# Thread: Vector calculus: simple surface integral problem

1. ## 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.

2. 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!

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!

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.

3. 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