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Disclaimer: This was given as a homework from college but the teacher didn't teach us anything about density or mass or anything related.

A lamina has the form of the region limited by the parabola $ y = x^2 $ and the straight line $ y = x $. The density varies as the distance from the $ X $ axis.

Find the mass and center of mass.

what i could find however is that the formula of mass is the following $$M = \int\int_R \rho(x,y)dA $$

so i tried doing something like this $$ \int_0^1\int_y^{\sqrt(y)} ? dxdy $$

the thing is that they say the density varies as the distance from the x axis, so i don't know what to replace for the density.. is it $ x + y $?

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    $\begingroup$ No, it would just be $ky$, where $k$ is some constant of proportionality that won't matter for the center of mass, but will (naturally) matter for the mass itself. $\endgroup$
    – Brian Tung
    May 5, 2015 at 2:29
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    $\begingroup$ (Since $y$ is the distance from the $x$-axis. Just making sure you see this correction.) $\endgroup$
    – Brian Tung
    May 5, 2015 at 2:30

2 Answers 2

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The mass density varies as the distance from the x-axis implies that $\rho =Ky$ where $K$ is a constant.

Now, the total mass is given by

$$M=\int_0^1 \int_y^{\sqrt{y}} (Ky) dx dy$$

The moment about x is

$$\frac{\int_0^1 \int_y^{\sqrt{y}} x(Ky) dx dy}{M}$$

$$\frac{\int_0^1 \int_y^{\sqrt{y}} y(Ky) dx dy}{M}$$

Can you complete? Notice that the moments are independent of $K$.

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  • $\begingroup$ i think i understand you but i have a couple of questions so everything comes clear: 1) why K? i mean is it part of a formula or something or what's the reason for the constant?. 2) if it were that the density varies as the distance from the y-axis, it would be $ \rho = Kx $? and 3) are the integration boundaries right? $\endgroup$
    – HardCodeStuds
    May 5, 2015 at 2:46
  • $\begingroup$ Great! On $1$, $K$ is a proportionality constant. Keep in mind dimensional analysis. If $M$ is mass in, say kilograms (kg), and $y$ is the distance from the $x$-axis in, say meters (m), then $K$ is a constant in kg/m. The greater $K$, the greater the mass density. On $2$, if you are at $(x,y)$ you are $x$ units away from the $x$-axis, and $y$ units away from the $x$-axis. On $3$, the integration limits that you used are indeed correct! $\endgroup$
    – Mark Viola
    May 5, 2015 at 2:54
  • $\begingroup$ Well thanks, it solved my questions and i learned a little more. $\endgroup$
    – HardCodeStuds
    May 5, 2015 at 2:56
  • $\begingroup$ You're welcome. My pleasure. I'm glad that this really helped!! $\endgroup$
    – Mark Viola
    May 5, 2015 at 2:57
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Mass of lamina is K/15 (Density=KY) On the vertical strip on the region of integration- Y varies from parabola to straight line i.e. x^2 to x And X varies from 0 to 1 Thus after double integration by the mass formula, we get mass of lamina=K/15

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