91Ó°ÊÓ

The function:

$$\begin{gathered} f\left(x\right)=x^{2}and \\ g\left(x\right)=64-x^2 \end{gathered}$$

Centroid of the region between two curves is given by the formula,

$(\stackrel{¯}{x},\stackrel{¯}{y})=(\frac{{∫}_a^{b}x\left|f\left(x\right)-g\left(x\right)\right|dx}{{∫}_a^b\left|f\left(x\right)-g\left(x\right)\right|dx},\frac{{\displaystyle \frac{1}{2}}{∫}_a^b\left|f{\left(x\right)}^2-g{\left(x\right)}^2\right|dx}{{∫}_a^b\left|f\left(x\right)-g\left(x\right)\right|dx})$

$$\begin{gathered} {∫}_a^b\left|f\left(x\right)-g\left(x\right)\right|dx={∫}_0^2\left|f\left(0\right)-g\left(0\right)\right|(0.5).dx+{∫}_2^4\left|f\left(2\right)-g\left(2\right)\right|(0.5).dx+{∫}_4^6\left|f\left(4\right)-g\left(4\right)\right|(0.5).dx+{∫}_6^8\left|f\left(6\right)-g\left(6\right)\right|(0.5).dx \\ =[{∫}_0^{2}64.dx+{∫}_2^{4}56.dx+{∫}_4^{6}32.dx+{∫}_6^{8}8.dx]0.5 \\ =\left[64\right(2)+56(2)+32(2)+8(2\left)\right]0.5 \\ =160 \\ {∫}_a^{b}x\left|f\left(x\right)-g\left(x\right)\right|dx={∫}_0^2\left|f\left(0\right)-g\left(0\right)\right|(0.5).dx+{∫}_2^4\left|f\left(2\right)-g\left(2\right)\right|(0.5).dx+{∫}_4^6\left|f\left(4\right)-g\left(4\right)\right|(0.5).dx+{∫}_6^8\left|f\left(6\right)-g\left(6\right)\right|(0.5).dx \\ =64{\left[\frac{x^2}{2}\right]}_0^2+56{\left[\frac{x^2}{2}\right]}_2^4+32{\left[\frac{x^2}{2}\right]}_4^6+16{\left[\frac{x^2}{2}\right]}_6^8 \\ =128+336+320+224 \\ =1008 \end{gathered}$$

$$\begin{gathered} {∫}_a^b\left|f{\left(x\right)}^2-g{\left(x\right)}^2\right|dx=\underset{n=0}{\overset{8}{0.5∑}}{∫}_0^2\left|{\left(x^2\right)}^2-{(64-x^2)}^2\right|dx \\ =0.5\underset{n=0}{\overset{8}{∑}}{∫}_0^2\left|(x^4-4096+x^4)\right|dx \\ =0.5\underset{n=0}{\overset{8}{∑}}{∫}_0^2\left|(2x^4-4096)\right|dx \\ =0.5\underset{n=0}{\overset{8}{∑}}{\left[\left|\frac{2}{5}{x}^5-4096x\right|\right]}_0^2=0.5×\frac{2}{5}[8160+7200+1140+16800] \\ =0.2(33,300) \\ =6660 \end{gathered}$$

$$\begin{gathered} (\stackrel{¯}{x},\stackrel{¯}{y})=(\frac{{∫}_a^{b}x\left|f\left(x\right)-g\left(x\right)\right|dx}{{∫}_a^b\left|f\left(x\right)-g\left(x\right)\right|dx},\frac{{\displaystyle \frac{1}{2}{∫}_a^b\left|f{\left(x\right)}^2-g{\left(x\right)}^2\right|dx}}{{∫}_a^b\left|f\left(x\right)-g\left(x\right)\right|dx}) \\ =(\frac{1008}{160},\frac{{\displaystyle \frac{1}{2}}×6660}{160}) \\ =(6.3,20.8125) \end{gathered}$$