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- Thread starter thojrie
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But the surface is a vector because its an orientated surface right (surface + direction) ?

I mean when you do your surface integrals, you don't just define the surface, you define direction of the normal vector of the differential elements of the surface surface

Supposing you have some surface dS

you parameterise it with two variables, say s and t

then we can trace out our surface with some function x(s,t)

and then we take the dot product of the surface normal with our vector field:

[tex]

\int_S {\mathbf v}\cdot \,d{\mathbf {S}} = \int_S ({\mathbf v}\cdot {\mathbf n})\,dS=\iint_T {\mathbf v}(\mathbf{x}(s, t))\cdot \left({\partial \mathbf{x} \over \partial s}\times {\partial \mathbf{x} \over \partial t}\right) ds\, dt.

[/tex]

The key part here is that

[tex]

d\mathbf{S} = \mathbf{n}dS

[/tex]

With regards to visualising it as a Riemann sum, this will need to be checked by some one with more mathematical knowledge, but treat the whole dot product

[tex] {\mathbf v}(\mathbf{x}(s, t))\cdot \left({\partial \mathbf{x} \over \partial s}\times {\partial \mathbf{x} \over \partial t}\right) [/tex]

As the Jacobian which maps your complicated surface integral of a vector field to a simple two dimensional integral which you can then use your standard Riemann sums on

I mean when you do your surface integrals, you don't just define the surface, you define direction of the normal vector of the differential elements of the surface surface

Supposing you have some surface dS

you parameterise it with two variables, say s and t

then we can trace out our surface with some function x(s,t)

and then we take the dot product of the surface normal with our vector field:

[tex]

\int_S {\mathbf v}\cdot \,d{\mathbf {S}} = \int_S ({\mathbf v}\cdot {\mathbf n})\,dS=\iint_T {\mathbf v}(\mathbf{x}(s, t))\cdot \left({\partial \mathbf{x} \over \partial s}\times {\partial \mathbf{x} \over \partial t}\right) ds\, dt.

[/tex]

The key part here is that

[tex]

d\mathbf{S} = \mathbf{n}dS

[/tex]

With regards to visualising it as a Riemann sum, this will need to be checked by some one with more mathematical knowledge, but treat the whole dot product

[tex] {\mathbf v}(\mathbf{x}(s, t))\cdot \left({\partial \mathbf{x} \over \partial s}\times {\partial \mathbf{x} \over \partial t}\right) [/tex]

As the Jacobian which maps your complicated surface integral of a vector field to a simple two dimensional integral which you can then use your standard Riemann sums on

Last edited:

- #3

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But the surface is a vector because its an orientated surface right (surface + direction) ?

I mean when you do your surface integrals, you don't just define the surface, you define direction of the normal vector of the differential elements of the surface surface

Supposing you have some surface dS

you parameterise it with two variables, say s and t

then we can trace out our surface with some function x(s,t)

and then we take the dot product of the surface normal with our vector field:

[tex]

\int_S {\mathbf v}\cdot \,d{\mathbf {S}} = \int_S ({\mathbf v}\cdot {\mathbf n})\,dS=\iint_T {\mathbf v}(\mathbf{x}(s, t))\cdot \left({\partial \mathbf{x} \over \partial s}\times {\partial \mathbf{x} \over \partial t}\right) ds\, dt.

[/tex]

The key part here is that

[tex]

d\mathbf{S} = \mathbf{n}dS

[/tex]

With regards to visualising it as a Riemann sum, this will need to be checked by some one with more mathematical knowledge, but treat the whole dot product

[tex] {\mathbf v}(\mathbf{x}(s, t))\cdot \left({\partial \mathbf{x} \over \partial s}\times {\partial \mathbf{x} \over \partial t}\right) [/tex]

As the Jacobian which maps your complicated surface integral of a vector field to a simple two dimensional integral which you can then use your standard Riemann sums on

I mean at the end of the day following the above steps turns your surface integral into a double integral (wrt to your parameterisation variables) of the form:

[tex] \int_{S} \mathbf{v}(s,t) \cdot d\mathbf{S} [/tex]

Hi thrillhouse86, thanks for the detailed reply! I don't think I phrased my question well enough though.

Because the dot product of two vectors is a scalar, when we find the flux through a surface the integrand and answer are both scalars. If instead, the integrand were a vector, the integral becomes,

[tex]\int_S \vec{F} dS[/tex]

which will give a vector answer. I assume this is analogous to flux (but with a direction), however I'm having trouble breaking it down and sussing out exactly what it means.

- #4

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If the field would represent something like "Surface current density" then it will give you the total/effective current on the surface.

- #5

HallsofIvy

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Writing [itex]\vec{F}= f\vec{i}+ g\vec{j}+ h\vec{k}[/itex], that integral would beHi thrillhouse86, thanks for the detailed reply! I don't think I phrased my question well enough though.

Because the dot product of two vectors is a scalar, when we find the flux through a surface the integrand and answer are both scalars. If instead, the integrand were a vector, the integral becomes,

[tex]\int_S \vec{F} dS[/tex]

which will give a vector answer. I assume this is analogous to flux (but with a direction), however I'm having trouble breaking it down and sussing out exactly what it means.

[tex]\int_S f dS\vec{i}+ \int_S g dS\vec{j}+ \int h dS\vec{k}[/itex].

- #6

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Hey thanks guys. That helped.

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