Centroid of a solid of revolution

[rock.freak667]

Homework Statement

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.

Homework Equations

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

[tex]\overline{x}=\frac{\int x dA}{\int dA}[/tex]

would be similar to

[tex]\overline{x}=\frac{\int x dV}{\int dV}[/tex]

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 [itex]dV= \pi x^2 dy[/itex]

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

 

[HallsofIvy]

Homework Statement

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.

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".

Homework Equations

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

[tex]\overline{x}=\frac{\int x dA}{\int dA}[/tex]

would be similar to

[tex]\overline{x}=\frac{\int x dV}{\int dV}[/tex]

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 [itex]dV= \pi x^2 dy[/itex]

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

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

 

[rock.freak667]

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

 

[HallsofIvy]

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
[tex]\overline{y}= \frac{\int y dV}{\int dV}[/tex]

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, [itex]x= r cos(\theta)[/itex], [itex]y= y[/itex], [itex]z= r sin(\theta)[/itex] and [itex]dV= r sin(\theta)drd\theta dy[/itex]. Alternatively, just replace "y" by "z" in the original equation and find the "z" component of the centroid.

 

[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.