<s>[INST] find the volume in the octant bounded by x+y+z=9,2x+3y=18 and x+3y=9 Find the volume in the first octant bounded by x+y+z=9,2x+3y=18 and x+3y=9.
it's obvious that z=9-x-y, but for range of x and y i am confused. from 2x+3y=18 and x+3y=9, i find that x=9-(3/2)y and x=9-3y, so 9-(3/2)y≤x≤9-3y. is that correct? also for y, 3≤y≤6( assume x=0).  
so volume=∫(3to6)∫(9-(3/2)y to 9-2y)(9-x-y)dx dy. answer is not correct(should be 81/2)i don't know which part i got wrong. 
 [/INST] Looking at the two constraints that don't involve $z$, we have the following:

Take the limits of $x$ as $0 \le x \le 9$
Then the lower bound of $y$ is given by $x+3y=9 \Rightarrow y=\frac {9-x}{3}$
and the lower bound of $y$ is given by $2x+3y=18 \Rightarrow y=\frac {18-2x}{3}$
The integral you want is $V=\int_{x=0}^{x=9} \int_{y=\frac {9-x}{3}}^{y=\frac {18-2x}{3}}(9-x-y) dy dx $
$V=\int_{x=0}^{x=9} [9y-xy-\frac 12y^2]_{y=\frac {9-x}{3}}^{y=\frac {18-2x}{3}} dx $
$V=\int_{x=0}^{x=9} [9\frac {18-2x}{3}-x\frac {18-2x}{3}-\frac 12(\frac {18-2x}{3})^2-9\frac {9-x}{3}+x\frac {9-x}{3}+\frac 12(\frac {9-x}{3})^2] dx $
$V=\int_{x=0}^{x=9} \frac 1{18}[54(18-2x)-6x(18-2x)-(18-2x)^2-54(9-x)+6x(9-x)+(9-x)^2] dx $
$V=\int_{x=0}^{x=9} \frac 1{18}[972-108x-108x+12x^2-324+72x-4x^2-486+54x+54x-6x^2+81-18x+x^2] dx $
$V=\int_{x=0}^{x=9} \frac 1{18}[243-54x+3x^2]dx $
$V=[\frac 1{18}(243x-27x^2+x^3)]_{x=0}^{x=9}  $
$V=[\frac 1{18}(243.9-27.81+729)]$
$V=\frac {729}{18}$
$V=\frac {81}{2}$
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