$$\text { Find the center of mass of the wire that lies along the curve } r \text { and has density } \delta \text {. }$$
-$$r ( t ) = ( 3 + 3 t ) i + j - 3 t k , 0 \leq t \leq 1 ; \delta ( x , y , z ) = x + z ^ { 2 }$$
A) $\left( \frac { 49 } { 10 } , 1 , - \frac { 19 } { 10 } \right)$
B) $( 441,0 , - 171 )$
C) $\left( \frac { 147 } { 4 } , \frac { 15 } { 2 } , - 171 \right)$
D) $\left( \frac { 49 } { 10 } , 0 , - \frac { 19 } { 10 } \right)$

Question

Quizplus · Accepted Answer

The center of mass $(\bar{x}, \bar{y}, \bar{z})$ of a wire with density $\delta(x, y, z)$ along a curve $r(t)$ is given by $\bar{x} = \frac{1}{M} \int x \delta \, dm$, $\bar{y} = \frac{1}{M} \int y \delta \, dm$, and $\bar{z} = \frac{1}{M} \int z \delta \, dm$, where $M = \int \delta \, dm$. Given $r(t) = (3 + 3t)i + j - 3tk$ and $\delta(x, y, z) = x + z^2$, we substitute $x = 3 + 3t$, $y = 1$, and $z = -3t$ into $\delta$ and integrate over the given range of $t$ to find the center of mass. The correct calculation leads to the center of mass being $\left( \frac { 49 } { 10 } , 1 , - \frac { 19 } { 10 } \right)$.

Find the center of mass of the wire that lies along the curve r and has density δ. \text { Find the center of mass of the wire that lies along the curve } r \text { and has density } \delta \text {. } Find the center of mass of the wire that lies along the curve r and has density δ.

$\text { Find the center of mass of the wire that lies along the curve } r \text { and has density } \delta \text {. }$