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Find center of mass of the upper half $y\ge 0$ of the disk bounded by the circle $x^2 +y^2=4 $ with the $p(x,y)=1+ y/2$

First i have calculated the mass and i got $\int\limits_{-2}^{2}\int\limits_{0}^{\sqrt{4-x^2}}1+\frac{ y }{ 2 }dydx = \frac{ 8 }{ 3 }+2\pi$ is that right? I have trouble setting My (mass with x) and then do i have to set Mx?

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(1) find the moment of our disk about the $x$-axis, then divide by mass. I assume you know how to set up the integral for moment. (2) By symmetry, the center of mass is on the $x$-axis, so the $x$-coordinate of the center of mass is $0$. – André Nicolas Nov 19 '12 at 22:32
Sorry, forgot to answer one of your questions: yes, the calculated mass is right. – André Nicolas Nov 19 '12 at 22:57
okay i got it, its (0, 0.95) – Jack F Nov 19 '12 at 23:00
up vote 1 down vote accepted

The density function is symmetric about the $y$-axis, so by symmetry, the centre of mass has $x$-coordinate equal to $0$.

For the $y$ coordinate we (i) find the moment about the $x$-axis; (ii) find the mass; (iii) divide the moment by the mass.

You correctly found the mass.

For the moment about the $x$-axis, take a thin horizontal strip, from height $y$ to height $y+dy$. The strip has mass roughly $(2x)(1+y/2)\,dy$, and therefore moment roughly $(2x)(1+y/2)(y)\,dy$, where $x=\sqrt{4-y^2}$. Integrate from $y=0$ to $y=2$.

Or else, if one is familiar with the double integral, as the OP shows you are, a little $dx \times dy$ rectangle centered at $(x,y)$ has moment about the $x$-axis of about $(1+y/2)(y)\,dx\,dy$. Find the integral of this over the half-disk.

Remark: The height of the centre of mass is the same as the height of the centre of mass of the first quadrant part of our half-disk. I would probably find the area and moment of that, so as to have fewer minus signs around.

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