Question Number 32506 by Tinkutara last updated on 26/Mar/18
Answered by mrW2 last updated on 26/Mar/18
![let BC=b Area of ΔABC=A=((bh)/2) M=ρA=((ρbh)/2) I=∫r^2 dM=∫(h−x)^2 ρdA=ρ∫_0 ^( h) (h−x)^2 ((xb)/h)dx =((ρb)/h)∫_0 ^( h) (h−x)^2 xdx =((ρb)/h)∫_0 ^( h) (h^2 x−2hx^2 +x^3 )dx =((ρb)/h)[h^2 (x^2 /2)−2h(x^3 /3)+(x^4 /4)]_0 ^h =((ρb)/h)[(h^4 /2)−((2h^4 )/3)+(h^4 /4)] =((ρbh^3 )/(12))=((ρbh)/2)×(h^2 /6) =((Mh^2 )/6)](Q32521.png)
$${let}\:{BC}={b} \\ $$$${Area}\:{of}\:\Delta{ABC}={A}=\frac{{bh}}{\mathrm{2}} \\ $$$${M}=\rho{A}=\frac{\rho{bh}}{\mathrm{2}} \\ $$$${I}=\int{r}^{\mathrm{2}} {dM}=\int\left({h}−{x}\right)^{\mathrm{2}} \rho{dA}=\rho\int_{\mathrm{0}} ^{\:{h}} \left({h}−{x}\right)^{\mathrm{2}} \frac{{xb}}{{h}}{dx} \\ $$$$=\frac{\rho{b}}{{h}}\int_{\mathrm{0}} ^{\:{h}} \left({h}−{x}\right)^{\mathrm{2}} {xdx} \\ $$$$=\frac{\rho{b}}{{h}}\int_{\mathrm{0}} ^{\:{h}} \left({h}^{\mathrm{2}} {x}−\mathrm{2}{hx}^{\mathrm{2}} +{x}^{\mathrm{3}} \right){dx} \\ $$$$=\frac{\rho{b}}{{h}}\left[{h}^{\mathrm{2}} \frac{{x}^{\mathrm{2}} }{\mathrm{2}}−\mathrm{2}{h}\frac{{x}^{\mathrm{3}} }{\mathrm{3}}+\frac{{x}^{\mathrm{4}} }{\mathrm{4}}\right]_{\mathrm{0}} ^{{h}} \\ $$$$=\frac{\rho{b}}{{h}}\left[\frac{{h}^{\mathrm{4}} }{\mathrm{2}}−\frac{\mathrm{2}{h}^{\mathrm{4}} }{\mathrm{3}}+\frac{{h}^{\mathrm{4}} }{\mathrm{4}}\right] \\ $$$$=\frac{\rho{bh}^{\mathrm{3}} }{\mathrm{12}}=\frac{\rho{bh}}{\mathrm{2}}×\frac{{h}^{\mathrm{2}} }{\mathrm{6}} \\ $$$$=\frac{{Mh}^{\mathrm{2}} }{\mathrm{6}} \\ $$
Commented by Tinkutara last updated on 27/Mar/18
$${Why}\:{dA}=\frac{{xb}}{{h}}{dx}? \\ $$
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
$${b}'=\frac{{x}}{{h}}×{b} \\ $$$${dA}={b}'\:{dx}=\frac{{b}}{{h}}{xdx} \\ $$