Answer to Question #342532 in Calculus for kate lim

Question

Find the centroid for the region bounded by y=3−e^-X, the x−axis, x=2, and the y−axis.

 

Answer 1

ANSWER The centroid of the region is $(x_{C},y_{C})$ : $x_{C}=\frac{5+3e^{-2}}{5+e^{-2}}\cong1.053, y_{C}=\frac{\frac{25}{4}+3e^{-2}-\frac{e^{-4}}{4}}{5+e^{-2}}\cong1.295$

EXPLANATION

The centroid of the plane region $D=\left\{(x,y): 0\leq x\leq 2, 0\leq y\leq 3-e^{-x}\right\}$ is $(x_{C},y_{C})$ , where $x_{C}=\frac{M_{x=0}}{A},\, y_{C}=\frac{M_{y=0}}{A}$ and $A=\int_{0}^{2}(3-e^{-x} ) dx, \, M_{x=0}=\int_{0}^{2}x\cdot(3-e^{-x} ) dx,\, M_{y=0}=\frac{1}{2}\int_{0}^{2} (3-e^{-x} )^{2} dy.$

We calculate these integrals.

1) $A=\int_{0}^{2}(3-e^{-x} ) dx = \left [ 3x+e^{-x} \right ]_{0}^{2}=6+e^{-2} -1=5+e^{ -2}\cong5.1353$ .

2) Let $x=u, \, (3-e^{-x} )dx=dv$ , then $dx=du,$ $v=3x+e^{-x}$ and

$M_{x=0}=\int_{0}^{2}x\cdot \left ( 3-e^{-x} \right )dx=\left [ x\cdot \left ( 3x+e^{-x} \right ) \right ]_{0}^{2}-\int_{0}^{2}\left ( 3x+e^{-x} \right )dx=\\=\left ( 12+2e^{-2} \right )-\left [ \frac{3x^{2}}{2}-e^{-x} \right ]_{0}^{2}= (12+2e^{-2})-[6-e^{-2}+1]=5+3e^{-2}\cong5.4060$

3) $M_{y=0}=\frac{1}{2}\int_{0}^{2}\left ( 3-e^{-x} \right )^{2} dx=\frac{1}{2}\int_{0}^{2}\left ( 9-6e^{-x} +e^{-2x}\right ) dx=\frac{1}{2}\left [ 9x+6e^{-x} -\frac{e^{-2x}}{2}\right ]_{0}^{2}=\frac{1}{2}\left ( 18+6e^{-2 } -\frac{e^{-4}}{2}-6+\frac{1}{2}\right )=\frac{25}{4}+3e^{-2}-\frac{e^{-4}}{4}\cong6.6514$