Volume of a Sphere: Calculation & Explanation

[mjordan2nd]

I was just thinking about how you would calculate the volume of a sphere last night, and tried it thusly, using the logic that a sphere is nothing but a bunch of infinitesimally thin circles with increasing radius from 0 to r, and then from decreasing radius r to 0:

\int^{R}_{0} \pi r^{2}dr= \frac{1}{3} \pi R^{3}

This, I thought should be one hemisphere, so I multiplied the result by 2, getting \frac{2}{3} \pi R^{3}. This, of course, is not the volume of a sphere. Why did my method not work?

 

[waht]

The surface area of a sphere is 4(pi)r^2

 

[mathwonk]

you have the radii growing linearly, so you are getting (twice) the volume of a right circular cone of base radius R and height R.

it seems you should integrate something like pi(R^2 - y^2) as y grows from 0 to R. and then double it. i.e. the thin disc at height y has radius sqrt(R^2-y^2) by pythagoras.by the way euclid also used the word "circle" to refer both to the one dimensional circumference and the 2 dimensional interior of a circle, so the flap about that usage is historically unjustified.

 

[bob1182006]

What you calculated was the area below the function \pi r^2 above some axis.

I don't think you can find the volume of a sphere in under 3 integrals since you need to account for the change in x/y/z.

The integral will be \int_0^{2\pi}d\theta \int_0^{\pi}sin\phi d\phi \int_0^R r^2 dr

Although it's a tripple integral since none of the limits of one variable depend on another variable you can treat it as the multiple of 3 integrals.
\int_0^{2\pi}d\theta
\int_0^{\pi}sin\phi d\phi
\int_0^R r^2 dr
After you multiply the results you should get the volume of the sphere \frac{4}{3}\pi R^3

 

[dst]

But clearly the starting integral was giving the area of a quadrant of the sphere?

 

[arildno]

Your proper integral should be, for the hemishere:
\int_{0}^{R}\pi{r}(z)^{2}dz
where z is the variable dsenoting at which height along the axis you are, while r(z) is the radius of the disk at that height.