How can I correctly solve for dA in Gauss's Law for a sphere?

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SUMMARY

The discussion focuses on solving for the differential area element (dA) in Gauss's Law for a spherical Gaussian surface surrounding a point charge. The correct expression for dA is derived as dA = r^2 sin(θ) dθ dφ, which leads to the integral Integral(1 dA) = 4πr^2. The participant initially miscalculated the integral, arriving at 2π²r² instead of the correct result. Understanding the integration limits for θ and φ is crucial for accurate calculations in spherical coordinates.

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Hey, I'm having some trouble solving for the dA portion of Guss's Law for a sphere as the Gaussian surface and a point charge on the inside.

According to my book, Integral(1dA) = 4(pi)r^2

So when I try to integrate it myself I get 2(pi^2)r^2
The integral I solve is \int\int r\phi r\theta where theta = [0,2pi] and phi = [0,pi]

Can anyone set me straight please?
 
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On the surface of a sphere, the element of area bounded by d\theta and d\phi has area (r d\theta)(r \sin \theta d\phi) = r^2 \sin \theta d\theta d\phi.
 

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