center of gravity

**kittycat** · September 2nd 2007, 06:03 PM

Find the center of gravity of the cube that is determined by the inequalities
0<or = x<=1 , 0<or = y<=1, 0<or = z<=1 if

a) the density is proportional to the square of the distance to the origin.

b) the density is proportional to the sum of the distances to the faces that lie in the coordinates planes.

**ThePerfectHacker** · September 2nd 2007, 06:18 PM

Originally Posted by kittycat

a) the density is proportional to the square of the distance to the origin.

How about I tell what the density function is. And you do the rest. Finding the desity function here is the key part.

First, do you what "(directly) proportional" means? It means (almost) that given two quantities $A$ and $B$ we say they are "directly proportional" when the ratio $\frac{A}{B}$ is constant.

Thus, since $\mbox{density}$ is proportional to $\mbox{distance}$ it means $\frac{ \mbox{density} }{ \mbox{distance} } = k \implies \mbox{density} = k \cdot \mbox{distance}$ .

Now $\mbox{density }=\delta(x,y)$ . And $\mbox{distance }= (\sqrt{x^2+y^2})^2 = x^2+y^2$ .

Thus, $\delta(x,y) = k(x^2+y^2)$ .

Now you continue ...

**kittycat** · September 2nd 2007, 07:15 PM

.

Why distance the above not equal to sqrt(x^2 + y^2)?

sorry, please ignore this post. I have forgotten to look back to the question before I posted this question.

**kittycat** · September 2nd 2007, 07:35 PM

okay,

I have Mass of the cube is k after calculation

and x bar = y bar = z bar =7/12k .

But the book suggested answer is 7/12. Can anyone tell me why?

Thank you very much.

**ThePerfectHacker** · September 2nd 2007, 07:39 PM

Originally Posted by kittycat

okay,

I have Mass of the cube is k after calculation

and x bar = y bar = z bar =7/12k .

But the book suggested answer is 7/12. Can anyone tell me why?

Thank you very much.

I think you forgot to divide the k's.

Because x bar = M_yz / M. But both integrals M_y and M contain a constant multiple k. Which can be factored out and cancelled.

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