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:


ρ=mV    m=ρV=ρspheredV=dρdxdydz\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=5r=\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=dxdydz=r2sinθdθdϕdr\,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⪕r5\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=ρdxdydz=ρ050π02πr2sinθdθdϕdrm = \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=ρ(05r2dr)(0πsinθdθ)(02πdϕ)m = \rho (\intop_{0}^{\sqrt{5}} r^2 dr)(\intop_{0}^{\pi} sin\theta d\theta)( \intop_{0}^{2\pi}d\phi)


m=ρ[r33]05[cosθ]0π[ϕ]02πm = \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)(53/230)((1)+1)(2π0)=(2)(2)(2π)(53/23)m = (2)( \frac{5^{3/2}}{3} - 0)(-(-1)+1)(2\pi-0) = (2)(2)(2\pi)(\frac{5^{3/2}}{3})


m=4053π93.664gm = \frac{40\sqrt5}{3}\pi ≈ 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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