moment of inertia of an area with respect to the line x = 2 Member for 10 years 1 month Submitted by Roydon Jude S… on Wed, 03/16/2016 - 16:48 The moment of inertia bounded by y=x^3 , y=2x , X + y= 6, with respect to x=2 pa help po until Friday. Plus ask ko Lang po ang pinakakapakipakinabang na calculus book for integration applications. Thanks. Tags area by integration Log in or register to post comments Re: moment of inertia Member for 17 years 6 months Jhun Vert Thu, 03/17/2016 - 06:41 $I_{\text{at }x = 2} = \Sigma \left[ {\displaystyle \int_{x_1}^{x_2}} (2 - x)^2 \, dA \right]$ $I_{\text{at }x = 2} = {\displaystyle \int_{\sqrt{2}}^{1.634}} (2 - x)^2(x^3 - 2x) \, dx + {\displaystyle \int_{1.634}^2} (2 - x)^2 [ \, (6 - x) - 2x \, ] \, dx$ $I_{\text{at }x = 2} = 0.0351 ~ \text{unit}^4$ Log in or register to post comments
Re: moment of inertia Member for 17 years 6 months Jhun Vert Thu, 03/17/2016 - 06:41 $I_{\text{at }x = 2} = \Sigma \left[ {\displaystyle \int_{x_1}^{x_2}} (2 - x)^2 \, dA \right]$ $I_{\text{at }x = 2} = {\displaystyle \int_{\sqrt{2}}^{1.634}} (2 - x)^2(x^3 - 2x) \, dx + {\displaystyle \int_{1.634}^2} (2 - x)^2 [ \, (6 - x) - 2x \, ] \, dx$ $I_{\text{at }x = 2} = 0.0351 ~ \text{unit}^4$ Log in or register to post comments
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Re: moment of inertia
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17 years 6 months$I_{\text{at }x = 2} = \Sigma \left[ {\displaystyle \int_{x_1}^{x_2}} (2 - x)^2 \, dA \right]$
$I_{\text{at }x = 2} = {\displaystyle \int_{\sqrt{2}}^{1.634}} (2 - x)^2(x^3 - 2x) \, dx + {\displaystyle \int_{1.634}^2} (2 - x)^2 [ \, (6 - x) - 2x \, ] \, dx$
$I_{\text{at }x = 2} = 0.0351 ~ \text{unit}^4$