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A disk of radius $5$ cm has density $10$ g/cm$^2$ at its center, density $0$ at its edge, and its density is a linear function of the distance from the center. Find the mass of the disk.

my answer: $157.08$g

$D=10-2x$

Double integration of $D$ from $y=0$ to $y=2\pi$ and $x=0$ to $x = 5$

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You computed the integral: $$\iint_{x\in[0,5],\,y\in[0,2\pi]}(10-2x)\,\mathrm{d}x\,\mathrm{d}y=50\pi\approx157.08.$$ Though your computation is correct, the result is not the total mass of the disc.

For a 2-dimensional object $\Sigma$ of density of mass $D$, the total mass $m$ of $\Sigma$ is $$m=\iint_\Sigma D\,\mathrm{d}A$$ where $\mathrm{d}A$ is the surface element. Since your object is a disc, and the way the density of mass is given, it makes sense to use polar coordinates. That's what you did. But in the process, you used the following wrong fact: $$\mathrm{d}A=\mathrm{d}r\,\mathrm{d}\theta\qquad\textbf{This is wrong!}$$ where $(r,\theta)$ are the polar coordinates, instead of: $$\mathrm{d}A=r\,\mathrm{d}r\,\mathrm{d}\theta\qquad\text{This is correct}.$$ So the total mass should instead be: $$m=\iint_{(r,\theta)\in[0,5]\times[0,2\pi]}(10-2r)\,r\,\mathrm{d}r\,\mathrm{d}\theta.$$ I'm sure you can take it from here. The answer is $m=\dfrac{250\pi}3\,\mathrm{g}\approx261.80\,\mathrm{g}$.

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Dimensional analysis tells me that is incorrect. $D$ has dimensions of $g/cm^2$ integrating with $dx$ gives it dimensions of $g/cm$ and not $g$ as required by the answer.

Solution

Consider the elemental area of the disc $dA=2\pi xdx$

Now mass=density*Area

$m=DdA$

$m=\int (10-2x)2\pi xdx$

$m=\int (10-2x)2\pi xdx$ over $x=0$ to $x=5$

$m=2\pi\int (10-2x) xdx$ over $x=0$ to $x=5$

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  • $\begingroup$ how can mass have units of g/cm? $\endgroup$ – user2789433 Oct 29 '14 at 3:00
  • $\begingroup$ Also I know my answer is wrong thats why I am stuck :/ $\endgroup$ – user2789433 Oct 29 '14 at 3:01
  • $\begingroup$ Is the correct answer $250\pi /3$ ? $\endgroup$ – Fahd Siddiqui Oct 29 '14 at 3:06
  • $\begingroup$ YES! Please explain how u got it. $\endgroup$ – user2789433 Oct 29 '14 at 3:26

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