Answer to Question #199246 in Calculus for Nikhil Rawat

Question #199246

Find the mass of the object, which is in the form of a sphere of radius √5cm, centred at the origin. The density at any point is given to be the constant 2.


1
Expert's answer
2021-05-31T00:49:41-0400

Since we have the density of the sphere (which is constant and given in g/cm3) and the radius we will have to use the information given to find the mass in grams:


"\\rho = \\frac{m}{V} \\implies m = \\rho*V = \\rho\\iiint_{sphere} dV = d\\rho\\iiint dx\\,dy\\,dz"


We also know that the radius is "r=\\sqrt{5}" and then we can establish that x2+y2+z2 = r2 = 5 (this also because the sphere was centered at the origin). The volume differential dV will be changed to polar coordinates to evaluate the integral easily:


The volume element in spherical coordinates is "\\,dV=dx\\,dy\\,dz =r^2 \\,sin \\theta \\, d \\theta \\, d\\phi \\, dr" and we also have to considerate the limits for the integration: (0⪕θπ), (0⪕ϕ ⪕2π) and (0⪕r"\\sqrt{5}" ). With this information we proceed to evaluate the integral for the volume and after we multiply it for the density we'll have the mass of the object:


"m = \\rho\\iiint dx\\,dy\\,dz = \\rho \\intop_{0}^{\\sqrt{5}} \\intop_{0}^{\\pi} \\intop_{0}^{2\\pi} r^2sin\\theta d\\theta d\\phi dr"


"m = \\rho (\\intop_{0}^{\\sqrt{5}} r^2 dr)(\\intop_{0}^{\\pi} sin\\theta d\\theta)( \\intop_{0}^{2\\pi}d\\phi)"


"m = \\rho \\cdot \\large [ \\frac{r^3}{3}]_{0}^{\\sqrt{5}} \\cdot \\large [ -cos\\theta]_{0}^{\\pi} \\cdot \\large [ \\phi]_{0}^{2\\pi}" (these are the integrals that have to be evaluated to find the mass)


"m = (2)( \\frac{5^{3\/2}}{3} - 0)(-(-1)+1)(2\\pi-0) = (2)(2)(2\\pi)(\\frac{5^{3\/2}}{3})"


"m = \\frac{40\\sqrt5}{3}\\pi \u2248 93.664\\,g"


In conclusion, the mass of the sphere is found as m = ρ*dV 93.664 g


References:

  • Castellan, G. W. (1983). Physical Chemistry. Ed.

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