-3y-2x)^4}{4(-3)3(-3)} \right ]\biggr|_{0}^{\dfrac{6-2x}{3}} \ dx$
$\int \int \int_D xyz \ dV = \dfrac{1}{8}\int_{0}^{3} x \left [y\dfrac{(6-3y-2x)^3}{3(-3)} -\dfrac{(6-3y-2x)^4}{4(-3)3(-3)} \right ]\biggr|_{0}^{\dfrac{6-2x}{3}} \ dx$
The terms are expanding, any way to find a solution in easy case. please help
 [/INST] The domain of integration is $D$, the set of $(x,y,z)$ with $x,y,z\ge 0$ and $2x+3y+2z\le 6$. Well, to have a simpler, symmetric formula, we can change variables. For instance $X=2x/6$, $Y=3y/6$, $Z=2z/6$. Then the domain $D$ is transformed into the simplex (cone) $E$ of all $(X,Y,Z)$ with $X,Y,Z\ge 0$ and $X+Y+Z\le 1$. Formally, $x=3X$, $y=2Y$, $z=3Z$, so $dx\; dy\; dz=3\cdot 2\cdot 3\; dX\; dY\; dZ$. We can compute in a few easy lines, explicitly, the integral $J$ (notation) from the OP:
$$
\begin{aligned}
J &=\iiint_D xyz\; dx\; dy\; dz\\
&=\iiint_E (3X)(2Y)(3Z)\cdot (3\cdot 2\cdot 3)\; dX\; dY\; dZ\\
&=324\iiint_{\substack{0\le X,Y,Z\\X+Y+Z\le 1}}XYZ\; dX\; dY\; dZ\\
&=324\int_0^1X\;dX\int_0^{1-X}Y\; dY\int_0^{1-X-Y}Z\; dZ\\
&=324\int_0^1X\;dX\int_0^{1-X}Y\; dY\cdot\left[\frac 12Z^2\right]_0^{1-X-Y}\\
&=324\cdot \frac 12\int_0^1X\;dX\int_0^{1-X}Y(1-X-Y)^2\; dY\\
&=324\cdot \frac 12\int_0^1X\;dX\int_0^{1-X}\Big(\ (1-X)(1-X-Y)^2-(1-X-Y)^3\ \Big)\; dY\\
&=324\cdot \frac 12\int_0^1X\;dX\left[
-\frac 13(1-X)(1-X-Y)^3+\frac 14(1-X-Y)^4\right]_0^{1-X}
\\
&=324\cdot \frac 12\int_0^1X\;dX\left(
\frac 13(1-X)^4-\frac 14(1-X)^4\right)
\\
&=324\cdot \frac 12\int_0^1X\cdot\frac 13\cdot \frac 14(1-X)^4\;dX
\\
&=324\cdot \frac 12\cdot\frac 13\cdot \frac 14\int_0^1(1-(1-X))\cdot(1-X)^4\;dX
\\
&=324\cdot \frac 12\cdot\frac 13\cdot \frac 14\int_0^1\Big((1-X)^4-(1-X)^5\Big)\;dX
\\
&=324\cdot \frac 12\cdot\frac 13\cdot \frac 14\left(\frac 15-\frac 16\right)\;dX
\\
&=324\cdot \frac 12\cdot\frac 13\cdot \frac 14\cdot \frac 15\cdot \frac 16\\
&=\frac{18^2}{6!}=\frac 9{20}\ .
\end{aligned}
$$
$\square$
Note: I saw no quick way to proceed using Stokes...
Homework: Try to find a generalization. What is for instance the integral of $XYZW$ w.r.t. volume $dX\; dY\; dZ\;dW$ on the simplex $0\le X,Y,Z,W\le X+Y+Z+W\le 1$?
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