Find the mass of the disk. - Double Integration Problem - Calculus 3

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 $$x = 5$$

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 $m=\dfrac{250\pi}3\,\mathrm{g}\approx261.80\,\mathrm{g}$.

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 $x=5$

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