Question Number 36613 by ajfour last updated on 03/Jun/18
Commented by ajfour last updated on 04/Jun/18
$${i}\:{just}\:{want}\:{to}\:{confirm}\:{with}\:{the} \\ $$$${right}\:{answer}.\:{please}\:{help}\:{finding}\:{it}. \\ $$
Commented by MrW3 last updated on 11/Jun/18
$${AB}=\mathrm{2}{a}\:{or}\:\mathrm{4}{a}? \\ $$
Commented by ajfour last updated on 12/Jun/18
$${AB}=\mathrm{2}{a} \\ $$
Commented by ajfour last updated on 12/Jun/18
Commented by ajfour last updated on 13/Jun/18
![eq. of ellipsoid : (((x−a)^2 )/a^2 )+(y^2 /b^2 )+(z^2 /c^2 )=1 y=y′sin ψ ; z=z′cos ψ (dI/dx)=ρ∫_(−z′) ^( z′) (x^2 +z^2 )(2y)dz =2ρ∫_π ^( 0) (x^2 +z′^2 cos^2 φ)(y′sin φ)(−z′sin φdφ) =2ρx^2 y′z′∫_0 ^( π) sin^2 φdφ +2ρy′z′^3 ∫_0 ^( π) cos^2 φsin^2 φdφ =ρx^2 y′z′∫_0 ^( π) (1−cos 2φ)dφ +(1/4)ρy′z′^3 ∫_0 ^( π) (1−cos 4φ)dφ =πρx^2 y′z′+(π/4)ρy′z′^3 y′=b(√(1−(((x−a)^2 )/a^2 ))) z′=c(√(1−(((x−a)^2 )/a^2 ))) I=πρbc∫_0 ^( 2a) x^2 [1−(((x−a)^2 )/a^2 )]dx +(π/4)ρbc^3 ∫_0 ^( 2a) [1−(((x−a)^2 )/a^2 )]^2 dx =πρbc∫_0 ^( 2a) (−(x^4 /a^2 )+((2x^3 )/a))dx +(π/4)ρbc^3 ∫_0 ^( 2a) (1+(((x−a)^4 )/a^4 )−((2(x−a)^2 )/a^2 ))dx =πρbca^3 (−((32)/5)+8) +((πρabc^3 )/4)(2+(2/5)−(4/3)) =πρabc[((8a^2 )/5)+(c^2 /4)(((16)/(15)))] =((4πρabc)/(15))(6a^2 +c^2 ) and with M=(4/3)πρabc I=(1/5)M(6a^2 +c^2 ) .](https://www.tinkutara.com/question/Q37415.png)
$${eq}.\:{of}\:{ellipsoid}\::\:\frac{\left({x}−{a}\right)^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }+\frac{{z}^{\mathrm{2}} }{{c}^{\mathrm{2}} }=\mathrm{1} \\ $$$${y}={y}'\mathrm{sin}\:\psi\:\:;\:\:{z}={z}'\mathrm{cos}\:\psi \\ $$$$\frac{{dI}}{{dx}}=\rho\int_{−{z}'} ^{\:\:{z}'} \left({x}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)\left(\mathrm{2}{y}\right){dz} \\ $$$$=\mathrm{2}\rho\int_{\pi} ^{\:\:\mathrm{0}} \left({x}^{\mathrm{2}} +{z}'^{\mathrm{2}} \mathrm{cos}\:^{\mathrm{2}} \phi\right)\left({y}'\mathrm{sin}\:\phi\right)\left(−{z}'\mathrm{sin}\:\phi{d}\phi\right) \\ $$$$=\mathrm{2}\rho{x}^{\mathrm{2}} {y}'{z}'\int_{\mathrm{0}} ^{\:\:\pi} \mathrm{sin}\:^{\mathrm{2}} \phi{d}\phi \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{2}\rho{y}'{z}'^{\mathrm{3}} \int_{\mathrm{0}} ^{\:\:\pi} \mathrm{cos}\:^{\mathrm{2}} \phi\mathrm{sin}\:^{\mathrm{2}} \phi{d}\phi \\ $$$$=\rho{x}^{\mathrm{2}} {y}'{z}'\int_{\mathrm{0}} ^{\:\:\pi} \left(\mathrm{1}−\mathrm{cos}\:\mathrm{2}\phi\right){d}\phi \\ $$$$\:\:\:\:\:\:\:\:\:\:\:+\frac{\mathrm{1}}{\mathrm{4}}\rho{y}'{z}'^{\mathrm{3}} \int_{\mathrm{0}} ^{\:\:\pi} \left(\mathrm{1}−\mathrm{cos}\:\mathrm{4}\phi\right){d}\phi \\ $$$$=\pi\rho{x}^{\mathrm{2}} {y}'{z}'+\frac{\pi}{\mathrm{4}}\rho{y}'{z}'^{\mathrm{3}} \\ $$$${y}'={b}\sqrt{\mathrm{1}−\frac{\left({x}−{a}\right)^{\mathrm{2}} }{{a}^{\mathrm{2}} }} \\ $$$${z}'={c}\sqrt{\mathrm{1}−\frac{\left({x}−{a}\right)^{\mathrm{2}} }{{a}^{\mathrm{2}} }} \\ $$$${I}=\pi\rho{bc}\int_{\mathrm{0}} ^{\:\:\mathrm{2}{a}} {x}^{\mathrm{2}} \left[\mathrm{1}−\frac{\left({x}−{a}\right)^{\mathrm{2}} }{{a}^{\mathrm{2}} }\right]{dx} \\ $$$$\:\:\:\:\:\:\:+\frac{\pi}{\mathrm{4}}\rho{bc}^{\mathrm{3}} \int_{\mathrm{0}} ^{\:\:\mathrm{2}{a}} \left[\mathrm{1}−\frac{\left({x}−{a}\right)^{\mathrm{2}} }{{a}^{\mathrm{2}} }\right]^{\mathrm{2}} {dx} \\ $$$$\:\:=\pi\rho{bc}\int_{\mathrm{0}} ^{\:\:\mathrm{2}{a}} \left(−\frac{{x}^{\mathrm{4}} }{{a}^{\mathrm{2}} }+\frac{\mathrm{2}{x}^{\mathrm{3}} }{{a}}\right){dx} \\ $$$$\:+\frac{\pi}{\mathrm{4}}\rho{bc}^{\mathrm{3}} \int_{\mathrm{0}} ^{\:\:\mathrm{2}{a}} \left(\mathrm{1}+\frac{\left({x}−{a}\right)^{\mathrm{4}} }{{a}^{\mathrm{4}} }−\frac{\mathrm{2}\left({x}−{a}\right)^{\mathrm{2}} }{{a}^{\mathrm{2}} }\right){dx} \\ $$$$=\pi\rho{bca}^{\mathrm{3}} \left(−\frac{\mathrm{32}}{\mathrm{5}}+\mathrm{8}\right) \\ $$$$\:\:\:\:\:+\frac{\pi\rho{abc}^{\mathrm{3}} }{\mathrm{4}}\left(\mathrm{2}+\frac{\mathrm{2}}{\mathrm{5}}−\frac{\mathrm{4}}{\mathrm{3}}\right) \\ $$$$=\pi\rho{abc}\left[\frac{\mathrm{8}{a}^{\mathrm{2}} }{\mathrm{5}}+\frac{{c}^{\mathrm{2}} }{\mathrm{4}}\left(\frac{\mathrm{16}}{\mathrm{15}}\right)\right] \\ $$$$=\frac{\mathrm{4}\pi\rho{abc}}{\mathrm{15}}\left(\mathrm{6}{a}^{\mathrm{2}} +{c}^{\mathrm{2}} \right) \\ $$$${and}\:{with}\:\:{M}=\frac{\mathrm{4}}{\mathrm{3}}\pi\rho{abc} \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{5}}{M}\left(\mathrm{6}{a}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)\:. \\ $$$$ \\ $$
Answered by MrW3 last updated on 12/Jun/18
![let′s say x_A =−a, x_B =a the intersection at x is an ellipse with semi minor axis b_x and c_x . ((x/a))^2 +((b_x /b))^2 =1 ((x/a))^2 +((c_x /c))^2 =1 ⇒b_x ^2 =[1−(x^2 /a^2 )]b^2 ⇒b_x =b(√(1−(x^2 /a^2 ))) ⇒c_x ^2 =[1−(x^2 /a^2 )]c^2 ⇒c_x =c(√(1−(x^2 /a^2 ))) dI_A =((ρπ)/4)b_x c_x ^3 dx+ρπb_x c_x (a+x)^2 dx =((ρπ)/4) (1−(x^2 /a^2 ))^2 bc^3 dx+ρπ(1−(x^2 /a^2 ))bc (1+(x/a))^2 a^2 dx =((ρbcπ)/4){c^2 (1−(x^2 /a^2 ))^2 +4a^2 (1−(x^2 /a^2 ))(1+(x/a))^2 }dx =((ρabcπ)/4){c^2 (1−(x^2 /a^2 ))^2 +4a^2 (1−(x^2 /a^2 ))(1+(x/a))^2 }d(x/a) I_A =((ρabcπ)/4)∫_(−a) ^a {c^2 (1−(x^2 /a^2 ))^2 +4a^2 (1−(x^2 /a^2 ))(1+(x/a))^2 }d(x/a) I_A =((ρabcπ)/4)∫_(−1) ^1 {c^2 (1−λ^2 )^2 +4a^2 (1−λ^2 )(1+λ)^2 }dλ I_A =((ρabcπ)/4)∫_(−1) ^1 {c^2 (1−2λ^2 +λ^4 )+4a^2 (1−λ^2 )(1+2λ+λ^2 )}dλ I_A =((ρabcπ)/4)∫_(−1) ^1 {c^2 (1−2λ^2 +λ^4 )+4a^2 (1+2λ−2λ^3 −λ^4 )}dλ I_A =((ρabcπ)/4){c^2 [λ−(2/3)λ^3 +(1/5)λ^5 ]_(−1) ^1 +4a^2 [λ+λ^2 −(1/2)λ^4 −(1/5)λ^5 ]_(−1) ^1 } I_A =((ρabcπ)/4){c^2 [2−(4/3)+(2/5)]+4a^2 [2−(2/5)]} I_A =((ρabcπ)/4){c^2 [((30−20+6)/(15))]+4a^2 [((10−2)/5)]} I_A =((4ρabcπ)/(15))(c^2 +6a^2 ) since M=ρV=((4ρπabc)/3) ⇒I_A =(M/5)(c^2 +6a^2 )](https://www.tinkutara.com/question/Q37387.png)
$${let}'{s}\:{say}\:{x}_{{A}} =−{a},\:{x}_{{B}} ={a} \\ $$$${the}\:{intersection}\:{at}\:{x}\:{is}\:{an}\:{ellipse} \\ $$$${with}\:{semi}\:{minor}\:{axis}\:{b}_{{x}} \:{and}\:{c}_{{x}} . \\ $$$$\left(\frac{{x}}{{a}}\right)^{\mathrm{2}} +\left(\frac{{b}_{{x}} }{{b}}\right)^{\mathrm{2}} =\mathrm{1} \\ $$$$\left(\frac{{x}}{{a}}\right)^{\mathrm{2}} +\left(\frac{{c}_{{x}} }{{c}}\right)^{\mathrm{2}} =\mathrm{1} \\ $$$$\Rightarrow{b}_{{x}} ^{\mathrm{2}} =\left[\mathrm{1}−\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\right]{b}^{\mathrm{2}} \Rightarrow{b}_{{x}} ={b}\sqrt{\mathrm{1}−\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }} \\ $$$$\Rightarrow{c}_{{x}} ^{\mathrm{2}} =\left[\mathrm{1}−\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\right]{c}^{\mathrm{2}} \Rightarrow{c}_{{x}} ={c}\sqrt{\mathrm{1}−\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }} \\ $$$${dI}_{{A}} =\frac{\rho\pi}{\mathrm{4}}{b}_{{x}} {c}_{{x}} ^{\mathrm{3}} \:{dx}+\rho\pi{b}_{{x}} {c}_{{x}} \:\left({a}+{x}\right)^{\mathrm{2}} \:{dx} \\ $$$$=\frac{\rho\pi}{\mathrm{4}}\:\left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\right)^{\mathrm{2}} {bc}^{\mathrm{3}} \:{dx}+\rho\pi\left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\right){bc}\:\left(\mathrm{1}+\frac{{x}}{{a}}\right)^{\mathrm{2}} {a}^{\mathrm{2}} \:{dx} \\ $$$$=\frac{\rho{bc}\pi}{\mathrm{4}}\left\{{c}^{\mathrm{2}} \:\left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\right)^{\mathrm{2}} +\mathrm{4}{a}^{\mathrm{2}} \left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\right)\left(\mathrm{1}+\frac{{x}}{{a}}\right)^{\mathrm{2}} \right\}{dx} \\ $$$$=\frac{\rho{abc}\pi}{\mathrm{4}}\left\{{c}^{\mathrm{2}} \:\left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\right)^{\mathrm{2}} +\mathrm{4}{a}^{\mathrm{2}} \left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\right)\left(\mathrm{1}+\frac{{x}}{{a}}\right)^{\mathrm{2}} \right\}{d}\frac{{x}}{{a}} \\ $$$${I}_{{A}} =\frac{\rho{abc}\pi}{\mathrm{4}}\int_{−{a}} ^{{a}} \left\{{c}^{\mathrm{2}} \:\left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\right)^{\mathrm{2}} +\mathrm{4}{a}^{\mathrm{2}} \left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\right)\left(\mathrm{1}+\frac{{x}}{{a}}\right)^{\mathrm{2}} \right\}{d}\frac{{x}}{{a}} \\ $$$${I}_{{A}} =\frac{\rho{abc}\pi}{\mathrm{4}}\int_{−\mathrm{1}} ^{\mathrm{1}} \left\{{c}^{\mathrm{2}} \:\left(\mathrm{1}−\lambda^{\mathrm{2}} \right)^{\mathrm{2}} +\mathrm{4}{a}^{\mathrm{2}} \left(\mathrm{1}−\lambda^{\mathrm{2}} \right)\left(\mathrm{1}+\lambda\right)^{\mathrm{2}} \right\}{d}\lambda \\ $$$${I}_{{A}} =\frac{\rho{abc}\pi}{\mathrm{4}}\int_{−\mathrm{1}} ^{\mathrm{1}} \left\{{c}^{\mathrm{2}} \:\left(\mathrm{1}−\mathrm{2}\lambda^{\mathrm{2}} +\lambda^{\mathrm{4}} \right)+\mathrm{4}{a}^{\mathrm{2}} \left(\mathrm{1}−\lambda^{\mathrm{2}} \right)\left(\mathrm{1}+\mathrm{2}\lambda+\lambda^{\mathrm{2}} \right)\right\}{d}\lambda \\ $$$${I}_{{A}} =\frac{\rho{abc}\pi}{\mathrm{4}}\int_{−\mathrm{1}} ^{\mathrm{1}} \left\{{c}^{\mathrm{2}} \:\left(\mathrm{1}−\mathrm{2}\lambda^{\mathrm{2}} +\lambda^{\mathrm{4}} \right)+\mathrm{4}{a}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{2}\lambda−\mathrm{2}\lambda^{\mathrm{3}} −\lambda^{\mathrm{4}} \right)\right\}{d}\lambda \\ $$$${I}_{{A}} =\frac{\rho{abc}\pi}{\mathrm{4}}\left\{{c}^{\mathrm{2}} \:\left[\lambda−\frac{\mathrm{2}}{\mathrm{3}}\lambda^{\mathrm{3}} +\frac{\mathrm{1}}{\mathrm{5}}\lambda^{\mathrm{5}} \right]_{−\mathrm{1}} ^{\mathrm{1}} +\mathrm{4}{a}^{\mathrm{2}} \left[\lambda+\lambda^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}\lambda^{\mathrm{4}} −\frac{\mathrm{1}}{\mathrm{5}}\lambda^{\mathrm{5}} \right]_{−\mathrm{1}} ^{\mathrm{1}} \right\} \\ $$$${I}_{{A}} =\frac{\rho{abc}\pi}{\mathrm{4}}\left\{{c}^{\mathrm{2}} \:\left[\mathrm{2}−\frac{\mathrm{4}}{\mathrm{3}}+\frac{\mathrm{2}}{\mathrm{5}}\right]+\mathrm{4}{a}^{\mathrm{2}} \left[\mathrm{2}−\frac{\mathrm{2}}{\mathrm{5}}\right]\right\} \\ $$$${I}_{{A}} =\frac{\rho{abc}\pi}{\mathrm{4}}\left\{{c}^{\mathrm{2}} \:\left[\frac{\mathrm{30}−\mathrm{20}+\mathrm{6}}{\mathrm{15}}\right]+\mathrm{4}{a}^{\mathrm{2}} \left[\frac{\mathrm{10}−\mathrm{2}}{\mathrm{5}}\right]\right\} \\ $$$${I}_{{A}} =\frac{\mathrm{4}\rho{abc}\pi}{\mathrm{15}}\left({c}^{\mathrm{2}} +\mathrm{6}{a}^{\mathrm{2}} \right) \\ $$$${since}\:{M}=\rho{V}=\frac{\mathrm{4}\rho\pi{abc}}{\mathrm{3}} \\ $$$$\Rightarrow{I}_{{A}} =\frac{{M}}{\mathrm{5}}\left({c}^{\mathrm{2}} +\mathrm{6}{a}^{\mathrm{2}} \right) \\ $$
Commented by ajfour last updated on 12/Jun/18
$${for}\:{a}={b}={c}\:\:={r} \\ $$$${I}=\frac{\mathrm{7}}{\mathrm{5}}{Mr}^{\mathrm{2}} \\ $$
Commented by MrW3 last updated on 12/Jun/18
$${I}\:{found}\:{a}\:{typo}\:{in}\:{my}\:{working}.\:{Now} \\ $$$${it}'{s}\:{fixed}.\:{Now}\:{it}\:{should}\:{be}\:{correct}. \\ $$
Commented by ajfour last updated on 12/Jun/18
$${yes}\:{sir},\:{this}\:{was}\:{the}\:{answer}\:{i}\:{had} \\ $$$${fetched},\:{but}\:{now}\:{i}\:{forget}\:{the} \\ $$$${substitution}\:{i}\:{made}.\:{Thank}\:{you}. \\ $$$${Your}\:{method}\:{is}\:{shorter}\:{and}\:{best}. \\ $$