# Thread: Vector Calculus (Surface Integral)

1. ## Vector Calculus (Surface Integral)

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 $0 \leq x \leq 1,$ $0 \leq z \leq 1$

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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.

2. ## Re: Vector Calculus (Surface Integral)

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.

3. ## Re: Vector Calculus (Surface Integral)

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?

4. ## Re: Vector Calculus (Surface Integral)

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.

5. ## Re: Vector Calculus (Surface Integral)

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.

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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.