## Essential University Physics: Volume 1 (3rd Edition)

$\frac{M}{4 \pi R^3(1-2e^{-1})}$
We know: $\rho(R)=\int_0^R \rho_0 e^{\frac{-r}{R}}dr$ We also know that $\rho=\frac{M}{4\pi R^3}$. Thus, we take the integral and simplify to obtain: $\rho=\frac{M}{4\pi R^3}=\rho_0(1-2e^{-1})$ $\rho_0=\frac{M}{4\pi R^3(1-2e^{-1})}$