1. Feb 1, 2006

### O.J.

I have always aked myself the question WHY does the area of a circle calculated by pi x r^2. Why is that? Can anyone illustrate it to me please?

2. Feb 1, 2006

### TheDestroyer

When you study Integration you will see that we devide any shape to a very small pieces and then take the some bounded by any function to get the volume or area and the result should be pi r^2, any wise try using mathematica which can teach you too many new things, or wait until you study calculus where this will be very clear for you

Last edited: Feb 1, 2006
3. Feb 1, 2006

### O.J.

i hav studied calculus in my Core math 1 (AS level). but there's more calculus to come in C2 so mayb it'll get clear later...

4. Feb 1, 2006

### JasonRox

You should get it in Calculus 2.

The best way to look at it for now is that it is the ratio of the circumference to the diameter.

Note: I believe that's right.

Note: This is why they used 3 in Ancient times because it was just about the ratio that came up all the time. Try it yourself using a string to calculate the circumference (you can use the formula, but then that ruins the purpose of not being dependent on Pi).

5. Feb 1, 2006

### arildno

Here's essentially Archimedes' argument:

1. $\pi$ is defined as the ratio between a circle's circumference P and its diameter D, and the circle's radius r satisfies D=2r
Thus, we have $P=2\pi{r}$

2. Now, draw N identical triangles in the following manner:
Let the apex of all triangles be the circle's centre, whereas the base of each triangle is the line segment between two points on the circle's circumference.
Thus, you will construct an N-gon whose circumference is approximately equal to the circle's circumference once N is a really big number.
(That is, the base of each triangle will be approximately P/N)

3. The height of each triangle is approximately equal to the circle's radius r, and once N is really big, even more so.

4. Thus, the area of each triangle is approximately (P*r)/(2*N), whereas the N-gon's area is $N*(P*r)(2*N)=P*r/2=\pi{r}*r$
As N goes to infinity, the area of the N-gon is indistinguishable from that of the circle, that is, $\pi{r}*r$ must be the area of the circle as well.

6. Feb 1, 2006

### O.J.

thanx mate. that was one beautiful argument. now i feel i have a much better grasp of it....

7. Feb 5, 2006

### robert Ihnot

arildo has shown that this argument was figured out much earlier than the calculus of Newton, though he presents a similar type of argument, exhausting the circle. This can be seen trigonometrically. We divide the circle into 2pi/N and then use the half angle formula to find the area of each segment:

$$\frac{R^2}{2}cos(\alpha)sin(\alpha)$$ where $$\alpha=\frac{\pi}{N}$$.

Then summing and reusing the half angle formula giving us:
$$\frac{NR^2}{2}sin(2\pi/N)$$. Now we put 1/N in the denominator, use L'Hopital's Rule, and take the derivative as N goes to infinity. (Or to look at it in a simpler way, forgoing rigor, the sin(x) goes to x in radients as x becomes small.)

The matter was handled by Euclid: Circles are to one another as the squares on their diameters. http://aleph0.clarku.edu/~djoyce/java/elements/bookXII/propXII2.html where it is explained that this also involves The Principal of Exhaustion.

Last edited: Feb 5, 2006
8. Feb 5, 2006

### mathwonk

just as a circle is a triangle with height equal to r, and base equal to C, so the area is (1/2)Cr = pi r^2, so also a sphere is a cone with height r and base equal to its surface area A, so the volume is
(1/3)Ar = (1/3)(4pi r^2)r = (4/3)pi r^3.

what do you suppose is the connection between the volume of a 4-sphere and its "surface area"