The center of mass of a lamina in the shape of a region in the xy-plane bounded by $r = 2 + \cos ( \theta ) , 0 \leq \theta \leq \pi$ and the polar axis with area density $\rho ( r , \theta ) = \sin \theta$ is

A) $\left( \frac { 15 } { 28 } , \frac { 5 } { 8 } \right)$
B) $\left( 0 , \frac { 21 } { 10 } \right)$
C) $\left( - \frac { 57 } { 44 } , 0 \right)$
D) $\left( \frac { 42 } { 65 } , \frac { 19 \pi } { 52 } \right)$
E) $\left( \frac { 8 } { 5 } , \frac { 8 } { 5 } \right)$

Question

The center of mass of a lamina in the shape of a region in the xy-plane bounded by

r = 2 + \cos ( \theta ) , 0 \leq \theta \leq \pi

and the polar axis with area density

\rho ( r , \theta ) = \sin \theta

is

A) $\left( \frac { 15 } { 28 } , \frac { 5 } { 8 } \right)$
B) $\left( 0 , \frac { 21 } { 10 } \right)$
C) $\left( - \frac { 57 } { 44 } , 0 \right)$
D) $\left( \frac { 42 } { 65 } , \frac { 19 \pi } { 52 } \right)$
E) $\left( \frac { 8 } { 5 } , \frac { 8 } { 5 } \right)$

Quizplus · Accepted Answer

The center of mass ($x_{cm}, y_{cm}$) of a lamina can be found using the formulas for the coordinates in polar coordinates, where $x_{cm} = \frac{1}{M} \int x dm$ and $y_{cm} = \frac{1}{M} \int y dm$, with $dm = \rho(r, \theta) r d\theta dr$. Given the density function $\rho(r, \theta) = \sin(\theta)$ and the bounds for $r$ and $\theta$, we integrate over the specified region. The correct answer must account for the given density and the bounds in polar coordinates, leading to the calculation of both $x_{cm}$ and $y_{cm}$ that matches the provided option.

The Center of Mass of a Lamina in the Shape r=2+cos⁡(θ),0≤θ≤πr = 2 + \cos ( \theta ) , 0 \leq \theta \leq \pir=2+cos(θ),0≤θ≤π

The Center of Mass of a Lamina in the Shape $r = 2 + \cos ( \theta ) , 0 \leq \theta \leq \pi$