Answer :
Integrate the density over the disk [tex]D[/tex] given by
[tex]D = \{(x,y) ~:~ x^2 + y^2 \le 2^2\}[/tex]
or more conveniently, in polar coordinates by
[tex]D' = \left\{(r,\theta) ~:~ 0 \le r \le 2 \text{ and } 0\le \theta\le2\pi\right\}[/tex]
Since the density depends on the distance from the origin, we have
[tex]\rho(r) = \left(1\dfrac{\rm lb}{\mathrm{in}^6}\right)r^3 - \left(2\dfrac{\rm lb}{\mathrm{in}^4}\right)r + 5\dfrac{\rm lb}{\mathrm{in}^3}[/tex]
(Note the units in each coefficient ensure that [tex]\rho[/tex] is measured in lb/in³.)
Then the mass (in lb) of the puck is
[tex]\displaystyle \iint_{D'} r\,\rho(r)\,dr\,d\theta = \int_0^{2\pi} \int_0^2 r (r^3 - 2r + 5) \, dr \, d\theta \\\\ ~~~~~~~~ = 2\pi \int_0^2 (r^4 - 2r^2 + 5r) \, dr \\\\ ~~~~~~~~ = 2\pi \left(\frac{2^5}5 - \frac{2\cdot2^3}3 + \frac{5\cdot2^2}2\right) = \boxed{\frac{332\pi}{15}}[/tex]