1. ## Stoke's Theorem Question

Evaluate $\iint_S \nabla \times \mathbf{F} \cdot dS$ Where S consists of the top and four sides of the cube with vertices $(\pm1,\pm1,\pm1)$ oriented outwards.

$\mathbf{F}(x,y,z) = xyz\ \mathbf{i} + xy\ \mathbf{j} + x^2yz\ \mathbf{k}$

pls help, never worked with cubes before.

2. ## Re: Stoke's Theorem Question

Do the six faces separately. On the face x= 1, $d\vec{S}= dydz\vec{i}$ oriented outward. y goes from -1 to 1, z goes from -1 to 1. $\nabla\times F(1, y, z)= z\vec{i}+ (-2yz)\vec{j}+ (y- z)\vec{k}$ so that $\nabla F(1, y, z)\cdot d\vec{S}= z dydz$ and the integral over that face is $\int_{y= -1}^1\int_{z= -1}^1 z dydz$.

Do the same for the other 5 faces, x= -1, y= 1, y= -1, z= 1, z= -1.

3. ## Re: Stoke's Theorem Question

wow i get it now thanks alot

4. ## Re: Stoke's Theorem Question

some of the terminology doesnt sit well with me.

when i did my curl i got $x^2z\mathbf{i} - (2xyz - xy)\mathbf{j} + (y - xz)\mathbf{k}$ which at $\mathbf{F}(1,y,z)$ is different to what you have, plus i dont understand where $\nabla\mathbf{F}(1,y,z) = zdydz$ comes from

5. ## Re: Stoke's Theorem Question

You are right. I accidently dropped the "-xy" term in $\vec{i}$i. $\nabla\times F= x^2z\vec{i}- (2xyz- xy)\vec{j}+ (y- xz)\vec{k}$ which, with x= 1 is $z\vec{i}- (2yz- y)\vec{j}+ (y- z)\vec{k}$

But it is not $\nabla\times F$ that is zdydz, it is $\nabla\times F \cdot d\vec{S}$. On the plane x= 1, $d\vec{S}= dydz\vec{i}$ and the dot product of that with $z\vec{i}- (2yz- y)\vec{j}+ (y- z)\vec{k}$ is zdydz.

6. ## Re: Stoke's Theorem Question

ahhh i see now, thankyou alot