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Question-213661




Question Number 213661 by Spillover last updated on 13/Nov/24
Commented by MathematicalUser2357 last updated on 14/Nov/24
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Answered by mahdipoor last updated on 14/Nov/24
My^− =∫_M ydm ⇒ρ=cte , dm=ρdV  Vy^− =∫_V ydV=∫_0 ^( 1) y(πz^2 dy)=π∫_0 ^( 1) y((y/4))dy=(π/(12))  V=∫_0 ^( 1) (πz^2 )dy=(π/8)  ⇒y^− = ((π/12)/(π/8))=(2/3)
$${M}\overset{−} {{y}}=\int_{{M}} {ydm}\:\Rightarrow\rho={cte}\:,\:{dm}=\rho{dV} \\ $$$${V}\overset{−} {{y}}=\int_{{V}} {ydV}=\int_{\mathrm{0}} ^{\:\mathrm{1}} {y}\left(\pi{z}^{\mathrm{2}} {dy}\right)=\pi\int_{\mathrm{0}} ^{\:\mathrm{1}} {y}\left(\frac{{y}}{\mathrm{4}}\right){dy}=\frac{\pi}{\mathrm{12}} \\ $$$${V}=\int_{\mathrm{0}} ^{\:\mathrm{1}} \left(\pi{z}^{\mathrm{2}} \right){dy}=\frac{\pi}{\mathrm{8}} \\ $$$$\Rightarrow\overset{−} {{y}}=\:\frac{\pi/\mathrm{12}}{\pi/\mathrm{8}}=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$ \\ $$

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