Please instruct on finding the center of mass given the vector coordinates in double integral form stated.

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vector coordinates of the center of mass of a lamina The attached image shows the vector coordinates of the center of mass of a lamina. Area is bounded by $y=a-x^2$ where $a>=0$ and by $y=0$.

I found the boundary is given by: $-sqrt(a)<=x<=sqrt(a)$ and $0<=y<=a-x^2$.

$p(x,y)$ is a density function but it can be treated as a constant $c$.

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Please instruct on how to go about solving for the center of mass formula (doing the integrals). Thank you very much.

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Start first by sketching out your feasible region. That is, the region bounded by, $y = a - x^2$ and $y = 0$ (noting $a$ is always positive).

You should get something that resembles a parabola with a maximum at $(0, a)$ and $x$ intercepts at $-\sqrt{a}$ and $\sqrt{a}$.

From there, you can choose your limits of integration appropriately. More specifically, choose (based on the graph) which values $x$ you want to integrate over and which values of $y$ you want to integrate over. Note that one integral should contain only constants.

For the denominator you should have,

$$\int_{-\sqrt{a}}^{\sqrt{a}}\int_{0}^{a - x^2} \ \rho(x,y) \ dy \ dx$$

Start by integrating the 'inside', $$\int_{0}^{a - x^2} \ \rho(x,y) \ dy\hspace{1cm}(1)$$

Then take what you computed in $(1)$ (we will call it $f$) and integrate it again, but this time wrt your limits of $x$,

$$\int_{-\sqrt{a}}^{\sqrt{a}} \ f \ dy\hspace{1cm}$$

Follow similar steps to compute the respective numerators.