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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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2 Answers

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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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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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