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Question Number 32506 by Tinkutara last updated on 26/Mar/18

Answered by mrW2 last updated on 26/Mar/18

$$letBC=b$$$$Areaof\mathrm{\Delta }ABC=A=\frac{bh}{2}$$$$M=\rho A=\frac{\rho bh}{2}$$$$I=\int r^{2}dM=\int {(h-x)}^2\rho dA=\rho {\int }_0^h{(h-x)}^2\frac{xb}{h}dx$$$$=\frac{\rho b}{h}{\int }_0^h{(h-x)}^{2}xdx$$$$=\frac{\rho b}{h}{\int }_0^h(h^{2}x-2{hx}^2+x^3)dx$$$$=\frac{\rho b}{h}{[h^2\frac{x^2}{2}-2h\frac{x^3}{3}+\frac{x^4}{4}]}_0^h$$$$=\frac{\rho b}{h}[\frac{h^4}{2}-\frac{2h^4}{3}+\frac{h^4}{4}]$$$$=\frac{\rho {bh}^3}{12}=\frac{\rho bh}{2}\times \frac{h^2}{6}$$$$=\frac{{Mh}^2}{6}$$

Commented by Tinkutara last updated on 27/Mar/18

$$WhydA=\frac{xb}{h}dx?$$

Commented by Tinkutara last updated on 27/Mar/18

Thank you very much Sir! ��������

Commented by mrW2 last updated on 27/Mar/18

Commented by mrW2 last updated on 27/Mar/18

$$b^{\prime }=\frac{x}{h}\times b$$$$dA=b^{\prime }dx=\frac{b}{h}xdx$$

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