# Thread:

1. ## Center of Mass problem

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

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!

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

3. Originally Posted by tttcomrader
Let $y=9-x^2, y=0, x=0$

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 $y=0$ and $y=h \le 9$ is:

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

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

CB

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

5. 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 $\int_{a}^{h} f(x)dx = \int_{h}^{b} f(x)dx$, 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.
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