Find the center of mass of the lamina that occupies the region $D$ and has the given density function, if $D$ is bounded by the parabola $y = 1 - x ^ { 2 }$ and the $x$-axis.
$\rho ( x , y ) = 4 y$
A) $( 8,4 )$
B) $( 4,46.44 )$
C) $( 0,0.57 )$
D) $( 12,0.57 )$
E) None of these

Question

Quizplus · Accepted Answer

The center of mass $(\bar{x}, \bar{y})$ of a lamina can be found using the formulas $\bar{x} = \frac{1}{M} \int_{a}^{b} x \rho(x,y) dA$ and $\bar{y} = \frac{1}{M} \int_{a}^{b} y \rho(x,y) dA$, where $M$ is the total mass of the lamina. For a density function $\rho(x, y) = 4y$ over the region bounded by $y = 1 - x^2$ and the x-axis, symmetry about the y-axis implies $\bar{x} = 0$. To find $\bar{y}$, we integrate over the given region, taking into account the given density. The result $\bar{y} = 0.57$ indicates the average y-coordinate of the lamina's mass, considering its density distribution. The symmetry and the calculation of $\bar{y}$ lead to the correct answer being $(0, 0.57)$.

Find the Center of Mass of the Lamina That Occupies DDD

Find the Center of Mass of the Lamina That Occupies $D$