Use the given transformation to evaluate the integral.
-$$\begin{aligned}
u = & 2 x + y - z , v = - x + y + z , w = - x + y + 2 z \
& \iiint ( 2 x + y - z ) ( z - x + y ) d x d y d z
\end{aligned}$$
where $R$ is the parallelepiped bounded by the planes $2 x + y - z = 2,2 x + y - z = 6 , - x + y + z = 3 , - x + y + z = 4$, $- x + y + 2 z = 6 , - x + y + 2 z = 8$
A) $\frac { 224 } { 3 }$
B) $\frac { 112 } { 3 }$
C) 336
D) 672

Question

Quizplus · Accepted Answer

Given the transformation and the bounds for $u, v, w$, the integral simplifies to integrating $u$ over the new bounds. The bounds for $u$ are from 2 to 6, for $v$ from 3 to 4, and for $w$ from 6 to 8. The integral becomes $\int_{2}^{6} \int_{3}^{4} \int_{6}^{8} u \, dw \, dv \, du$, which simplifies to $\frac{112}{3}$ after performing the integration.