# Divergence Theorem gets me the correct answer, but the alternative method doesn't

1. Nov 13, 2014

### goraemon

1. The problem statement, all variables and given/known data
The problem is given in the attached file.

2. Relevant equations
Divergence theorem, flux / surface integral

3. The attempt at a solution

As you can see I got the question correct using Divergence theorem. But I wanted to make sure that I could arrive at the same answer using the standard method for surface integrals, so I tried the following:

Given S: x+y+z=4, bounded by the axes.
->dS = $\sqrt3 dx dy$
-> the normal vector = $\frac{i + j + k}{\sqrt3}$
SO, F * n dS becomes -> (6xy + 2z) + (y^2 + 1) - (x + y) dx dy
= 6xy + 2(4 - x - y) + y^2 + 1 - x - y dx dy

The region's bounds for the double integral is: 0 <= x <= 4, and 0 <= y <= 4 - x.

Solving the double integral gets me 280 / 3...which is inconsistent with the correct answer I got using Divergence Theorem.

Where did I go wrong? And sorry in advance for the less-than-stellar formatting.

#### Attached Files:

• ###### div.pdf
File size:
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Views:
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2. Nov 13, 2014

### vela

Staff Emeritus
You only integrated over one face. You have to integrate over the entire surface of the volume.

3. Nov 13, 2014

### goraemon

OK, now I'm really confused then. Because sometimes using the surface integrals get me the correct answer but using Divergence Theorem doesn't, and vice versa. Below is a VERY similar question that I solved correctly using surface integrals...and I can't for the life of me see what is substantively different between the problem I posted in the OP and the problem below:

Solve the surface integral of F*n dS, where n is the outer unit normal of S, and F = (x+y)i + (9x - z)j + yk, and S is the tetrahedron formed by the coordinate planes and the plane z + 2x + 2y = 8.

I found the correct answer via the following method:
dS = 3 dx dy.
n = $\frac{2i+2j+k}{3}$

So, F*n dS = <x+y, 9x-z, y>*<2, 2, 1> dy dx = (20x + 3y - 2z) dy dx -> substituting z = 8 - 2x - 2y -> (24x + 7y - 16) dy dx

Double integrating the above over the bounds 0 <= x <= 4 and 0 <=y <= 4 - x gets me the correct answer of: 608 / 3.

So what is it that I did for the problem in the OP that was substantively different from what I did for the problem above? Why am I getting the OP's problem wrong, but this problem correct?

4. Nov 14, 2014

### vela

Staff Emeritus
You didn't get the correct answer for the second problem either if the point was to get a result that's supposed to be equal to $\int \nabla\cdot\vec{F}\,dv$. The divergence is equal to 1, so the volume integral is equal to the volume of the tetrahedron, which is 64/3, not 608/3.

5. Nov 14, 2014

### vela

Staff Emeritus
I should add that you are doing the flux calculation correctly for the one face in both cases. The divergence theorem, however, says that
$$\oint_S \vec{F}\cdot\hat{n}\,dS = \int_V \nabla\cdot\vec{F}\,dV.$$ The integral on the left is over a closed surface S, so you have to integrate over the surface that completely encloses the volume. In both problems, you've neglected the contribution to the integral from the other three faces.

6. Nov 14, 2014

### Zondrina

Using the divergence theorem for this problem is just fine.

$$\iint_S \vec{F} \cdot d \vec{S} = \iiint_V \vec{\nabla} \cdot \vec{F} dV = \int_{0}^{4} \int_{0}^{4-y} \int_{0}^{4 - x - y} 8y \space dzdxdy$$

If you want to do it the hard way, you need to do each surface separately.

Last edited: Nov 14, 2014