Surface integral

Hi, it's been a while since ive done the divergence theorem and surface integrals; if you had an integral ∫ u . n z dS over a boundary dK where n is the unit normal, what would be equivelant in one dimension?

Would it just be u*z?

Reply 1

atsruser

11

Original post by MaloneJ

Hi, it's been a while since ive done the divergence theorem and surface integrals; if you had an integral ∫ u . n z dS over a boundary dK where n is the unit normal, what would be equivelant in one dimension?

Would it just be u*z?

It's not clear what your integral is. What is z? You mention the divergence theorem, but you seem to be integrating over a surface, not a volume.

It looks like you want to apply Stoke's theorem, but for that, you need to be integrating a curl over a surface, and you don't seem to be doing that. Unless u can be expressed as the curl of another field, you're probably out of luck.

Reply 2

poorform

10

Post a photo of the question or type it up fully.

BTW Divergence theorem relates the flux through a closed surface to the divergence integral inside the surface and Stokes' theorem relates a line integral (in a vector field) over it's boundary to the surface integral of the curl inside the boundary.

I'm not sure about this next part but I don't see why you couldn't use the divergence theorem to calculate it possibly although you haven't made it clear what you want to calculate.

Post it clearer in latex and someone will help you I'm sure.

Reply 3

DFranklin

18

I'm not sure what you're asking, but in one dimension, the divergence theorem is just the fundamental theorem of calculus:

\displaystyle \int_a^b \dfrac{df}{dx}\,dx = f(b) - f(a)

.

Reply 4

jsMath

8

INTEGRAL OF A VECTOR FILED -> SURFICE INTEGRAL

To Refresh your mamory on the topic...

Either

\hat{n}

is a norm of a parametric surface or an explicit one.

if

\hat{n}

is applied to an explicit surface then

\hat{n}

=

\frac{\nabla{\vec{V}}}{\| \mathbf{\vec{V}} \|}

if

\hat{n}

is applied to a parametric surface then

\hat{n}

=

\frac{{\vec{p}_u \times \vec{p}_v}}{\|\vec{p}_u \times \vec{p}_v \|}

where

{\vec{p}_u , \vec{p}_v}

are partial derivatives with respect to u and v.

Apply the surface integral

Unparseable latex formula:

\iint\limits_S (\vec{V}\cdot \hat{n}) \dif dS

where dS is

{\vec{p}_u \times \vec{p}_v}dudv \in\mathbb{R}\times\mathbb{R}

V(x,y,z)=\langle V_1,V_2,V_3\rangle \in\mathbb{R}\times\mathbb{R}[br]\times\mathbb{R}

The idea is to position the vectors flowing through a surface in the direction of the unit vectors as components are increased partially. in order to calculate and measure the flux using surface integral methodology.

(edited 8 years ago)

Surface integral

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you