# Vector calculus: simple surface integral problem

• May 24th 2011, 02:43 AM
Jnix
Vector calculus: simple surface integral problem
At least, it should be simple!

I am working my way through div, grad, curl and all that by H. M. Schey and have just finished the chapter on surface integrals and the divergence. I don't seem able to get the right answer for one of the problems, which suggests it should be possible to evaluate the surface integral without going through long-winded calculations.

I need to find
$\iint_{S} \mathbf{F}\cdot \hat{\mathbf{n}} dS$,
the surface integral over surface S.

The surface S is open and consists of 3 squares with side length b, forming half a cube at the origin. The 3 squares are adjacent and there is one in each of the xy, xz, and yz planes.

The vector function is:
$\mathbf{F}=\hat{\mathbf{i}}x+\hat{\mathbf{j}}y+\ha t{\mathbf{k}}z$.

My thinking is as follows: the three squares are identical, so I need only find the surface integral for one square and multiply it by 3.

I start with $S_{1}$, the square in the xy plane. The unit normal vector, $\hat{\mathbf{n}}$, is equal to $\hat{\mathbf{k}}$ and the surface integral becomes:

$\iint_{S_{1}} \mathbf{F} \cdot \hat{\mathbf{k}} dS =\int_{x=0}^{x=b}\int_{y=0}^{y=b} z dx dy = b^{2} z$

I know the surface integral should overall evaluate to zero, but repeating this for the 3 surfaces, I get

$\iint_{S} \mathbf{F} \cdot \hat{\mathbf{n}} dS = b^{2}(x+y+z)$.

I'm fairly sure I've missed something here, but not sure what. Any suggestions?

Thanks for taking the time to read this.
• May 24th 2011, 04:00 AM
HallsofIvy
Quote:

Originally Posted by Jnix
At least, it should be simple!

I am working my way through div, grad, curl and all that by H. M. Schey and have just finished the chapter on surface integrals and the divergence. I don't seem able to get the right answer for one of the problems, which suggests it should be possible to evaluate the surface integral without going through long-winded calculations.

I need to find
$\iint_{S} \mathbf{F}\cdot \hat{\mathbf{n}} dS$,
the surface integral over surface S.

The surface S is open and consists of 3 squares with side length b, forming half a cube at the origin. The 3 squares are adjacent and there is one in each of the xy, xz, and yz planes.

The vector function is:
$\mathbf{F}=\hat{\mathbf{i}}x+\hat{\mathbf{j}}y+\ha t{\mathbf{k}}z$.

The "t" in that last is an artifact of the LaTeX. I'm not sure how it got there- I can't find an error in the code!

Quote:

My thinking is as follows: the three squares are identical, so I need only find the surface integral for one square and multiply it by 3.

I start with $S_{1}$, the square in the xy plane. The unit normal vector, $\hat{\mathbf{n}}$, is equal to $\hat{\mathbf{k}}$ and the surface integral becomes:

$\iint_{S_{1}} \mathbf{F} \cdot \hat{\mathbf{k}} dS =\int_{x=0}^{x=b}\int_{y=0}^{y=b} z dx dy = b^{2} z$
Have you forgotten that this is in the xy plane? z= 0 in the xy plane!

Quote:

I know the surface integral should overall evaluate to zero, but repeating this for the 3 surfaces, I get

$\iint_{S} \mathbf{F} \cdot \hat{\mathbf{n}} dS = b^{2}(x+y+z)$.

I'm fairly sure I've missed something here, but not sure what. Any suggestions?

Thanks for taking the time to read this.
• May 25th 2011, 01:05 AM
Jnix
Oh good point!

I think the mysterious t is something to do with the \hat{} tag not being processed correctly on the k. No idea why, though.

I supposed this is true for all the other faces, too, as in each case, x,y or z=0 for the entire surface. Thanks :)