1
$\begingroup$

Problem : A disk of radius $6 \;cm$ has density $10 \; g\;cm^{-2}$ at it's center, density $0$ at its edge, and its density is a linear function of the distance from the center. Find the mass of the disc.

My Thought Process : I formulated this

$$\int_{\theta=0}^{2\pi}\int_{r=0}^6(10r-2r^2)\, \mathrm{dr} \mathrm\,{d\theta}$$

I got $72\pi$ as the answer, but the system is saying that it is incorrect.

$\endgroup$
1

1 Answer 1

1
$\begingroup$

$$\rho (r)=ar+b $$

$$\rho (0)=10=b $$ $$\rho (R)=aR+b=0$$ thus

$$\rho (r)=b (1-\frac {r}{R} )$$ with $b=10g/cm^2$ and $R=6cm $.

the mass is $$\int_0^R2 \pi r \rho (r)dr=$$ $$2\pi b\int_0^R (r-\frac {r^2}{R})dr=$$ $$2\pi b \Bigl [\frac {r^2}{2}-\frac {r^3}{3R}\Bigr]_0^R=$$ $$2\pi b R^2(\frac {1}{2}-\frac {1}{3})=$$ $$\frac {\pi b R^2}{3}=120\pi g$$

$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .