- #1

physlosopher

- 30

- 3

$$\int_0^{2\pi} R sin(\theta)\, d\phi = 2\pi R sin(\theta).$$

This is meant to illustrate that the surface is curved, because the circumference is less than ##2\pi R##. But isn't this the wrong R? The R in the integral is the distance from the origin of the coordinate system; the radius of the circle that's meant to illustrate the curvature would have to be the distance from the center of the circle as measured on the surface of the sphere itself - so the point where the sphere intersects the z-axis. (Which should be ##R\theta##, no?) Doesn't the integral (without comparing it to ##R\theta##) just illustrate that the circle is smaller by ##sin(\theta)## than the circle lying in the xy-plane, rather than illustrating curvature? Is this a flaw in the example, or am I overlooking something?