Mass and Center of Mass of Surface

[burt]

I am working on the following problem:


As you can see, I did not finish working through it. I just would like to know if I am on the right track.

What I still need to do - solve the moment integrals. Take the \(\frac{\text{moment}}{\text{mass}}\) to get the \(\bar x\)etc. The center of mass is the \((\bar x,\bar y,\bar z)\).
Is this correct? And, is my work this far correct?

 

[HallsofIvy]

It's not clear to me what you are trying to do. You say "surface" and your integral is "dA" but you have a triple integral. Are you integrating over a surface or over a three dimensional region?

 

[burt]

It's not clear to me what you are trying to do. You say "surface" and your integral is "dA" but you have a triple integral. Are you integrating over a surface or over a three dimensional region?

I don't know ... I must be misunderstanding something here - all I know for sure is the information in the question which is in my original post.
Maybe you can shed some light on the correct way to solve this?

 

[HallsofIvy]

The surface is the plane x+ 2y+ z= 4. So z= 4- x- 2y. A tangent vector in the x-direction is i- k A tangent vector in the y-direction is j- 2k. The cross product of those vectors
\(\displaystyle \left|\begin{array}{ccc} i & j & k \\ 1 & 0 & -1 \\ 0 & 1 & -2\end{array}\right|= -2i- j- 2k\) has length \(\displaystyle \sqrt{4+ 1+ 4}= \sqrt{9}= 3\). The differential of surface area" is 3dxdy.

(dxdy because I solved for z in terms of x and y. If I had solved for y= 2- x/2- z/2 I would have used dxdz. If I had solved for x= 4- 2y- z I would have used dydz.

The region here lies above the region in the xy-plane above \(\displaystyle y= x^2\) and below \(\displaystyle y= 1\). Those two curves intersect at x= -1 and x= 1 so to integrate over it take x from -1 to 1 and, for each x, y from \(\displaystyle x^2\) to 1. With density y the mass is \(\displaystyle \int_{x= -1}^1 \int_{x^2}^1 y(3dydx)= 3\int_{-1}^1\int_{x^2}^1 y dydx\).

The three moments are
\(\displaystyle M_{yz}= 3\int_{-1}^1\int_{x^2}^1 xy dydx\)
\(\displaystyle M_{xz}= 3\int_{-1}^1\int_{x^2}^1 y^2 dydx\)
\(\displaystyle M_{xy}= 3\int_{-1}^1\int_{x^2}^1 yz dydx= 3\int_{-1}^1\int_{x^2}^1 y(4- x- 2y) dydx\).

 

[burt]

A tangent vector in the x-direction is i- k A tangent vector in the y-direction is j- 2k. The cross product of those vectors

Can you explain why you did this?


Is it because what I really should be doing to look for the mass is \(\iint\rho(x,y,z)\ dS\) and dS is the surface area integral - which means we need to put in the norm?

Why is this a surface integral problem?

 

[burt]

@HallsofIvy Can't I use \(\sqrt{f_x^2+f_y^2+1} dA\) instead? Why or why not?

 

[burt]

If I do the work for the mass like a surface integral, I would have this:

 

[burt]

I think I got it!

Is that right?

It looks right on a graph: