Find and describe the domain of $ g$
let$$g(x,y,z)= x^3y^2z \sqrt{\ 10-x-y-z} $$
my approach was by considering
$$ 10-x-y-z \geq 0 $$
but I can't seem to know how to describe it by a $3D$ domain sketching representation, in addition I need to check my answer of the domain approach by getting intersection between the functions in
$$g(x,y,z)$$
By considering the domain of the following function we get the intersection of all the functions such as $$ x^3 , Df=R$$ $$ y^2,Df=R$$ $$ z, Df=R$$
$$ \sqrt {\ 10-x-y-z} $$ $Df$ of the root : $$ 10-x-y-z \geq 0$$ $$x+y+z \leq 10 $$ the intersecting domain between all previous function will be $$Df ={(x,y,z) ; x+y+z \leq 10}$$
$x+y+z \leq 10 $ which is the equation of plane :
$$ ax+by+cz=d$$
for easier sketching than the known triangle intersecting the three axis one can consider drawing the $ 3D$ function by each two planes that is $XY$ plane , $XZ$ plane ,$YZ$plane .
$ @ XY plane , z=0 $
$\therefore x+y \leq 10$
$ y \leq 10-x$
that's a straight line
$ @XZ plane , y=0 $
$\therefore x+z \leq 10$
$ z \leq 10-x$
that's a straight line
$ @ YZ plane , x=0 $
$\therefore y+z \leq 10$
$ z \leq 10-y$
that's a straight line
by connecting the three points a plane is formed and $ \ hence $ by substituting points @ each 2 dimensional plane we check this point satisfies or not with the condition we get to know where the area from the line : above or below , does the domain lie which will be in that case the area between the triangular lines causing the 3D dimensional triangular plane.