Answer to Question #142085 in Calculus for Promise Omiponle

Question

Find the mass and center of mass of the cube given by 0≤x≤a, 0≤y≤a, 0≤z≤a for which the density is ρ(x,y,z) =x^2+y^2+z^2.

Answer 1

$$Solution$$

Mass of the cube

$m=\intop \intop \intop_E \rho (x,y,z) dv= \intop_0^a \intop_0^a \intop_0^a (x^2+y^2+z^2) \delta z\ \delta y\ \delta x\\ =\intop_0^a \intop_0^a [ (x^2+y^2)(z)_0^a+(\frac {z^3}{3})_0^a]\delta y\ \delta x\\ =\intop_0^a \intop_0^a [ (x^2+y^2)a+\frac {a^3}{3}]\delta y\ \delta x\\ =\intop_0^a [ax^2(y)_0^a+a(\frac {y^3}{3})_0^a+\frac {a^3}{3}(y)_0^a] \delta x\\ =\intop_0^a [a^2x^2+\frac {a^4}{3}+\frac {a^4}{3}] \delta x\\ =\intop_0^a [a^2x^2+\frac {2}{3}a^4] \delta x=a^2(\frac {x^3}{3})_0^a+\frac {2}{3}a^4(x)_0^a\\ m=\frac{a^5}{3}+\frac {2}{3}a^5 = a^5------------>Mass\ of\ the\ cube$

$\bar x=\frac1m\intop \intop \intop_E x\rho (x,y,z) dv=\frac1m \intop_0^a \intop_0^a \intop_0^a x(x^2+y^2+z^2) \delta z\ \delta y\ \delta x\\ =\frac1m\intop_0^a \intop_0^a [ (\frac{x^4}{4})_0^a+(y^2+z^2)(\frac{x^2}{2})_0^a]\delta y\ \delta z\\ =\frac1m\intop_0^a \intop_0^a [ \frac{a^4}{4}+\frac{a^2}{2}(y^2+z^2)_0^a]\delta y\ \delta z\\ =\frac1m\intop_0^a [ \frac{a^4}{4}(y)_0^a+\frac{a^2}{2}(\frac{y^3}{3})_0^a)+\frac{a^2}{2}\cdot z^2(y)_0^a] \delta z\\ =\frac1m\intop_0^a [ \frac{a^5}{4}+\frac{a^5}{6}+\frac{a^3}{2}\cdot z^2] \delta z\\ =\frac1m\intop_0^a [ \frac{5}{12}a^{12}+\frac{a^3}{2}\cdot z^2] \delta z=\frac1m [ \frac{5}{12}a^{12}(z)_0^a+\frac{a^3}{2}(\frac{z^3}{3})_0^a]\\ =\frac1m [ \frac{5}{12}a^{6}+\frac{1}{6}a^{6}]=\frac{1}{a^5}\cdot\frac{7}{12}\cdot a^6=\frac{7}{12}a\\ \implies \bar x=\frac{7}{12}a$

$\bar y=\frac1m\intop \intop \intop_E y\rho (x,y,z) dv=\frac1m \intop_0^a \intop_0^a \intop_0^a y(x^2+y^2+z^2) \delta z\ \delta y\ \delta x\\ =\frac1m\intop_0^a \intop_0^a [(x^2+z^2)(\frac{y^2}{2})_0^a+ (\frac{y^4}{4})_0^a]\delta x\ \delta z\\ =\frac1m\intop_0^a \intop_0^a [ \frac{a^2}{2}(x^2+z^2)_0^a+\frac{a^4}{4}]\delta x\ \delta z\\ =\frac1m\intop_0^a [ \frac{a^4}{4}(x)_0^a+\frac{a^2}{2}(\frac{x^3}{3})_0^a)+\frac{a^2}{2}\cdot z^2(x)_0^a] \delta z\\ =\frac1m\intop_0^a [ \frac{a^5}{6}+\frac{a^5}{4}+\frac{a^3}{2}\cdot z^2] \delta z\\ =\frac1m\intop_0^a [ \frac{5}{12}a^{12}+\frac{a^3}{2}\cdot z^2] \delta z=\frac1m [ \frac{5}{12}a^{12}(z)_0^a+\frac{a^3}{2}(\frac{z^3}{3})_0^a]\\ =\frac1m [ \frac{5}{12}a^{6}+\frac{1}{6}a^{6}]=\frac{1}{a^5}\cdot\frac{7}{12}\cdot a^6=\frac{7}{12}a\\ \implies \bar y=\frac{7}{12}a$

Similarly By Symmetry

$$\bar z=\frac{7}{12}a$$

$Center\ of\ mass\ (\bar x, \bar y, \bar z )=(\frac{7}{12}a, \frac{7}{12}a, \frac{7}{12}a)$