Find the center of mass of the lamina bounded by the graphs of the equations $y = e ^ { x } , y = 0 , x = 0$, and $x = 20$ for the density $\rho = k$
A) $\left( \frac { 1 - 21 e ^ { 21 } } { e ^ { 20 } - 1 } , \frac { e ^ { 21 } + 1 } { 2 } \right)$
B) $\left( \frac { 1 + 19 e ^ { 20 } } { e ^ { 20 } - 1 } , \frac { e ^ { 20 } + 1 } { 4 } \right)$
C) $\left( \frac { 1 - 19 e ^ { 20 } } { e ^ { 20 } + 1 } , \frac { e ^ { 20 } - 1 } { 4 } \right)$
D)$\left( \frac { 1 + 21 e ^ { 21 } } { e ^ { 20 } - 1 } , \frac { e ^ { 21 } - 1 } { 4 } \right)$
E) $\left( \frac { 1 + 19 e ^ { 20 } } { e ^ { 20 } + 1 } , \frac { e ^ { 20 } + 1 } { 2 } \right)$

Question

Find the center of mass of the lamina bounded by the graphs of the equations $y = e ^ { x } , y = 0 , x = 0$, and $x = 20$ for the density $\rho = k$
A) $\left( \frac { 1 - 21 e ^ { 21 } } { e ^ { 20 } - 1 } , \frac { e ^ { 21 } + 1 } { 2 } \right)$
B) $\left( \frac { 1 + 19 e ^ { 20 } } { e ^ { 20 } - 1 } , \frac { e ^ { 20 } + 1 } { 4 } \right)$
C) $\left( \frac { 1 - 19 e ^ { 20 } } { e ^ { 20 } + 1 } , \frac { e ^ { 20 } - 1 } { 4 } \right)$
D)$\left( \frac { 1 + 21 e ^ { 21 } } { e ^ { 20 } - 1 } , \frac { e ^ { 21 } - 1 } { 4 } \right)$ 
E) $\left( \frac { 1 + 19 e ^ { 20 } } { e ^ { 20 } + 1 } , \frac { e ^ { 20 } + 1 } { 2 } \right)$

Quizplus · Accepted Answer

The center of mass ($x_{\text{cm}}, y_{\text{cm}}$) of a lamina can be found using the formulas $x_{\text{cm}} = \frac{1}{M} \int x \rho(x,y) \,dA$ and $y_{\text{cm}} = \frac{1}{M} \int y \rho(x,y) \,dA$, where $M$ is the total mass of the lamina and $\rho(x,y)$ is the density function. Given $\rho = k$, a constant, the density function simplifies the integrals since it can be factored out. The bounds of integration are from $x = 0$ to $x = 20$ and $y = 0$ to $y = e^x$. Calculating these integrals and solving for $x_{\text{cm}}$ and $y_{\text{cm}}$ with the given bounds yields the coordinates in choice B.

Find the Center of Mass of the Lamina Bounded by the Graphs