Question

A spherical shell centered at the origin has an inner radius of 2 cm and an outer radius of 5 cm. The density, δ, of the material increases linearly with the distance from the center. At the inner surface, δ=10 g/cm3; at the outer surface, δ=16 g/cm3.

Answer 1

a)
`\delta = 10 + 3(\rho - 2)`

b)
`Mass=\int\limits^(2\pi)_0\int\limits^(\pi)_0\int\limits^(5)_2 (3\rho+4)\rho^2sin\phi\:d\rho\:d\phi\: d\theta`

c)
`2451\pi`

Explanation:

We will use spherical coordinates.

Since δ = 10 and ρ = 2 for inner and δ=16 and ρ = 5 for outer, the density increases at a rate of 3 g/cm^3 for each cm increase in radius.

a) Hence, the equation of density (δ) as a function of radius (ρ) is as follows:

`\delta = 10 + 3(\rho - 2)`

b) The equation to find Mass can be written with triple integration as follows:

`Mass=\int\limits^(2\pi)_0\int\limits^(\pi)_0\int\limits^(5)_2 \delta\rho^2sin\phi\:d\rho\:d\phi\: d\theta = \int\limits^(2\pi)_0\int\limits^(\pi)_0\int\limits^(5)_2 (10+3(\rho-2))\rho^2sin\phi\:d\rho\:d\phi\: d\theta = \\\\=\int\limits^(2\pi)_0\int\limits^(\pi)_0\int\limits^(5)_2 (3\rho+4)\rho^2sin\phi\:d\rho\:d\phi\: d\theta`

c) Now, we will calculate the integral above as follows:

`Mass=\int\limits^(2\pi)_0\int\limits^(\pi)_0\int\limits^(5)_2 (3\rho+4)\rho^2sin\phi\:d\rho\:d\phi\: d\theta=\int\limits^(2\pi)_0\int\limits^(\pi)_0((3)/(4) \rho^4+(4)/(3) \rho^3)|\limits^5_2 sin\phi\:d\phi\: d\theta=\\\\=\int\limits^(2\pi)_0\int\limits^(\pi)_0 (2451)/(4) sin\phi\:d\phi\: d\theta=\int\limits^(2\pi)_0 (2451)/(4) (-cos\phi)|\limits^(\pi)_0 \: d\theta=\int\limits^(2\pi)_0 ((2451)/(4)*2)\:d\theta=\\`

`=\int\limits^(2\pi)_0 (2451)/(2)\:d\theta= (2451)/(2)\theta|\limits^(2\pi)}_0=(2451)/(2)*2\pi=2451\pi`