# Homework Help: Volume integral

1. Feb 20, 2012

### aaaa202

1. The problem statement, all variables and given/known data
For a particular material the density varies with position as ρ(r)=x2yz
Find the total mass of a unit cube with one edge in the origin made by a such a material.

2. Relevant equations
We have dm = ρ(r)dV = ρ(r)dxdydz
So we want to calculate the volume integral (all from 0 to 1):
∫∫∫x2yz dxdydz = 1/12

First of all: Is this correct?

Now if so, my problem is just that I don't find the approach quite intuitive. You want to sum up all small volume contributions. What is that then makes you able to split the integral into integration over 3 directions? Can you explain to me what happens intuitively?

2. Feb 20, 2012

### Staff: Mentor

Isn't it "one vertex at the origin"?
The base of the cube is on the square whose vertices are at (0, 0), (0, 1), (1, 1), and (1, 0). The square is divided into a grid of squares, each of which is Δx by Δy. The interval along the z-axis is divided into intervals of length Δz, so now we have layers of grids stacked on top of each other.

Assuming that 1/Δx = m, and 1/Δy = n, and 1/Δz = p, the cube is divided into m*n*p small cubes, each of which has a volume of Δx * Δy * Δz.

If the density happened to be constant (which for this problem it isn't), the mass would be density * volume = density * 1.

Since the density varies by position, we have to take that into account. In your integral, you are first integrating with respect to x, then wrt y, and finally wrt z.

The first integration sums the cubes in a particular layer, running from x = 0 to x = 1. This gives you the mass for the cubes in a straight line.

The next integration sums the lines in a particular layer, running now from y = 0 to y = 1. This gives the the mass for the cubes in one of the layers.

The final integration sums the layers from z = 0 to z = 1, which gives you the total mass of the larger cube.

3. Feb 20, 2012

### aaaa202

well yeh, problem is just that it technially isnt the mass of a layer since we are multiplying by a length not a volume?

And yes, I meant vertex - my english is not always precise :)

4. Feb 20, 2012

### Staff: Mentor

No. In each of the steps I talked about in the previous post we're getting mass. In the first step (integration wrt x), we're getting the mass of a line of small cubes. In the second step (integration wrt y), we're getting mass of a layer of small cubes. In the final step (integration wrt z), we're getting the mass of all of the layers.

Last edited: Feb 20, 2012
5. Feb 20, 2012

### aaaa202

I see it now :) Thank you!

6. Feb 20, 2012

### aaaa202

no wait.... I still dont get it :(
In the first integral you are essentially multiplying each little density with a infinitesimal length dx. That does NOT give you a mass, as far as I can see.

7. Feb 20, 2012

### Staff: Mentor

I believe that it does, as your density function is in units of mass/unit length.

BTW, I get 1/12 as the answer as well.