S is the part of the plane x+2y+3z=6 in the first octant; the density at a point P on S is directly proportional to the square of the distance between P and the yz plane.
My attempt so far: the distance between P and the yz plane is x, so density=kx^2
f(g(y,z),y,z) where g(y,z)=6-2y-3z.
partial derivative gy=-2
partial derivative gz=-3
z limits: 0 to 6
y limits: 0 to 1
and then I'm about to plug it in to the surface integral equation.
But I dont think I'm doing this right...
Equation of plane $$ z = 2 - \frac x3 -\frac{2y}3$$ $y$ can range from $0$ to $3-\frac x2$ (solve for $y$ where $z=0$)
$x$ ranges from $0$ to $6$
So $$ M = k\int_0^6\int_0^{3-\frac x2}x^2 dy\; dx $$
You could also do
$x$ ranges from $0$ to $6-2y$
$y$ ranges from $0$ to $3$
So $$ M = k\int_0^3\int_0^{6-2y}x^2 dx\; dy $$
you should get the same answer in either case ( I think it works out to $M=54k$)