Volumes in the 4th spatial dimension
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Mar3-05, 01:31 AM
Now how could you calculate the surface area of a sphere? If you get a basket ball or something you can see that the surface area of a sphere is the infinite sum of circles which starting from one pole of the surface of the sphere, get bigger, until one reaches the equator then shrink back to zero radius at the other pole. Assuming your sphere has radius 1, you'll find the circumference of your circle r units away from a pole is 2*Pi*sin(r). Integrate that between 0 and Pi and you'll get 4*Pi, which is the surface area of your sphere. Since the surface area of a sphere of radius R has units R^2, then the Surface area of a general sphere of radius R is 4*Pi*R^2.
I'm already lost here, lol. How did you get 4*pi when you integrate 2*pi*sin(r)? Shouldn't it be 2*pi*(-cos r) if you take the antiderivative?
And what do you mean by "integrate between 0 and pi)?