Solve the problem.
-Find the center of mass of the rectangular solid of density $\delta ( x , y , z ) = x y z$ defined by $0 \leq x \leq 10,0 \leq y \leq 6,0 \leq z \leq 8$.
A) $\bar { x } = \frac { 20 } { 3 } , \bar { y } = 4 , \bar { z } = \frac { 16 } { 3 }$
B) $\bar { x } = \frac { 5 } { 2 } , \bar { y } = \frac { 3 } { 2 } , \bar { z } = 2$
C) $\bar { x } = 5 , \bar { y } = 3 , \bar { z } = 4$
D) $\bar { x } = \frac { 10 } { 3 } , \bar { y } = 2 , \bar { z } = \frac { 8 } { 3 }$

Question

Quizplus · Accepted Answer

The center of mass $(\bar{x}, \bar{y}, \bar{z})$ of a solid with density function $\delta(x, y, z) = xyz$ over the given limits can be found using the formulas for the coordinates of the center of mass: $\bar{x} = \frac{1}{M}\int\int\int_V x\delta(x,y,z) dV$, $\bar{y} = \frac{1}{M}\int\int\int_V y\delta(x,y,z) dV$, and $\bar{z} = \frac{1}{M}\int\int\int_V z\delta(x,y,z) dV$, where $M$ is the total mass of the solid. Given the symmetry and the linear increase of density with $x$, $y$, and $z$, the center of mass will be at the geometric center of the solid, which is at $\bar{x} = \frac{20}{3}$, $\bar{y} = 4$, and $\bar{z} = \frac{16}{3}$, considering the integration over the specified limits.

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