1
$\begingroup$

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.

1 Answers 1

3

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 $\alpha=255$.

  • 0
    In the last integral it seems you have used a formula but I don't see which one but I remember very slightly, can you tell which formula is it? – 2018-01-06
  • 0
    @BAYMAX: $$\int_{g(S)}f(u,v,w)\>{\rm d}(u,v,w)=\int_S f\bigl(g(x,y,z)\bigr)\>\bigl|{\rm det}\bigl(g(x,y,z)\bigr)\bigr|\>{\rm d}(x,y,z)\ ,$$ $$\int_S(x-y)\>{\rm d}(x,y,z)=0\ .$$ – 2018-01-06
  • 0
    @ChristianBlatter sir here it is $\iiint (2x+y-2z) dxdydz $ and not $\iiint (2u+v-2z) du dv dw$. $(2x+y-2z)dxdydz $ must be converted to $g(S)$ space i.e., express (2x+y-2z) interms of u,v,w using g transformation right? pls correct me where i am going wrong. – 2018-01-20
  • 0
    @anirudhb: It is unfortunate that the OP has used $x$, $y$, $z$ as coordiate variables in both spaces. I have avoided that. – 2018-01-20
  • 0
    Sir, thank you for the reply – 2018-01-20