Answer:
a) [tex]\delta = 10 + 3(\rho - 2)[/tex]
b) [tex]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[/tex]
c) [tex]2451\pi[/tex]
Step-by-step 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:
[tex]\delta = 10 + 3(\rho - 2)[/tex]
b) The equation to find Mass can be written with triple integration as follows:
[tex]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[/tex]
c) Now, we will calculate the integral above as follows:
[tex]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(\frac{3}{4} \rho^4+\frac{4}{3} \rho^3)|\limits^5_2 sin\phi\:d\phi\: d\theta=\\\\=\int\limits^{2\pi}_0\int\limits^{\pi}_0 \frac{2451}{4} sin\phi\:d\phi\: d\theta=\int\limits^{2\pi}_0 \frac{2451}{4} (-cos\phi)|\limits^{\pi}_0 \: d\theta=\int\limits^{2\pi}_0 (\frac{2451}{4}*2)\:d\theta=\\[/tex]
[tex]=\int\limits^{2\pi}_0 \frac{2451}{2}\:d\theta= \frac{2451}{2}\theta|\limits^{2\pi}}_0=\frac{2451}{2}*2\pi=2451\pi[/tex]
