Change of Variables — Question 8

Question 8 – Solution

Strategy. A nonsingular linear map sends the unit cube to a parallelepiped with constant volume scale.

Step 1: Jacobian and geometry

$$J=\frac{\partial(x,y,z)}{\partial(u,v,w)} =\begin{vmatrix}1&1&0\\0&1&1\\1&0&1\end{vmatrix}=2.$$ Because $J\ne 0$, the linear map is invertible and covers its image once.

Step 2: Transform the integrand We have $x+y+z=2(u+v+w)$ and $dV=2\,du\,dv\,dw$. Therefore $$\begin{align*} I&=\int_0^1\int_0^1\int_0^1 4(u+v+w)\,dw\,dv\,du\\ &=4\left(\frac 12+\frac 12+\frac 12\right) =\boxed{6}. \end{align*}$$

Verification The image volume is $|J|$ times the cube volume, hence $2$. The cube centroid $(1/2,1/2,1/2)$ maps to $(1,1,1)$, where $x+y+z=3$. A linear function averages to its centroid value, so $I=(2)(3)=6$.