# Center of Mass problem

• Dec 1st 2008, 11:50 AM
Center of Mass problem
Let $\displaystyle y=9-x^2, y=0, x=0$

Find a number h such that y=h would cut the region into two equal areas.

Now, I found out that the center of mass is (1,3.6), but how do I find the line? Do I just find the line that cut across (0,0) and (1,3.6)? The problem is the hint says that I need to integrate the area with respect to y, but I don't know how that is going to help.

Thanks!
• Dec 1st 2008, 01:06 PM
vincisonfire
What you really seach for is the mean of y.
The definition of the mean is
$\displaystyle \frac{\int yf(y)dy}{\int f(y)dy}$*
In your case $\displaystyle h = \frac{\int_0^9 y\sqrt{9-y}dy}{\int_0^9 \sqrt{9-y}dy}=\frac{324}{5}\frac{1}{18}=\frac{18}{5}$
You must do the top part using part integration.
$\displaystyle \int y\sqrt{9-y}dy = -\frac{12}{5}(9-y)^{3/2}-\frac{2}{5}y(9-y)^{3/2}$
• Dec 2nd 2008, 04:47 AM
CaptainBlack
Quote:

Let $\displaystyle y=9-x^2, y=0, x=0$

Find a number h such that y=h would cut the region into two equal areas.

Now, I found out that the center of mass is (1,3.6), but how do I find the line? Do I just find the line that cut across (0,0) and (1,3.6)? The problem is the hint says that I need to integrate the area with respect to y, but I don't know how that is going to help.

Thanks!

The area between $\displaystyle y=0$ and $\displaystyle y=h \le 9$ is:

$\displaystyle A(h)=\int_{y=0}^h \sqrt{9-y}\ dy=\frac{2}{3}[(9-h)^{3/2}-27]$

So the total area is $\displaystyle A(9)=18$ , so we look for $\displaystyle h$ such that $\displaystyle A(h)=9$

CB
• Dec 2nd 2008, 04:52 AM
CaptainBlack
Quote:

Originally Posted by vincisonfire
What you really seach for is the mean of y.
The definition of the mean is
$\displaystyle \frac{\int yf(y)dy}{\int f(y)dy}$*
In your case $\displaystyle h = \frac{\int_0^9 y\sqrt{9-y}dy}{\int_0^9 \sqrt{9-y}dy}=\frac{324}{5}\frac{1}{18}=\frac{18}{5}$
You must do the top part using part integration.
$\displaystyle \int y\sqrt{9-y}dy = -\frac{12}{5}(9-y)^{3/2}-\frac{2}{5}y(9-y)^{3/2}$

For this to be correct would require that median = mean (that is we seek the median not the mean and these are not in generaly equal)

CB
• Dec 2nd 2008, 04:59 AM
Jameson
I used to do these problems by simply finding the bounds of the area of integration - in this case you need to know the y min and max values of the region. Then you just say $\displaystyle \int_{a}^{h} f(x)dx = \int_{h}^{b} f(x)dx$, where a and b are your min and max.

This is essentially the same thing that CB suggested, but to me it is easier this way.