We have $g:\mathbb R^3\to\mathbb R^3$,is defined by $g(x,y,z)=(3x+4z,2x-3z,x+3y)$
and $S=\{(x,y,z)\in\mathbb R^3:0\leq x\leq 1,0\leq y\leq 1,0\leq z\leq 1\}$ and it is given that$$\displaystyle\iiint_{g(S)}(2x+y-2z)dxdydz=\alpha\iiint_Sz\,dxdydz$$.How can we find the value of $\alpha?$
MY TRY:I know how to solve these integrals but problem is that ,i'm confused with $g(S)$.Thank you.
The set $S$ is the unit cube in $(x,y,z)$-space, and $g(S)$ is the image of this cube in $(u,v,w)$-space produced by the linear map $$g:\quad (x,y,z)\mapsto(u,v,w):=(3x+4z,2x-3z,x+3y)\ .$$ One computes $\det (g)=51$. It follows that $$2u+v-2w=6x-6y+5z\ ,$$ so that $$\int_{g(S)}(2u+v-2w)\>{\rm d}(u,v,w)=51\int_S(6x-6y+5z)\>{\rm d}(x,y,z)=255\int_S z\>{\rm d}(x,y,z)\ ,$$ since the terms $6x$ and $6y$ give the same contribution. It follows that $\alpha=255$.