Question

Mass In Exercises 23 and 24, use spherical coordinates to find the mass of the sphere \(x^{2}+y^{2}+z^{2}=a^{2}\) with the given density. The density at any point is proportional to the distance between the point and the origin.

Short answer

The mass of the sphere is \(2πka^{5}/5\), where k is the constant of proportionality for the density.

Step-by-step solution

Set up the Integral for the Mass

The mass of the sphere can be represented as the triple integral of the density function over the whole volume of the sphere. Using spherical coordinates, the sphere’s equation becomes \(r^{2}=a^{2}\). This gives the limits for the integral. Represent density using a proportionality constant \(k\).\nTherefore the integral to solve is \(\iiint_{V} kr^{2}r^{2}\sin(φ) dr dφ dθ\).

Compute the Integral

Next, it's time to compute the integral. The integral splits into three separate integrals, which can be solved individually. The limits for r are 0 to a, for \(φ\) are 0 to \(π\) and for \(θ\) are 0 to 2\(π\). Calculating each integral separately yields: \(\int_{0}^{a} r^{4} dr \), \(\int_{0}^{π} sin(φ) dφ \), \(\int_{0}^{2π} dθ \).\nSolving these integrals gives: [(\(a^{5}/5\)), -cos(φ) evaluated from 0 to \(π\) giving 2, and \(θ\) evaluated from 0 to 2\(π\) giving 2\(π\).\nThe total integral then becomes \(2πka^{5}/5\).

Find the Mass

The mass of the sphere is the result of the integral which we’ve computed as \(2πka^{5}/5\). Therefore the mass is \(2πka^{5}/5\) where k is the constant of proportionality for the density.