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
 A: 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 $0\le X,Y,Z,W\le X+Y+Z+W\le 1$?
A: The domain of integration is a tetrahedron with vertices at $(0,0,0),(3,0,0),(0,2,0),(0,0,3)$. The integrand is symmetric with respect to permuting variables. Thus we are allowed to rotate the domain of integration so that slices perpendicular to the $z$-axis are right isosceles triangles and the integral is equal to
$$\int_0^2\int_0^{(6-3z)/2}\int_{(6-3z-2y)/2}xyz\,dx\,dy\,dz$$
We move the $z$ as far out as it will go:
$$\int_0^2z\int_0^{(6-3z)/2}\int_{(6-3z-2y)/2}xy\,dx\,dy\,dz$$
Then rewrite to make the slices of right isosceles triangles clearer. $T(k)$ is the triangle with vertices at $(0,0),(k,0),(0,k)$:
$$\int_0^2z\iint_{T((6-3z)/2)}xy\,dA\,dz$$
Now let's work out what the inner integral is before we substitute $(6-3z)/2$. In other words, what is
$$\iint_{T(k)}xy\,dA=\int_0^k\int_0^{x-k}xy\,dy\,dx$$
This evaluates to
$$\int_0^k x[y^2/2]_0^{x-k}\,dx=\frac12\int_0^k x(x-k)^2\,dx=\frac12[x^4/4-(2/3)kx^3+k^2x^2/2]_0^k=k^4/24$$
So our original integral becomes
$$\frac1{24}\int_0^2z\left(\frac{6-3z}2\right)^4\,dz=\frac{81}{24×16}\int_0^2z(2-z)^4\,dz$$
$$=\frac{27}{128}[8z^2-(32/3)z^3+6z^4-(8/5)z^5+z^6/6]_0^2$$
$$=\frac{27}{128}×\frac{32}{15}=\frac9{20}$$
