Surface Integral using Divergence Theorem

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Homework Help Overview

The problem involves evaluating a surface integral using the Divergence Theorem, specifically for a vector field defined as \(\vec{F} = \). The surface \(S\) is bounded by a paraboloid and a plane, with an outward normal orientation.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the application of the Divergence Theorem and the evaluation of the divergence of the vector field. There is a focus on integrating the divergence over the volume and the subsequent surface integral. Questions arise regarding the correct evaluation of integrals and the use of polar coordinates.

Discussion Status

Some participants are exploring the integration process and questioning the correctness of the steps taken. There is a suggestion to use polar coordinates, and some guidance has been provided regarding the evaluation of integrals. However, there is no explicit consensus on the approach yet.

Contextual Notes

Participants are navigating through the integration limits and the application of the Divergence Theorem, with some expressing uncertainty about the correct factors to include in their calculations.

Saladsamurai
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Homework Statement



Evaluate the surface Integral I=\int\int_S\vec{F}\cdot\vec{n}\,dS

where \vec{F}=<z^2+xy^2,x^100e^x, y+x^2z>

and S is the surface bounded by the paraboloid z=x^2+y^2

and the plane z=1; oriented by the outward normal.

The Attempt at a Solution



I=int\int_S\vec{F}\cdot\vec{n}\,dS=\int\int\int_E(div\vec{F})dV

(div\vec{F})=y^2+x^2

\Rightarrow I=\int\int_D(\int_{z=x^2+y^2}^1(x^2+y^2)\,dz)\,dA

\Rightarrow I=\int\int_D(1-(x^2+y^2)\,dA

Is it just Polar Coordinates all the way home now?

Thanks,
Casey
 
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No, I don't think that's right. The integral of f(x,y)dz is z*f(x,y), isn't it? Try evaluating that between the limits again.
 
I am sorry, where are you referring to?
The divergence of F= x^2+y^2 That is f(x,y), now if I integrate that wrt z
ohh! I forgot to include the factor of f(x,y) so it should be

I=\int\int_D(x^2+y^2)*(1-(x^2+y^2)\,dA

yes? Now Polar?
 
Now polar.
 
Polar bears. Polar beers. Sweet numchuck skills Dick. Thanks.
 

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