Solve the problem.
-Rewrite the integral
$\int _ { 0 } ^ { 1 / 4 } \int _ { 0 } ^ { ( 1 - 4 z ) / 8 } \int _ { 0 } ^ { ( 1 - 8 y - 4 z ) / 6 } d x d y d z$
in the order $\mathrm { dz } \mathrm { dy } \mathrm { dx }$.
A) $\int _ { 0 } ^ { 1 / 6 } \int _ { 0 } ^ { ( 1 - 6 x ) / 8 } \int _ { 0 } ^ { ( 1 - 6 x - 8 y ) / 4 } d z d y d x$
B) $\int _ { 0 } ^ { 1 / 6 } \int _ { 0 } ^ { ( 1 - 6 x ) / 8 } \int _ { 0 } ^ { ( 1 - 8 x - 6 y ) / 4 } d z d y d x$
C) $\int _ { 0 } ^ { 1 / 4 } \int _ { 0 } ^ { ( 1 - 4 z ) / 8 } \int _ { 0 } ^ { ( 1 - 8 y - 4 z ) / 6 } d z d y d x$
D) $\int _ { 0 } ^ { 1 / 4 } \int _ { 0 } ^ { ( 1 - 6 x ) / 8 } \int _ { 0 } ^ { ( 1 - 4 x - 8 y ) / 6 } d z d y d x$

Question

Quizplus · Accepted Answer

To change the order of integration from $dx dy dz$ to $dz dy dx$, we need to re-evaluate the limits for each variable based on the new order. The original limits for $x$ are from $0$ to $(1 - 8y - 4z)/6$, for $y$ from $0$ to $(1 - 4z)/8$, and for $z$ from $0$ to $1/4$. When rearranging to $dz dy dx$, we need to solve the inequalities for $z$ in terms of $x$ and $y$, and similarly for $y$ in terms of $x$, to find the new limits. The correct limits for $x$ are from $0$ to $1/6$, for $y$ from $0$ to $(1 - 6x)/8$, and for $z$ from $0$ to $(1 - 6x - 8y)/4$, which matches choice A.

Solve the Problem ∫01/4∫0(1−4z)/8∫0(1−8y−4z)/6dxdydz\int _ { 0 } ^ { 1 / 4 } \int _ { 0 } ^ { ( 1 - 4 z ) / 8 } \int _ { 0 } ^ { ( 1 - 8 y - 4 z ) / 6 } d x d y d z∫01/4​∫0(1−4z)/8​∫0(1−8y−4z)/6​dxdydz

Solve the Problem $\int _ { 0 } ^ { 1 / 4 } \int _ { 0 } ^ { ( 1 - 4 z ) / 8 } \int _ { 0 } ^ { ( 1 - 8 y - 4 z ) / 6 } d x d y d z$