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Math Help - Rotational area of polar functions?

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    Rotational area of polar functions?

    Hello, I am working on a project to rotate polar functions along another axis (the analog of the z axis in the Cartesian coordinate plane) but technically another r-axis.

    Here is a circle shown in a typical polar coordinate graph:

    Now, if one thinks of those two r axes as the "x" and "y" axes, then imagine if a new "z" r-axis was made, and a polar graph could be rotated on that axis, creating a solid. This is what it looks like if the sphere r=1 was rotated: The point x can be shown as x' when the equation is rotated pi/2 radians on this "z" r-axis.

    I knew I needed to look at spherical sectors since to find the area of polar functions it is necessary to look at circular sectors as such:

    \[\pi*R^2*\frac{\Delta\theta}{2\pi}=\frac{R^2}{2}\De  lta\theta\]<br />

    thus giving rise to the integral:

    \[\int_{a}^{b}\frac{f(\theta)^2}{2} d\theta\]

    So, utilizing a spherical sector (or rather, subtracting that from a half of a sphere) I get the volume of a rotated "slice of pie."

    So:

    Volume of a rotated wedge/slice of pie with polar function r=f(theta) and slice/wedge of angle theta sub 1=
    \frac{2}{3}\pi f(\theta_1)^3 - \frac{2}{3}\pi f(\theta_1)^2 h

    If h is just \[r-rsin(\theta_1)=r(1-sin(\theta_1))\], then we can plug into the equation to find the volume

    \frac{2}{3}\pi f(\theta_1)^3 - \frac{2}{3}\pi f(\theta_1)^3 (1-sin(\theta_1)

    However, here's my dilemma. To find the volume for one rotation of a wedge (or 2 pi radians around the "y" r-axis), I need the volumes of infinitesimally small wedges ( d\theta) rotated, not just of any theta sub 1. If one could help me find how to find that and express it in an integral, that would really help me out.

    Any help would be greatly appreciated. If what I said didn't make sense, please tell me (I wont take it personally), I'll try to post some pictures of what I'm saying.

    Thanks!
    Last edited by progressive; December 19th 2010 at 07:30 PM. Reason: New pictures to better help understand problem.
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