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I'd like some help here because I can't get the right answer. The lamina we are working with is defined by: $B={(x,y);x^2+y^2\le1,0\le y}$. Also the density function is "proportional to the distance of the point (x,y) to the x-axis"

I think here is where my mistake lies, for me density function is: $p(x,y)=y$

Acording to the exercise center of mass should be: $(Xc=0,Yc=\frac{3pi}{32})$

So I did this:

To find mass M I converted to polar coordinates. $x=rcos(\theta),y=rsin(\theta)$

$M=\int\int_Bp(x,y)dA => M=\int_0^{\pi}\int_0^1rsin(\theta)rdrd\theta=2/3$.

Now with the mass we can find $Xc$ and $Yc$. It's easy to note that $Xc=0$ with symmetry

$Yc = \frac{1}{M}\int\int_Byp(x,y)da = > Yc=\frac{3}{2}\int_0^\pi\int_0^1rsin(\theta)rsin(\theta)rdrd\theta=\frac{3\pi}{16}$

But Yc should be $\frac{3\pi}{32}$, so I'm close but I'm doing something wrong.

Electrolite
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    Your solution looks correct to me. Is there a mistake in the textbook? – mjw Apr 16 '19 at 02:58
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    Well... it would not be the first mistake. This is the second one, worse part is that I rely on these answers since I'm not good at math so I keep hunting what I did wrong for quite some time... – Electrolite Apr 16 '19 at 03:40
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    I checked as well, but I also did not see anything you did incorrectly either. – John Omielan Apr 16 '19 at 03:47

1 Answers1

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$X=0$ by symmetry.

$Y= {\int_0^1 2 y^2 \sqrt{1-y^2}dy \over \int_0^1 2 y \sqrt{1-y^2}dy } = {3 \pi \over 16}$.

copper.hat
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