Flux through a section of a sphere

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SUMMARY

The discussion focuses on calculating the flux of the vector field F = through a spherical surface defined by ρ = 2, specifically above the plane z = 1. The initial attempt yielded an incorrect integral result of \(\frac{4\pi}{3}(\sqrt{32} - 1)\), while the correct flux calculation is \(\frac{14\pi}{3}\). The correct approach involves using polar coordinates and integrating the function \(r\sqrt{4 - r^{2}}\) over the specified limits, leading to the final answer through a proper substitution method.

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Homework Statement


Find the flux of F=<y,-x,z> through the piece of ρ=2 that lies above z=1 and is oriented up.


Homework Equations





The Attempt at a Solution



S = &lt; x, y, \sqrt{4-x^{2}-y^{2}} &gt;

Take Find Sx and Sy, cross them and end up with:

dS = &lt; \frac{x}{\sqrt{4-x^{2}-y^{2}}}, \frac{y}{\sqrt{4-x^{2}-y^{2}}}, 1 &gt;

Z is positive, orientation is OK.

F dot dS = \sqrt{4-x^{2}-y^{2}}

Therefore the integral should be

\int^{2\pi}_{0}\int^{\sqrt{3}}_{0} r\sqrt{4-r^{2}}drd\theta

= \frac{4\pi}{3} (\sqrt{32} -1)

Incorrect.
The correct answer is

\frac{14\pi}{3}
 
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All I can say is that you have clearly integrated wrong. Since you don't show how you did that integral, I cannot say more.
 
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\int^{2\pi}_{0} \int^{\sqrt{3}}_{0}r \sqrt{4-r^{2}}dr d\theta

= 2\pi \int^{\sqrt{3}}_{0}r\sqrt{4-r^{2}}dr

=\pi \int^{4}_{1}u^{1/2}du

= \frac{2\pi}{3}(4^{3/2}-1)

I see what I did. Thanks.
 

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