- #1

Tony11235

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[tex] \int_{Q} \bigtriangledown} \cdot G dxdydz = \int_{\partial Q} G \cdot n dS [/tex]

I am not sure how to evaluate the right side. Any help would be good.

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- Thread starter Tony11235
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- #1

Tony11235

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[tex] \int_{Q} \bigtriangledown} \cdot G dxdydz = \int_{\partial Q} G \cdot n dS [/tex]

I am not sure how to evaluate the right side. Any help would be good.

- #2

arildno

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Well, you've got six sides on your surface, right?

Treat the contribution from each side separately.

Treat the contribution from each side separately.

- #3

Tony11235

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What would say..the first integral of the six look like? I just need one example.

- #4

arildno

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Here, the normal vector is +i.

G(1,y,z)=(y,e^z+3y,y^3sinx)

Forming the dot product between G and the normal vector yields the integrand "y".

This is easy to integrate over the y,z-square (yielding 1/2 in contribution)

Okay?

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Tony11235

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- #6

Hurkyl

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- #7

arildno

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That's correct.Tony11235 said:

Remember that the integral operation has the additive property; this entails that you may split up the surface in an arbitrary number of sub-surfaces, calculate the individual contributions and then add the individual contributions together to gain the correct value.

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