1. ## Centroid

Find the centroid of the region bounded by
y=3x^2+6x, y=0 and x=3.

So what I've done so far
0<x<3
0<y<3x^2+6x
those are the bounds
and i take the double integral using those bound to find m
but I don't know what function to use I originally used 3x^2+6x but thats not correct.

2. Originally Posted by koalamath
Find the centroid of the region bounded by
y=3x^2+6x, y=0 and x=3.

So what I've done so far
0<x<3
0<y<3x^2+6x
those are the bounds
and i take the double integral using those bound to find m
but I don't know what function to use I originally used 3x^2+6x but thats not correct.
I don't have the time right now to answer, but a centroid problem [where I have worked out double integrals] was asked here...

--Chris

3. my problem isn't the actual process of integration, I don't know what to integrate

4. Originally Posted by koalamath
my problem isn't the actual process of integration, I don't know what to integrate

$\bar{x}=\frac{1}{\int_0^3\left(3x^2+6x\right)\,dx} \int_0^3\int_0^{3x^2+6x}x\,dy\,dx$ and

$\bar{y}=\frac{1}{\int_0^3\left(3x^2+6x\right)\,dx} \int_0^3\int_0^{3x^2+6x}y\,dy\,dx$

where $\int_0^3\left(3x^2+6x\right)\,dx$ is the area of the region $R$

Do you see how I got these? It's pretty much "plug and chug" into the formula for $\bar{x}$ and $\bar{y}$.

--Chris

5. no i don't
the formula for m is the double integral of p(x,y) where p is the density function.

Your m doesn't make sense to me how did you find 3x^2+6x to be the density function and why isn't a double integral ?

6. Originally Posted by koalamath
no i don't
the formula for m is the double integral of p(x,y) where p is the density function.

Your m doesn't make sense to me how did you find 3x^2+6x to be the density function and why isn't a double integral ?
Are you trying to find the center of mass or the centroid?

They are two different definitions...

Center of mass requires a density function, whereas a centroid doesn't.

--Chris

7. aren't they the same thing?