Ištrintas turinys Pridėtas turinys
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:<math>=\int_0^{2\pi} \left( \int_0^R \rho (R^2-\rho^2) \mathbf{d}\rho \right) \mathbf{d}\phi .</math>
:Pasinaudodami [http://integrals.wolfram.com/index.jsp?expr=x+%28R%5E2+-+x%5E2%29%5E%283%2F2%29&random=false internetiniu integratoriumi], gauname, kad
:<math> \int_0^R \rho\sqrt{ (R^2-\rho^2)^3} \mathbf{d}\rho =-\frac{1}{5}(int_0^R (\rho R^2-\rho^3) \mathbf{d}\rho=(R+\frac{\rho))^2 R^2}{5/2}-\frac{\rho^4}{4})|_0^R=</math>
:<math>=-\frac{1}{5}((R-^2 \cdot R)(R+R))^2}{5/2}-\left( -\frac{1R^4}{54}((=\frac{R-0)(R+0))^4}{5/24}\right)=.</math>
:<math>V_x=0+\fracint_0^{12\pi} \left( \int_0^R \rho \sqrt{5}((R^2-0)(R+0\rho^2)^3} \mathbf{d}\rho \right) \mathbf{d}\phi =\int_0^{5/2\pi}= \frac{1R^4}{54}( \mathbf{d}\phi =\frac{2\pi R^2)^4}{5/24}=\frac{\pi R^54}{52}; .</math>
:<math>V_x=\int_0^{2\pi} \left( \int_0^R \rho \sqrt{(R^2-\rho^2)^3} \mathbf{d}\rho \right) \mathbf{d}\phi =\int_0^{2\pi} \frac{R^5}{5} \mathbf{d}\phi =\frac{2\pi R^5}{5}.</math>
:Kadangi reikia dviejų rutulio pusrutulių (teigiama ir neigiama ''Ox'' kryptimi), tai
:<math>V_X=2 V_x=2\cdot \frac{2\pi R^54}{52}=\frac{4\pi R^5}{5}4.</math>
==Taip pat skaitykite==
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