Correct Application of Divergence Theorem?

[TranscendArcu]

Homework Statement

http://img593.imageshack.us/img593/5713/skjermbilde20111204kl11.png

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

 

[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.

 

[LCKurtz]

Homework Statement

http://img593.imageshack.us/img593/5713/skjermbilde20111204kl11.png

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

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.

\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?

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.

 

[TranscendArcu]

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

 

[Char. Limit]

That sounds more likely to be correct, yes.

 

[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?