Question

Assume a planet having a radius R and its density is expressed as = \(\frac{R+r}{2r}D’\).

(a) \(\frac{5\pi D’R^3}{2}\)

(b) \(\frac{4\pi D’R^3}{3}\)

(c) \(\frac{5\pi D’R^3}{3}\)

(d) \(\frac{5\pi D’R^3}{12}\)

I have been asked this question in an online quiz.

The query is from Applications of Triple Integral in portion Multiple Integrals of Engineering Mathematics

Answer 1

The correct answer is (c) \(\frac{5\pi D’R^3}{3}\)

For explanation I would say: Consider the case of r=R

Then,

D=D’

Where D’ is the surface density of the planet

As D → ∞, r → 0

For finding the mass of the planet, we use the triple integration formula

M=∫∫∫ dV

Converting into spherical co-ordinates, we get

M=\(\int\int\int D’r^2 sin\theta \frac{R+r}{2r} dr d\theta d\theta\)

Applying the limits

0 to π

0 to 2π

0 to R

Solving the Triple Integral we get,

M=\(\frac{5\pi D’R^3}{3}\)

Thus, the mass of the planet is \(\frac{5\pi D’R^3}{3}\).