Vector Calculus (Surface Integral)

**PedroMinsk** · June 24th 2011, 10:14 AM

Calculate $\iint\limits_S \vec{F} .\vec{n} dS$ :

$\vec{F} (x, y, z)$ : (x, y, 2z - x - y)

S: y - z = 0 $0 \leq x \leq 1,$ $0 \leq z \leq 1$
y + z = 0 $0 \leq x \leq 1,$ $0 \leq z \leq 1$

__________________________

The answer is 0 (zero), so i'm wondering if there is a trick to solve this problem.

If not, I know how to aplly the definition to solve it.

**HallsofIvy** · June 24th 2011, 10:25 AM

Originally Posted by PedroMinsk

Calculate $\iint\limits_S \vec{F} .\vec{n} dS$ :

$\vec{F} (x, y, z)$ : (x, y, 2z - x - y)

S: y - z = 0 $0 \leq x \leq 1,$ $0 \leq z \leq 1$
y + z = 0 $0 \leq x \leq 1,$ $0 \leq z \leq 1$

?? The conditions on x and z are identical for these two equations.

The answer is 0 (zero), so i'm wondering if there is a trick to solve this problem.

If not, I know how to aplly the definition to solve it.

**PedroMinsk** · June 24th 2011, 10:29 AM

Originally Posted by HallsofIvy

?? The conditions on x and z are identical for these two equations.

It means that the region is symmetric. But with this fact in hands I can say that the answer is 0?

**ojones** · June 24th 2011, 05:28 PM

The surface $S$ looks like a right triangular prism. If the double integral is zero it means that the total flux of the vector field through this surface is zero. It's not obvious from the expression for $\mathbf{F}$ why this should be the case and so you're going to have to do the calculation.

**HallsofIvy** · June 25th 2011, 05:45 AM

Originally Posted by PedroMinsk

Calculate $\iint\limits_S \vec{F} .\vec{n} dS$ :

$\vec{F} (x, y, z)$ : (x, y, 2z - x - y)

S: y - z = 0 $0 \leq x \leq 1,$ $0 \leq z \leq 1$
y + z = 0 $0 \leq x \leq 1,$ $0 \leq z \leq 1$

So these are two separate planes? And there is no bound on x?
On the first, y= z so we can write $\vec{r}(x, y)= x\vec{i}+ y\vec{j}+ y\vec{k}$
$\vec{r}_x= \vec{j}$
$\vec{r}_y= \vec{j}+ \vec{k}$

$\vec{n}dS= d\vec{S}= \vec{r}_x\times\vec{r}_y dxdy= (\vec{j}+ \vec{k})dxdy$

On that plane, $\vec{F}= x\ve{i}+ y\vec{j}+ (2z- x- y)\vec{k}= x\vec{i}+ y\vec{j}+ (y- x)\vec{k}$

$\vec{F}\cdot\vec{n}dS= (2y- x) dx dy$
Since there are no bounds on x, that integral does not exist.

__

________________________

The answer is 0 (zero), so i'm wondering if there is a trick to solve this problem.

If not, I know how to aplly the definition to solve it.

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