Converting Area to Volume: A Spherical Coordinate Approach

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To find the infinitesimal area element dA in spherical coordinates for a hemispherical surface, one must consider the relationships between the spherical coordinates (r, θ, φ) and the geometry of the hemisphere. The discussion emphasizes the connection between area and volume elements, suggesting that understanding the volume element from previous studies can aid in deriving the area element. The area element dA can be expressed in terms of the angles θ and φ, with the normal vector n pointing outward from the surface. The participants highlight the importance of integrating over the appropriate limits to accurately represent the surface area. This approach is crucial for calculating the electric flux through the hemispherical surface in the given electric field.
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Homework Statement



A hemispherical surface of radius b = 61 m is fixed in a uniform electric field of magnitude E0 = 3 V/m as shown in the figure. The x-axis points out of the screen.

Enter the general expression for an infinitesimal area element dA in spherical coordinates (r, θ, φ) using n as your outward-pointing normal vector. In these coordinates θ is the polar angle (from the z-axis) and φ is the azimuthal angle (from the x-axis in the x-y plane).
Mcu4dpw.png

Homework Equations


φ(flux) = ∫ E dA

The Attempt at a Solution


I understand that we're supposed to be looking for a small piece of area, but I don't know what makes up that area. We had a triple integral last semester of ∫∫∫r2dr sin(θ)dθ dφ. Is that related? We used that to integrate the volume of a sphere. Is there a similar process one can use for area?
 

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I guess you can either look up the area element in spherical coordinates or work it out for yourself from first principles.
 
brioches said:
∫∫∫r2dr sin(θ)dθ dφ. Is that related?
Very much so. Suppose you had an area element dA for a shell radius r within the sphere. How would you turn it into a volume element? Compare that with the integrand above.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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