Solve the problem.
-by the plane $\frac { x } { 8 } + \frac { y } { 3 } + \frac { z } { 5 } = 1$.
A)
$\int _ { 0 } ^ { 5 } \int _ { 0 } ^ { 8 ( 1 - y / 3 ) } \int _ { 0 } ^ { 8 ( 1 - y / 3 - z / 5 ) } d x d y d z$
B) $\int _ { 0 } ^ { 5 } \int _ { 0 } ^ { 1 - z / 5 } \int _ { 0 } ^ { 1 - y / 3 - z / 5 } d x d y d z$
C) $\int _ { 0 } ^ { 5 } \int _ { 0 } ^ { 3 ( 1 - z / 5 ) } \int _ { 0 } ^ { 8 ( 1 - y / 3 - z / 5 ) } d x d y d z$
D) $\int _ { 0 } ^ { 5 } \int _ { 0 } ^ { 1 - y / 3 } \int _ { 0 } ^ { 1 - y / 3 - z / 5 } d x d y d z$

Question

Quizplus · Accepted Answer

The correct setup for the triple integral bounded by the plane $\frac{x}{8} + \frac{y}{3} + \frac{z}{5} = 1$ involves integrating in the correct order and limits that respect the equation of the plane. Choice C correctly sets the limits for $z$ from 0 to 5, for $y$ from 0 to $3(1 - z/5)$ which accounts for the decreasing limit of $y$ as $z$ increases, and for $x$ from 0 to $8(1 - y/3 - z/5)$ which accounts for the decreasing limit of $x$ as both $y$ and $z$ increase, all in accordance with the given plane equation.

Solve the Problem x8+y3+z5=1\frac { x } { 8 } + \frac { y } { 3 } + \frac { z } { 5 } = 18x​+3y​+5z​=1

Solve the Problem $\frac { x } { 8 } + \frac { y } { 3 } + \frac { z } { 5 } = 1$