double integral of mass density of the disk

Question

Would some kind soul give me a step by step solution to a fundamental double integral of $$\int_0^{2 π} \int_0^R \left(ρ_0 + \frac{(ρ_1 - ρ_0) r}R\right) r \,dr \,d\theta = \frac13 π R^2 (ρ_0 + 2 ρ_1),$$ I understand it's true, but I several steps in computing it trip me up. It is meant for integrating the mass density in polar coordinates to find the total mass of the disk. Please, help!

Answer 1

The variable $\theta$ doesn't included on integrand, so you can separate the integral $\int_0^{2\pi}d\theta$ from other parts.

Then, the integral changes to... $$\int_0^{2 π} \int_0^R \left(ρ_0 + \frac{(ρ_1 - ρ_0) r}R\right) r dr d\theta=\int_0^{2 π} d\theta \int_0^R \left(ρ_0 + \frac{(ρ_1 - ρ_0) r}R\right) r dr$$ $$=2\pi\int_0^R \left(ρ_0r + \frac{ρ_1 - ρ_0 }R r^2\right)dr=2\pi\left[\frac{\rho_0}2r^2+ \frac{\rho_1-\rho_0}{3R}r^3 \right]_0^R=2\pi\left( \frac{\rho_0 R^2}{2} + \frac{(\rho_1-\rho_0)R^2}{3} \right)$$$$=\frac13 π R^2 (ρ_0 + 2 ρ_1)$$