Answer to Question #176687 in Calculus for Joshua

Question

Find the center mass of the region bounded by y=e^2x and y=-cos(πx) between - 1/2 ≤x≤1/2.

Answer 1

assuming uniform mass density so that $ρ(x,y)=k$ for k a constant, the center of mass coincides with the centroid of the region;

we have

total mass; $m=ρ∫_{-1/2}^{1/2} [h(x)-g(x)]dx$

$=k∫_{-1/2}^{1/2}[e^2x-(-cosπx)dx]$

$=k[\frac12 e^2x+\frac1π sin⁡πx ]_{-1/2}^{1/2}$

$=[(\frac12 e+\frac 1π sin⁡ \fracπx )-(\frac 12 e^{-1}+ \frac 1π sin \frac {-π}{2 } )]k$

$=(\frac 12 e-\frac12 e^{-1} +\frac 2π)k …(i)$

the moments of the region about the x- axis equal;

$m_x=\frac k 2 ∫_{-1/2}^{1/2}[(e^2x )^2+(-cos⁡πx )^2] dx$

$=\frac 12 k ∫_{-1/2}^{1/2}(e^4π+cos^2⁡πx)dx$

$=\frac12k ∫_{-1/2}^{1/2}[e^4x+\frac 12 (1+cos⁡2πx )]dx$

$=\frac12 k [\frac14 e^2+\frac 12 x+ \frac14π sin⁡2πx ]_{-1/2}^{1/2}$

$=\frac 12 k [(\frac14 e^2+\frac14+\frac14π sin⁡π )-(\frac 14 e^{-2}-\frac14+\frac14π sin⁡(-π))]$

$=\frac12 k [\frac14 e^2-\frac14 e^{-2}+\frac12]=k[\frac {e^4-1+2e^2 }{4e^2 }]…(ii)$

and moments of the region about y-axis ;

$=k∫_{-1/2}^{1/2} (xe^2x+xcos πx)dx$

$=k[\frac12 xe^2x-\frac14 e^2x+\frac1π x sin⁡πx+\frac1{π^2} cos⁡πx ]_{-1/2}^{1/2}$

$=k(\frac12 e^{-1} )=\frac12 ke^{-1}$

the center of mass, ( $\bar x$ , $\bar y$ ) is the point

$\bar x=m_y/m$ and $\bar y=m_x/m$

$\bar x=\frac {(\frac12 e^{-1})}{(k(\frac12 e-1/2 e^{-1}+\frac2π)}=\fracπ{πe^2-π+4 e}=0.102$

$\bar y =\frac{\frac12 k [\frac14 e^2-\frac14 e^{-2}+\frac12]} {k[\frac 12 e-- \frac 12 e^{-1}+\frac 2 \pi ))}= \frac {(πe^4-π+2πe^2)} {(4πe^3-4πe+16 e^2 )}=0.638$

$(0.102,0.638)$ is the center of mass of the region