What's the area of this volume?

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?

arildno

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]

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Jin314159	What's the area of this volume? Tweet 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?

arildno Recognitions: Gold Member Homework Help Science Advisor	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]