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Question Number 95773 by Fikret last updated on 27/May/20

Answered by Fikret last updated on 27/May/20

$$$$$$findthegeometricalcentersof$$$$thefollowingsections$$

Answered by john santu last updated on 27/May/20

$$\mathrm{shaded}\mathrm{area}={\left(\mathrm{xy}\right)}^2-\frac{1}{4}\pi {\left(\mathrm{xy}\right)}^2$$$$=\left(\frac{4-\pi }{4}\right){\left(\mathrm{xy}\right)}^2$$

Commented by Fikret last updated on 27/May/20

$$wantedgeommetrikcenter$$

Answered by mr W last updated on 27/May/20

Commented by mr W last updated on 27/May/20

$$a=XY$$$$y=\sqrt{a^2-x^2}$$$$x_G=\frac{{\int }_0^a(a-\sqrt{a^2-x^2})xdx}{{\int }_0^a(a-\sqrt{a^2-x^2})dx}$$$$=\frac{\frac{a^3}{2}+\frac{1}{3}{\left[{(a^2-x^2)}^{\frac{3}{2}}\right]}_0^a}{a^2-\frac{1}{2}{[a^2{\sin }^{-1}\frac{x}{a}+x\sqrt{a^2-x^2}]}_0^a}$$$$=\frac{\frac{a^3}{2}-\frac{a^3}{3}}{a^2-\frac{\pi a^2}{4}}$$$$=\frac{2a}{3(4-\pi )}$$$$x_G^{{}^{\prime }}=a-x_G=\frac{(10-3\pi )a}{3(4-\pi )}\approx 0.2234a$$$$y_G^{{}^{\prime }}=x_G^{{}^{\prime }}=\frac{(10-3\pi )a}{3(4-\pi )}\approx 0.2234a$$

Answered by mr W last updated on 27/May/20

Commented by mr W last updated on 27/May/20

$$a=XY$$$$(a^2-\frac{\pi a^2}{4})x_G^{{}^{\prime }}+\frac{\pi a^2}{4}(a-\frac{4a}{3\pi })=a^2\times \frac{a}{2}$$$$\Rightarrow x_G^{{}^{\prime }}=\frac{(10a-3\pi )a}{3(4-\pi )}=y_G^{{}^{\prime }}$$

Commented by mr W last updated on 27/May/20

Commented by Fikret last updated on 27/May/20

$$thankyou$$

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