Surface Integral using Divergence Theorem

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The discussion focuses on evaluating the surface integral I=\int\int_S\vec{F}\cdot\vec{n}\,dS using the Divergence Theorem. The vector field is defined as \vec{F}=<z^2+xy^2,x^{100}e^x,y+x^2z>, with the surface S being bounded by a paraboloid and a plane. Participants clarify the divergence of \vec{F} as x^2+y^2 and discuss the integration process, emphasizing the need to correctly evaluate the integral with respect to z. There is a consensus on switching to polar coordinates for the final integration step. The discussion concludes with a light-hearted exchange, maintaining focus on the mathematical process.
Saladsamurai
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Homework Statement



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

where \vec{F}=&lt;z^2+xy^2,x^100e^x, y+x^2z&gt;

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.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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