1. ## Stokes Theorem

$\displaystyle F = 3yi - 2yj + {z}^{2 }k$ S the hemisphere$\displaystyle {x}^{2 } + {y}^{2 } + {z}^{2 }=1$ $\displaystyle z\geqslant 0$ So i ahve done $\displaystyle \nabla x F = 0i + 0j -3k$ Now my integral is $\displaystyle \iiint -3{r}^{ 2} sin\phi$ $\displaystyle dr d\phi d\theta$ where r is between 1 and 0, $\displaystyle d\phi$ is between $\displaystyle \frac{\pi }{ 2}$ and 0 and $\displaystyle d\theta$ is between $\displaystyle 2\pi$ and 0

But this isnt gettign me the right answer what have i done wrong

2. Are you sure that your question is related to the Stokes Theorem?

3. Originally Posted by FernandoRevilla
Are you sure that your question is related to the Stokes Theorem?
Verify Stokes theorem in the following cases

4. The answer is $\displaystyle -3\pi$ by the way

$\displaystyle F = 3yi - 2yj + {z}^{2 }k$ S the hemisphere$\displaystyle {x}^{2 } + {y}^{2 } + {z}^{2 }=1$ $\displaystyle z\geqslant 0$ So i ahve done $\displaystyle \nabla x F = 0i + 0j -3k$ Now my integral is $\displaystyle \iiint -3{r}^{ 2} sin\phi$ $\displaystyle dr d\phi d\theta$ where r is between 1 and 0, $\displaystyle d\phi$ is between $\displaystyle \frac{\pi }{ 2}$ and 0 and $\displaystyle d\theta$ is between $\displaystyle 2\pi$ and 0

But this isnt gettign me the right answer what have i done wrong
You seem to be mixing Stokes Theorem with the divergence theorem.

Given a Vector field and a surface we know that

$\displaystyle \iint_{S}\nabla \times \mathbf{F} \cdot d\mathbf{S}=\oint_{\partial S}\mathbf{F}\cdot d\mathbf{s}$

The boundary of this curve is the unit circle in the xy plane.

My question is what is the integral you started with and what are you trying to compute? Are you trying to compute the flux of

$\displaystyle \mathbf{F}$

though the hemisphere?

Edit: Just typing too slow

Verify Stokes theorem in the following cases

Then, what has to do the triple integral?. You have to prove

$\displaystyle \displaystyle\int_C \vec{F}\cdot d\vec{r}=\displaystyle\iint_{S}(\textrm{rot} \vec{F})\cdot \vec{n}\;dS$

Edited: Sorry, I didn't see TheEmptySet's post.

7. Aaaahhh right, so diveregnce you can use a triple, stokes you cant?

Aaaahhh right, so diveregnce you can use a triple, stokes you cant?
Still a bit stuck

$\displaystyle \iint -3{z}^{ 2}$ $\displaystyle z = r$ so $\displaystyle \iint -3{r}^{ 2} r dr d\theta$

Is that right?

Aaaahhh right, so diveregnce you can use a triple, stokes you cant?

Yes, the Divergence Theorem states

$\displaystyle \displaystyle\iint_{S} \vec{F}\cdot \vec{n}\;dS=\displaystyle\iiint_{V}\textrm{div}( \vec{F})\;dxdydz$

Still a bit stuck

$\displaystyle \iint -3{z}^{ 2}$ $\displaystyle z = r$ so $\displaystyle \iint -3{r}^{ 2} r dr d\theta$

Is that right?
These aren't spherical co-ordinates, how do i use spherical co-ordinates?

11. Use the divergence theorem as hinted by others. The volume you're dealing with is a hemisphere which is easy to deal with in spherical coordinates.

Spherical coordinate system - Wikipedia, the free encyclopedia

The limits for a hemisphere with radius=1 is:

12. Originally Posted by Mondreus
Use the divergence theorem as hinted by others. The volume you're dealing with is a hemisphere which is easy to deal with in spherical coordinates.

Spherical coordinate system - Wikipedia, the free encyclopedia

The limits for a hemisphere with radius=1 is:
Is $\displaystyle \nabla \cdot F = -2 + 2z$