Prove that volume of a sphere with radius $r$ is $V=\frac43r^3\pi$

Question

Prove that volume of a sphere with radius $r$ is $V=\frac43r^3\pi$.

I know to prove this in following way:
If I rotate graph of the function $y=\sqrt{r^2-x^2}$ around $x$-axis, it will result a sphere with radius $r$. So, it can be computed as $$V=\int_{-r}^ry^2\pi dx=\int_{-r}^r(r^2-x^2)\pi dx=\frac43r^3\pi$$ My question is: can we prove it without rotating anything? Can we just integrate a formula for sphere? From formula I get $$ x^2+y^2+z^2=r^2\\ z=\pm\sqrt{r^2-x^2-y^2} $$ Can we get a volume as $$2\int_{-r}^r\int_{-r}^r\sqrt{r^2-x^2-y^2}\,dx\,dy$$ I tried to integrate this, but it looks very complicated.

Answer 1

The limits are wrong. It should be

$$2\int^r_{-r}\int^{\sqrt{r^2-x^2}}_{-\sqrt{r^2-x^2}}\sqrt{r^2-x^2-y^2}\,dy\,dx$$

It is easier to use spherical coordinates though:

$$\int^R_0 \int^{2\pi}_0\int^{\pi}_0r^2\sin{\theta}\,d\theta \,d\phi \,dr$$

Answer 2

Complete re-edit:

What you had described integration over a square in the plane.

Since KittyL beat me to it with the right limits, let me point out that by far the most efficient way to get the volume of a sphere is to do a one-variable integration of the cross-sectional area of the sphere when you cut it with a vertical plane through the point $(x,0,0)$. So the integral becomes $$ \int_{-r}^rA(x)dx\,, $$ where $A(x)$ is the aforementioned area, that of a circle of radius $\sqrt{r^2-x^2}$.

Answer 3

$x^2 + y^2 + z^2 = R^2$ is the equation of a sphere of radius $R$, centered at the origin. Its volume is given by:

$$V = \int_{-R}^R\int_{-\sqrt{R^2 - x^2}}^{\sqrt{R^2 - x^2}}\int_{-\sqrt{R^2 - x^2 - y^2}}^{\sqrt{R^2 - x^2 - y^2}}dzdydx$$

That's because the projection of the sphere on the $xy$ plane is the circle centered at the origin, having radius $R$. Its equation is $x^2 + y^2 = R^2$. $x$ moves between $-R$ and $R$ while $y$ moves between $-\sqrt{R^2 - x^2}$ and $\sqrt{R^2 - x^2}$. This is obtained from the equation of the circle, after taking the domain to be regular in $x$ (We could have things reversed if we take the domain to be regular in $y$). Finally, $z$'s behaviour is obtained from the equation of the sphere.

Answer 4

That way works if you get the bounds right. You have $$ \int_{-r}^r \Big( \cdots \Big) \, dy $$ and then for any particular $y$ between $\pm r$ you have $x$ running between $\pm\sqrt{r^2-y^2}$, so it's $$ \int_{-r}^r \left(\int_{-\sqrt{r^2-y^2}}^{\sqrt{r^2-y^2}} \cdots \,dx\right) \, dy $$ and so on.

If you then let $\sqrt{r^2-y^2}\,\sin\alpha=x$, so that $dx=\sqrt{r^2-y^2}\,\cos\alpha\,d\alpha$, then you get $\sqrt{r^2-y^2-x^2}$ $= \sqrt{r^2-y^2}\sqrt{1-\sin^2\alpha}$ $= \sqrt{r^2-y^2}\,\cos\alpha$, etc., and the inside integral becomes $\displaystyle\int_{-\pi/2}^{\pi/2}$.

And so on.

That's one of many ways.

Answer 5

Archimedes argued that with respect to volume, a sphere is equivalent to a cone whose base is equal to the surface area of the sphere, and whose altitude is equal to the radius.