# Correct Application of Divergence Theorem?

1. Dec 4, 2011

### TranscendArcu

1. The problem statement, all variables and given/known data
http://img593.imageshack.us/img593/5713/skjermbilde20111204kl11.png [Broken]

3. The attempt at a solution
I thought it seemed appropriate to use divergence theorem here: I have,
$$div F = 0 + 1 + x = 1+x$$
I let that 0≤z≤c. If,

$$x/a + y/b = 1$$then y=b(1-x/a)
$$x/a +z/c = 1$$then x=a(1-z/c)

I have,

$$\int_0^a \int_0^{c(1-x/a)} \int_0^{b(1-x/a)} (1+x) dydzdx$$
$$\int_0^a \int_0^{c(1-x/a)} (x+1) b(1-x/a) dzdx$$
$$\int_0^a (x+1) b(1-x/a) c(1-x/a) dx$$

But here I get stuck because I can't integrate this. Do I have the wrong bounds or am I missing something?

Last edited by a moderator: May 5, 2017
2. Dec 4, 2011

### Char. Limit

You can integrate this just fine, once you distribute everything. Your integral at the end is:

$$bc \int_0^a \left(1+x\right) \left(1 - \frac{x}{a}\right)^2 dx$$

That's going to factor out to a cubic, which is easily integrable.

3. Dec 4, 2011

### LCKurtz

Your upper limit on the dy integral is wrong. y should go from y = 0 to the y on the plane. Solve the equation of the plane for y. It should have both x and z in it.
It is wrong because of the above comment, but of course you should be able to integrate something like that. It is just a polynomial. All you have to do is multiply it out first.

Last edited by a moderator: May 5, 2017
4. Dec 4, 2011

### TranscendArcu

So does y=b(1- z/c - x/a)?

5. Dec 4, 2011

### Char. Limit

That sounds more likely to be correct, yes.

6. Dec 5, 2011

### TranscendArcu

Okay. Let's see if I can do this. As above, I'll integrate in the order dydzdx.

$$\int_0^{b(1-z/c-x/a)}(1+x) dy = (x+1) b(-x/a-z/c+1)$$
$$= (1+x)*b \int_0^{c(1-x/a)} (-x/a-z/c+1) dz$$
$$= (1+x)*b*[(1/2)*c*(x^2/a^2-1)+c*(1-x/a)]$$
$$\int_0^{a} (1+x)*b*[(1/2)*c*(x^2/a^2-1)+c*(1-x/a)] dx= (1/24)*a*(a+4)*b*c$$

Hmm. I still feel like I missed something. Does it look about right?

Last edited: Dec 5, 2011