I Deriving the volume of a sphere using semi-circles

[FeynmanFtw]

Apologies if this question has already been asked, but is it not possible to derive the formula of a sphere by imagining a circle sliced in two, then rotating this semi-circle about the flat side (imagine the flat side is stuck to a skewer) by 2Pi, so as to sum up the semi-circles to "create" the sphere?

I mean, by using the area of the semi-circle and using that to create an infinitesimally thin wedge, and summing those, surely one should arrive at the volume expression? For some reason I'm having trouble formulating the maths, unless this is not the way to go about it?

TIA

 

[BvU]

Hi FF,
You mean the expression for the volume ?
What would be the volume of such a thin wedge ?
(expressed in the angle you rotate over)

 

[julian]

Apologies if this question has already been asked, but is it not possible to derive the formula of a sphere by imagining a circle sliced in two, then rotating this semi-circle about the flat side (imagine the flat side is stuck to a skewer) by 2Pi, so as to sum up the semi-circles to "create" the sphere?

I mean, by using the area of the semi-circle and using that to create an infinitesimally thin wedge, and summing those, surely one should arrive at the volume expression? For some reason I'm having trouble formulating the maths, unless this is not the way to go about it?

TIA

Theorem of Pappus: "If a plane area revolves about an axis in its plane not intersecting it, the volume of revolution is equal to the area multiplied by the lenghth of the path of its centre of mass."

Proof:

Say we are rotating about the $x-$axis, then the centre of mass is the distance given by

$\bar{y} = lim_{\delta A \rightarrow 0} \dfrac{\sum y \delta A}{\sum \delta A}$

i.e.

$\bar{y} = \dfrac{\int y dA}{A}$

where the integral extends over the whole area.

But the volume swept out by $\delta A$ in a whole revolution $= 2 \pi y \delta A$. As such

Volume of revolution $= 2 \pi \int y d A = 2 \pi \bar{y} A$
= Area $\times$ length of path of centre of mass.

In order to find the centre of mass of a semi-circle use:

$\delta A = 2 r \cos \theta \delta y$ and $y = r \sin \theta$.

The answer you should get is $\bar{y} = 4r / 3 \pi$ on the radius of symmetry ($x=0$). And hence the volume of revolution is:

${1 \over 2} \pi r^2 \times 2 \pi \cdot {4r \over 3 \pi} = {4 \over 3} \pi r^3$.

 

[FeynmanFtw]

Theorem of Pappus: "If a plane area revolves about an axis in its plane not intersecting it, the volume of revolution is equal to the area multiplied by the lenghth of the path of its centre of mass."

Proof:

Say we are rotating about the $x-$axis, then the centre of mass is the distance given by

$\bar{y} = lim_{\delta A \rightarrow 0} \dfrac{\sum y \delta A}{\sum \delta A}$

i.e.

$\bar{y} = \dfrac{\int y dA}{A}$

where the integral extends over the whole area.

But the volume swept out by $\delta A$ in a whole revolution $= 2 \pi y \delta A$. As such

Volume of revolution $= 2 \pi \int y d A = 2 \pi \bar{y} A$
= Area $\times$ length of path of centre of mass.

In order to find the centre of mass of a semi-circle use:

$\delta A = 2 r \cos \theta \delta y$ and $y = r \sin \theta$.

The answer you should get is $\bar{y} = 4r / 3 \pi$ on the radius of symmetry ($x=0$). And hence the volume of revolution is:

${1 \over 2} \pi r^2 \times 2 \pi \cdot {4r \over 3 \pi} = {4 \over 3} \pi r^3$.

Thank you! This is brilliant!

 

[julian]

The other theorem of Pappus applies to the area of surface of revolution of an arc:

Area of surface of revolution = Length of arc $\times$ length of path of the centre of mass of the arc.

Here's a question for you which can easily be answered by using the theorems of Pappus:

A grove of semi-circular section, radius $b$, is cut round a cylinder of radius $a$, find the volume removed and the area of surface of the groove.

 

[FeynmanFtw]

I'll give it a go at some point over the holidays, but as I'm incredibly busy with Xmas preparations and having family over...