Centroids and finding moments: y = x, y = 6 - x^2

[johnboy]

I'm having centroid issues. I can do them when they are shapes like triangles, rectangles, circles, etc. But when they aren't like that, I'm getting confused. An example that I'm working on is:

. . .y = x
. . .y = 6 - x^2

I'm pretty sure that X moment just equals (-3+2/2) = -1/2. But I'm not sure how to calculate Y moment. I was trying to use the integral of 0.5*(x(6 - x^2))^2, but my answer was coming out way wrong. Is anyone familiar with these things? Thanks in advance.

 

[pka]

First, did you graph the region?

Next find the area of the region: A.
Then use the rules.
\(\displaystyle \
\begin{array}{l}
A = \int\limits_{ - 3}^2 {\left[ {\left( {6 - x^2 } \right) - \left( x \right)} \right]dx} \\
\bar x = \frac{1}{A}\int\limits_{ - 3}^2 {x\left[ {\left( {6 - x^2 } \right) - \left( x \right)} \right]dx} \\
\bar y = \frac{1}{A}\int\limits_{ - 3}^2 {\frac{1}{2}\left[ {\left( {6 - x^2 } \right)^2 - \left( x \right)^2 } \right]dx} \\
\end{array}\)