Solve the problem.
-Rewrite the integral
$\int _ { 0 } ^ { 10 } \int _ { 0 } ^ { 5 ( 1 - z / 10 ) } \int _ { 0 } ^ { 6 ( 1 - y / 5 - z / 10 ) } d x d y d z$
in the order $\mathrm { dz } \mathrm { dy } \mathrm { dx }$.
A) $\int _ { 0 } ^ { 10 } \int _ { 0 } ^ { 5 ( 1 - x / 6 ) } \int _ { 0 } ^ { 6 ( 1 - x / 6 - y / 5 ) } d z d y d x$
B) $\int _ { 0 } ^ { 6 } \int _ { 0 } ^ { 5 ( 1 - z / 10 ) } \int _ { 0 } ^ { 10 ( 1 - y / 5 - z / 10 ) } d z d y d x$
C) $\int _ { 0 } ^ { 6 } \int _ { 0 } ^ { 5 ( 1 - x / 6 ) } \int _ { 0 } ^ { 10 ( 1 - x / 6 - y / 5 ) } d z d y d x$
D) $\int _ { 0 } ^ { 10 } \int _ { 0 } ^ { 5 ( 1 - z / 10 ) } \int _ { 0 } ^ { 6 ( 1 - y / 5 - z / 10 ) } d z d y d x$

Question

Quizplus · Accepted Answer

We start with the innermost integral with respect to $x$. The limits of the integral with respect to $x$ are $0$ and $6(1-y/5-z/10)$. This suggests that $x$ only depends on $y$ and $z$. Hence, we have the following:

\begin{align*}
\int _ { 0 } ^ { 10 } \int _ { 0 } ^ { 5 ( 1 - z / 10 ) } \int _ { 0 } ^ { 6 ( 1 - y / 5 - z / 10 ) } d x d y d z &= \int _ { 0 } ^ { 10 } \int _ { 0 } ^ { 5 ( 1 - z / 10 ) } \left[\int _ { 0 } ^ { 6 ( 1 - y / 5 - z / 10 ) } d x ight] d y d z \
&= \int _ { 0 } ^ { 10 } \int _ { 0 } ^ { 5 ( 1 - z / 10 ) } 6(1-y/5-z/10) dy dz\
&= \int _ { 0 } ^ { 6 } \int _ { 0 } ^ { 5 ( 1 - x / 6 ) } \int _ { 0 } ^ { 10 ( 1 - x / 6 - y / 5 ) } d z d y d x
\end{align*}

Therefore, the answer is $\boxed{	extbf{(C)}}$.

Solve the Problem ∫010∫05(1−z/10)∫06(1−y/5−z/10)dxdydz\int _ { 0 } ^ { 10 } \int _ { 0 } ^ { 5 ( 1 - z / 10 ) } \int _ { 0 } ^ { 6 ( 1 - y / 5 - z / 10 ) } d x d y d z∫010​∫05(1−z/10)​∫06(1−y/5−z/10)​dxdydz

Solve the Problem $\int _ { 0 } ^ { 10 } \int _ { 0 } ^ { 5 ( 1 - z / 10 ) } \int _ { 0 } ^ { 6 ( 1 - y / 5 - z / 10 ) } d x d y d z$