MHB Center of Mass Calculus 2 Integration

[Dustinsfl]

Uniform semi circle of radius R whose diameter is on the x axis.
Since it is uniform and symmetric on the x axis, x = 0.
For y, we have
$$
y_{cm} = \frac{\int y\sigma dA}{\frac{\pi R^2}{2}}
$$
In polar, $dA = rdrd\phi$.
So the integral becomes
$$
\int_0^{\pi}\int_0^R y^2\sin\phi drd\phi
$$
How did we end up with an additionally $y$ and a $\sin\phi$?

 

[Chris L T521]

Uniform semi circle of radius R whose diameter is on the x axis.
Since it is uniform and symmetric on the x axis, x = 0.
For y, we have
$$
y_{cm} = \frac{\int y\sigma dA}{\frac{\pi R^2}{2}}
$$
In polar, $dA = rdrd\phi$.
So the integral becomes
$$
\int_0^{\pi}\int_0^R y^2\sin\phi drd\phi
$$
How did we end up with an additionally $y$ and a $\sin\phi$?

In this case,

\[\int y\,dA = \int_0^{\pi}\int_0^R (r\sin\phi)r\,dr\,d\phi =\int_0^{\pi}\int_0^Rr^2\sin\phi\,dr\,d\phi.\]

I don't see why you should have $y$ in your integral once you've converted everything to polar.

 

[Dustinsfl]

In this case,

\[\int y\,dA = \int_0^{\pi}\int_0^R (r\sin\phi)r\,dr\,d\phi =\int_0^{\pi}\int_0^Rr^2\sin\phi\,dr\,d\phi.\]

I don't see why you should have $y$ in your integral once you've converted everything to polar.

Maybe it was a typo. I was solving $y = r\sin\phi$ for r and couldn't get anything like that.

 

[MarkFL]

I came to the same conclusion as Chris L T521, so my vote is for typo too.

 

[Chris L T521]

Maybe it was a typo. I was solving $y = r\sin\phi$ for r and couldn't get anything like that.

If they wanted it to have y terms in there, the integral would have to be $\displaystyle\int_0^{\pi}\int_0^R y^2\csc\phi\,dr\,d\phi$, not $\displaystyle\int_0^{\pi}\int_0^R y^2\sin\phi\,dr\,d\phi$.

Either way, there's a typo somewhere in their computation.