# Homework Help: Divergence Theorem

1. Aug 31, 2010

### vorcil

1. The problem statement, all variables and given/known data
Check the divergence theorem using the function:

$$\mathbf{v} = y^2\mathbf{\hat{x}} + (2xy + z^2) \mathbf{\hat{y}} + (2yz)\mathbf{\hat{z}}$$

2. Relevant equations

$$\int_\script{v} (\mathbf{\nabla . v }) d\tau = \oint_\script{S} \mathbf{v} . d\mathbf{a}$$

3. The attempt at a solution

taking the dot product it becomes

$$\frac{\partial}{\partial x} y^2 \mathbf{\hat{x}} + \frac{\partial}{\partial y} ( 2xy + z^2) \mathbf{\hat{y}} + \frac{\partial}{\partial z} (2yz)\mathbf{\hat{z}}$$

so by only differentiating the certain parts:

i get y^2 + 2x + z^2 + 2y,
however the dot product of del and my vector v, should've been 2(x+y)
how come I've got y^2 and z^2?

does $$\frac{\partial}{\partial x} y^2$$ not equal y^2???

2. Aug 31, 2010

### Office_Shredder

Staff Emeritus
Your divergence should not actually include any vectors. It's just a number. So you should have
$$\frac{\partial}{\partial x} y^2 + \frac{\partial}{\partial y} ( 2xy + z^2) + \frac{\partial}{\partial z} (2yz)$$

Once you take the dot product

Then it looks like you differentiated your unit vectors as though they were variables

$$\frac{\partial}{\partial x} y^2$$ is 0, not y2

3. Aug 31, 2010

### lanedance

this isn't correct, you applied a dot product, it should be
$$\frac{\partial}{\partial x} y^2 + \frac{\partial}{\partial y} ( 2xy + z^2) + \frac{\partial}{\partial z} (2yz)$$

to see where you went wrong, what is
$$\frac{\partial}{\partial y} z^2$$

4. Aug 31, 2010

### vorcil

Is it 0? is that because z^2 is a constant since it's being differentiated with respect to y?
and d/dx of a constant is 0?

-

now knowing that $$\frac{\partial}{\partial y} z^2 = 0$$
I think I can solve the rest,
$$\int_\script{v} (\mathbf{\nabla . v }) d\tau = \oint_\script{S} \mathbf{v} . d\mathbf{a}$$
$$\frac{\partial}{\partial x} y^2 + \frac{\partial}{\partial y} ( 2xy + z^2) + \frac{\partial}{\partial z} (2yz)$$
so
$$\oint_{\script{V}} 2(x+y) d\tau$$

taking the constant out and making tau dxdydz = $$2 \int_0^1 \int_0^1 \int_0^1 (x+y) dxdydz$$

separating the three integrals out, (i think I can do that?)

dx part: $$\int_0^1 (x+y)dx = \left[ \frac{x^2}{2}+y \right]_0^1 = (\frac{1}{2} + y)$$

dy part, using the result of the dx part:
$$\int_0^1 (\frac{1}{2} + y) dy = \left[ \frac{1}{2} + \frac{y^2}{2} \right]_0^1 = (\frac{1}{2} + \frac{1}{2})$$

dz part using the result of the dy part:
$$\int_0^1 (1)dz = \left[ 1*z\right]_0^1 = 1$$

bringing in the constant 2, of what I brought outside the integral,

$$\int_{\script{V}} \mathbf{\nabla . v} d\tau = 2$$

so that means the integral of a derivative over the vector is equal to 2,
and to confirm the divergence theorem,

$$\int_\script{v} (\mathbf{\nabla . v }) d\tau = \oint_\script{S} \mathbf{v} . d\mathbf{a}$$

I also need to integrate over a closed surface for the cube?

that means i need to solve for 6 integrals,

5. Aug 31, 2010

### lanedance

yes that is the definition of a partial derivative, the other variables are held constant

6. Aug 31, 2010

### hunt_mat

It does indeed require you to compute 6 double integrals over the surface of the unit cube.

7. Aug 31, 2010

### lanedance

not sure why you would separate it, anyway though you get the right answer (luck), the integral isn't correct. y is treated as a constant, what is teh integral of a constant?

8. Aug 31, 2010

### vorcil

oh right, I should've separated and differentiated each bit individually,
so \int x dx, |_0^1
then added \int y dy |_0^1 to it
and then multiplied by int dz,

9. Aug 31, 2010

### lanedance

no i meant
$$\int_0^1 dx y = yx|_0^1 = y(1-0) = y$$

10. Aug 31, 2010

### lanedance

just treat it as
$$2 \int_0^1 \left(\int_0^1 \left(\int_0^1 (x+y) dx\right)dy\right)dz$$