Question

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.

Answer 1

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⪕θ⪕π), (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)

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

$m = \frac{40\sqrt5}{3}\pi ≈ 93.664\,g$

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

References:

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