Why does this way of calculating volume of sphere give result two times smaller than it should be? Integrals/onion method.

Question

I'm high school student, and they never taught us how to use integrals, but I tried to learn it on my own. So I don't understand most of the wikipedia explanations. I'd be great if someone could give me an explanation that doesn't use high level maths, so I can understand it. There are tons of sources on the internet, but I don't understand most of them.

I started by trying to compute area of circle. I started with a one-fourth of a circle, and given the definition of radian, I came up with this:

So for one-fourth of a circle of diameter $R$, it's arc is $\frac{\pi}{2}R$. My reasoning is, that we can get area of this one-fourth of a circle, if we sum it's arc $r$ times, or $\int_{0}^{r} \frac{\pi}{2}R dR$, which correctly evaluates to $\frac{\pi r^{2}}{4}$. Lowercase $r$ is diameter of the target circle that we want to find area of, and uppercase $R$ is generally diameter of the circle I'm measuring (in this case going from $0$ to $r$.)

Please let me know if this is correct or if this is just a coincidence, but it looks like this is a way how to compute it.

Then I tried to extend this method to a sphere, but it stopped working. If we imagine one-eighth of a sphere, one of it's sides is $\frac{\pi r^{2}}{4}$. If we extend the side and sum it $r$ times, we should get volume of one-eighth of a sphere.

$$\int_{0}^{r} R^{2} \frac{\pi}{4} dR = r^3 \frac{\pi}{4 \cdot 3}$$

If we multiply this by $8$, to get volume of whole sphere (as this is one eighth of a sphere), we get $r^{2}\frac{2\pi}{3}$, which is, according to online sources, only one half of a sphere, not a full sphere.

Where did I make a mistake? It should be multiplied by $2$, but why? Did I forget something or is this method just a crap that doesn't work and the area of circle was just a coincidence?

Thanks, Peter

Answer 1

Your method of integration for the circle worked because the area of the quadrant is the sum of the areas of arcs. Now for a sphere, this does not work because the radius does not decrease the same way. What you have calculated is the one-fourth portion of a cone. The correct way to do this would be as follows-

Let the radius of the sphere be $R$. At a particular height $x$, let the radius be $r$. From Pythagoreas theorem, $r^2=R^2-x^2$

$$V={\int_0^R}{{\pi}\over 4}r^2\,dx$$ $$V={{\pi}\over 4}{\int_0^R}(R^2-x^2)\,dx$$ $$V={{\pi}\over 4}\Bigl[R^2x-{x^3\over 3}\Bigl]_0^R$$ $$V={{\pi}\over 4}\Bigl(R^3-{R^3\over 3}-0+0\Bigl)$$ $$V={{\pi}R^3\over 6}$$