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Question Number 151899 by liberty last updated on 24/Aug/21

Answered by john_santu last updated on 24/Aug/21

$$\mathrm{A}={\int }_0^6(\frac{-{\mathrm{x}}^3}{6}+{\mathrm{x}}^2)\mathrm{dx}={[-\frac{{\mathrm{x}}^4}{24}+\frac{{\mathrm{x}}^3}{3}]}_0^6$$$$\mathrm{A}=-\frac{6^4}{24}+\frac{6^3}{3}=18$$$$\mathrm{xA}={\int }_0^6(-\frac{{\mathrm{x}}^4}{6}+{\mathrm{x}}^3)\mathrm{dx}={[-\frac{{\mathrm{x}}^5}{30}+\frac{{\mathrm{x}}^4}{4}]}_0^6$$$$\mathrm{xA}=-\frac{6^5}{30}+\frac{6^4}{4}=\frac{324}{5}$$$$\stackrel{-}{\mathrm{x}}=\frac{\mathrm{xA}}{\mathrm{A}}=\frac{324}{90}=\frac{18}{5}$$$$\mathrm{yA}={\int }_0^6\frac{1}{2}{(\frac{-{\mathrm{x}}^3}{6}+{\mathrm{x}}^2)}^2\mathrm{dx}$$$$\mathrm{yA}=\frac{1}{2}{\int }_0^6(\frac{{\mathrm{x}}^6}{36}-\frac{{\mathrm{x}}^5}{3}+{\mathrm{x}}^4)\mathrm{dx}$$$$\mathrm{yA}=\frac{1}{2}{[\frac{{\mathrm{x}}^7}{252}-\frac{{\mathrm{x}}^6}{18}+\frac{{\mathrm{x}}^5}{5}]}_0^6$$$$\mathrm{yA}=\frac{6^7}{504}-\frac{6^6}{36}+\frac{6^5}{10}=\frac{1296}{35}$$$$\stackrel{-}{\mathrm{y}}=\frac{\mathrm{yA}}{\mathrm{y}}=\frac{1296}{630}=\frac{72}{35}$$$$\mathrm{therefore}\mathrm{centroid}\mathrm{C}(\frac{18}{5},\frac{72}{35})$$

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