Is My Double Integral Formula for Rotating a Function Around a Circle Correct?

[viciousp]

Homework Statement

This isn't really a homework question but I was wondering if the equation I wrote makes any sense. Basically if you have a function f(x) from x=a to x=b and rotate it around a circle with radius a. At first the height of f(x) is at full height but when its at 2pi radians it has height zero, and as result must decrease at every change in dtheta.

The radius a is not necessarly equal to the a in the interval [a,b] for the function f(x)

Homework Equations

The Attempt at a Solution

I used a double integral as shown below:

\int_{0}^{2\pi }\int_{x=a}^{x=b}(\frac{1}{2}a^{2}(f(x)-\frac{f(x)\theta}{2\pi }))dxd\thetadoes this make any sense?

 

[HallsofIvy]

Homework Statement

This isn't really a homework question but I was wondering if the equation I wrote makes any sense. Basically if you have a function f(x) from x=a to x=b and rotate it around a circle with radius a.

I assume you mean you are rotating the graph of the function, but about what center?

At first the height of f(x) is at full height but when its at 2pi radians it has height zero, and as result must decrease at every change in dtheta.

The radius a is not necessarly equal to the a in the interval [a,b] for the function f(x)

Homework Equations

The Attempt at a Solution

I used a double integral as shown below:

\int_{0}^{2\pi }\int_{x=a}^{x=b}(\frac{1}{2}a^{2}(f(x)-\frac{f(x)\theta}{2\pi }))dxd\theta


does this make any sense?

 

[viciousp]

I was rotating the graph throughout the circle so the width of the function is the length of the radius, and its being rotated in a circular motion from zero radians to 2pi radians.

The easiest way I can explain what I'm doing is if I have a 2x4 rectangle sticking out of the circle base, and at zero radians it will have width 2 and a height of 4, then at pi/2 radians it will have a width of 2 and a height of 3 and so on till its is at 2pi, when it has a height of zero and width of 2.