# Centroid of a solid of revolution

1. Oct 31, 2008

### rock.freak667

1. The problem statement, all variables and given/known data
The curve 3x2+2y2-12y=32 is rotated about the x-axis and forms a solid hemisphere.

Verify that the weight is 8cm from the bottom of the hemisphere.

2. Relevant equations

3. The attempt at a solution

Now, I can only do a little bit in centroids but that is for just plane surfaces. I do not think I know how to do it when it is rotated.

I thought it would be similar to the formula for centroids like

$$\overline{x}=\frac{\int x dA}{\int dA}$$

would be similar to

$$\overline{x}=\frac{\int x dV}{\int dV}$$

when rotated. I need some help in doing it.

EDIT:

I considered a cylindrical element of radius x and width dy.
So that the volume of this small element is $dV= \pi x^2 dy$

Then I should have to take the first moment of volume about the x-axis? (like the first moment of area)
But I am not sure how to take this moment of volume

Last edited: Oct 31, 2008
2. Nov 1, 2008

### HallsofIvy

Staff Emeritus
This makes no sense at all. Rotating 3x2+ 2y2- 12y= 32 around the x-axis does NOT form a "solid hemisphere" and there is no "weight".

No, it wouldn't be. If a figure is rotated around the x-axis then it would be $\pi y^2 dx$.

3. Nov 1, 2008

### rock.freak667

sorry...I meant rotated around the y-axis.

4. Nov 1, 2008

### HallsofIvy

Staff Emeritus
Rotating that around the y-axis still does not give a "solid hemisphere", it gives an ellipsoid.

It shoule be obvious, by symmetry, that the x and z components of the centroid are 0. Yes, the y component of the centroid is given by
$$\overline{y}= \frac{\int y dV}{\int dV}$$

I would recommend doing the integration in cylindrical coordinates except altered to use polar coordinates in the xz-plane rather than the xy-plane. That is, $x= r cos(\theta)$, $y= y$, $z= r sin(\theta)$ and $dV= r sin(\theta)drd\theta dy$. Alternatively, just replace "y" by "z" in the original equation and find the "z" component of the centroid.

5. Nov 1, 2008

### rock.freak667

I have never done double/triple integrals as dv is in terms of three differentials. But the question just showed me a picture of a kind of semi-circle and said that was the curve.