Surface Integral (Integral Setup)

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SUMMARY

The discussion focuses on setting up a surface integral for a sphere intersecting with the plane z=2. The limits for the angle phi are established as ranging from 0 to arccos(2/3), while theta spans from 0 to 2π. The differential surface area element dS is correctly identified as 9sin(phi) dθ dφ. Participants confirm the integration limits and provide clarity on the substitution of Cartesian coordinates for the integral setup.

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Ronaldo95163
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Homework Statement


I'm just required to setup the integral for the question posted below

Homework Equations

The Attempt at a Solution


So solving for phi @ the intersection of the sphere and the plane z=2:

z = pcos(phi)
2 = 3cos(phi)
phi = arccos(2/3)

so my limits for phi would go from 0 to arccos(2/3)

and my limits for theta would go from 0 to 2pi

for the surface integral dS would become 9sin(phi)

The rest of the integral I can obtain from just substituting x = psin(phi)cos(theta) y = psin(phi)sin(theta) and z = pcos(phi)

What I really want to be sure of is limits of integration.

Thanks in advance guys
 

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Hi,
Ronaldo95163 said:
for the surface integral dS would become 9sin(phi)
I hope you mean ##9 \sin\phi \;
d\theta d\phi## :rolleyes:
The rest of the integral I can obtain from just substituting x = psin(phi)cos(theta) y = psin(phi)sin(theta) and z = pcos(phi)
The latter for the numerator. The denominator is simply 9 -- easy !
 

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