Mass of a sphere of varying density

[Just a nobody]

Homework Statement

Given that the density of a sphere with respect to radius is $$\rho(r) = \rho_0 \left( 1 - \frac{\alpha r}{R_0} \right)$$ (where $$\rho_0$$, $$\alpha$$, and $$R_0$$ are constants), find the total mass of the sphere. The radius of the sphere is equal to $$R_0$$.

$$\rho_0 = 5320 \, \frac{\mathrm{kg}}{\mathrm{m}^3}$$
$$\alpha = 0.13$$
$$R_0 = 89.13 \cdot 10^6 \, \mathrm{m}$$

Homework Equations

$$\rho(r) = \rho_0 \left( 1 - \frac{\alpha r}{R_0} \right)$$
$$Mass = \int_V dm = \int_V \! \rho(r) \, dV$$

The Attempt at a Solution

I have attempted the mass integral and found it to be:
$$Mass = \int_0^{2 \pi} \! \int_0^\pi \! \int_0^{R_0} \rho(r) r^2 \sin \phi \, dr \, d\phi \, d\theta = 4 \pi \rho_0 R_0^3 \left( \frac{1}{3} - \frac{\alpha}{4} \right) = 1.42 \cdot 10^{28} \, \mathrm{kg}$$

I also tried a different form of the integral and came to the same solution:
$$Mass = \int_0^{R_0} \! \rho(r) (4 \pi r^2) \, dr = 4 \pi \rho_0 R_0^3 \left( \frac{1}{3} - \frac{\alpha}{4} \right) = 1.42 \cdot 10^{28} \, \mathrm{kg}$$

I verified my integrals with a CAS to be sure they're correct. However, my solution still doesn't match the solution in my school's online homework system. This leads me to believe my logic is incorrect. Can someone lead me in the right direction on this problem?

 

[fantispug]

I've checked it all, but I get the same answers as you. I'd be suspicious of your school's online homework system.

 

[Just a nobody]

I've checked it all, but I get the same answers as you. I'd be suspicious of your school's online homework system.

Thanks for checking my work. I'll probably talk to the professor on Monday.