What's the area of this volume?

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SUMMARY

The volume of the surface generated by rotating a circle defined by the equation z² + (x-a)² = a² around the z-axis is calculated to be V = (π²/2)a³. This conclusion is based on the application of Cavalieri's principle, which leads to the integral V = ∫₀^(π/2) πa² a dθ. The area function for the rotation is f(θ) = πa², confirming that the volume represents a quarter of a torus.

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Jin314159
Consider a 3d coordinate system with axis x,y and z.

We are given a circle on the x-z plane with function [tex]z^2 + (x-a)^2 = a^2[/tex]. We rotate this circle 90 degrees around the z-axis. What's the volume of the resulting surface?
 
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Well, shouldn't that be a quarter of a torus?
I'll opt for that and say:
[tex]V=\pi{a}^{2}*\frac{\pi}{2}a=\frac{\pi^{2}}{2}a^{3}[/tex]

Edit:
The area function for a given angle measured relative to the x-axis (and with the origin as the pole) is [tex]f(\theta)=\pi{a}^{2}[/tex]

By Cavalieri's principle, we have:
[tex]V=\int_{0}^{\frac{\pi}{2}}\pi{a}^{2}ad\theta[/tex]
 
Last edited:

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