trying to derive area of circle....?

Question

This is not homework just out of general intrest.
I figured a cirle is just a shape with a constant radius and a side that is always perpendicular to its centre. So i figured you could break it down into tiny triangles and take the sum of the infinatly small integrals, ie integrate. But i am integrating 1/2 f(x) .dx And thats clearly not going to get the answer.
It seems logical in my head but it doesnt take into account pi, so if its the wrong approach how do you derive the area of a circle.

Answer 1

Consider a circle of radius r
Within the circle there is a concentric strip of radius x and width δx
Area of this strip is (2πx)(δx)
Area of circle = A (sum of strips)
A = ∫ 2 π x dx between limits 0 to r
A = 2 π x ² / 2 between 0 and r
A = π r ² , the area of a circle.

Answer 2

You have the right idea but just need to go a little bit further in analyzing what your variables mean.

try this. f(x) = r (the radius of the circle)
dx = rdT where T is the angle in radians between the current r and the x-axis. This is ok since dx is very small and approximately equal to the arc length. So;

(1/2) x dx = (1/2) r (rdT) = (1/2)r^2 dT

Integrate from T = 0 to 2pi

(1/2)r^2[2pi - 0] = pi r^2

Answer 3

Eqn of the circle with radius r and centre at the origin is;

x²+y²= r²

y =± √(r²-x²)

Since the eqn contains only even powers of x and y it is symmetric about both the axes

In first quadrant;
y=√(r²-x²)

Area of circle=4 * area in 1 quadrant

=4 integral(0 to r) ydx

=4 int (0 to r) √(r²-x²)dx

=4 [(x/2) √(r²-x²)+ (r²/2)sin^-1 (x/r)] (0 to r)

=4 [0+(r²/2)(π /2)]

=π r²

Answer 4

The seqment of a circle is just like a triangle with angle(dtheta) as rhe vertex at the centre of the circle. now 3sides of the triangles are r (=radius), another r and r(dtheta) where theta is in Radians

area of triangle is 1/2 base x height
base = r(dtheta)
height = r.....take it as r only because triangle is tiny segment

then area = 1/2 x r x r(dtheta)
Now Integarte (hilimit is 2pi radians & low limit is 0 radians)
=1/2 r^2 x integrate( dtheta)
=1/2 r^2 x theta
= 1/2r^2 (2pi-0)
=Pi r^2

Answer 5

The area of a triangle = 1/2 ab sinC

So if you have broken the circle into N triangle then the total area = N(r^2)/2 sin 2pi/N (a=b=r the radius)
as N becomes large then sin (2pi/N) --> 2pi/N
Substitute

Area = N(r^2)/2 * 2pi/N = pi*r^2

Answer 6

Actually, there is a technique that gets you there. It requires intense computation.