Question

Find the mass of a two-dimensional, centered at the origin, oversized hockey puck a radius 2 in. with a radial density of p(x)=x^3-2x+5lb/in^2

Answer 1

Integrate the density over the disk
`D` given by

`D = \{(x,y) ~:~ x^2 + y^2 \le 2^2\}`

or more conveniently, in polar coordinates by

`D' = \left\{(r,\theta) ~:~ 0 \le r \le 2 \text{ and } 0\le \theta\le2\pi\right\}`

Since the density depends on the distance from the origin, we have

`\rho(r) = \left(1\frac{\rm lb}{\mathrm{in}^6}\right)r^3 - \left(2\frac{\rm lb}{\mathrm{in}^4}\right)r + 5\frac{\rm lb}{\mathrm{in}^3}`

(Note the units in each coefficient ensure that
`\rho` is measured in lb/in³.)

Then the mass (in lb) of the puck is

`\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{(332\pi)/(15)}`