Please integrate the square root of (R - x squared) with resect to x, where R is a constant.?

Question

No its not homework, I'm trying to understsand the calculus-based derivation of Pi. By "Pythagoras'" theorm, a circle centred on the origin represents this relationship: y = the function above. By taking strips of the circle from from x = -R to x = +R, the area may be composed. By reducing them towards zero width whilst multiplying towards infinity in number, the area may become accurate. Newtons integration makes the calulation exact. The integrated funtion = pi times R squared, we are told. Can anyone help me to understand how this works please?

Thank you for answering. Why is Pi = 3.14...?

The point of the Q is that I want to derive 3.14 using calculus. I'll use the string for now thanks!

Answer 1

As someone else indicated the equation for a circle is:
x^2 + y^2 = R^2
=>
y = +- sqrt(R^2 - x^2)

You can focus on the positive half of the circle, calculate the area under than curve (i.e. between it and the x-axis) and then double it to get the area of the full circle. In fact simpler still is to focus on the first quadrant and then multiply by 4.
So CircleArea = 4 * Integral { x = 0 to R } sqrt(R^2 - x^2) dx
One way to compute the integral is to make the following change of variables:
x^2 = R^2 cos^2 u
x = R cos u
dx/du = - R sin u
dx = - R sin u du
Substituting:
Integral { u = Pi/2 to 0 } sqrt(R^2 - R^2 cos^2 u) (-R) sin u du
= Integral { u = Pi/2 to 0 } -R^2 sqrt(1 - cos^2 u) sin u du
= Integral { u = Pi/2 to 0 } -R^2 sqrt(sin^2 u) sin u du
= Integral { u = Pi/2 to 0 } -R^2 sin^2 u du
= Integral { u = Pi/2 to 0 } -R^2 ( 1 - cos 2u ) /2 du
= [ - (1/2) R^2 ( u - (1/2) sin 2u) ] (Pi2 to 0)
= [ 0 + Pi/4 R^2]
= Pi/4 R^2
Thus CircleArea = 4 * (Pi/4 R ^2) = Pi R^2 .

Answer 2

Our geometric intuition is that the circumference of a circle is linear in the radius and the area of a circle is quadratic in the radius, i.e., there are positive real constants c and k such that if the radius of the circle is R then:

circumference = c R
area = k R^2

Calculus gets the same result, for example substitute "x = R t" into your integral for the area of the upper half of the circle and you get the area of the upper half circle is R^2 integral { t from -1 to 1 } sqrt(1 - t^2) dt and so the constant k = 2 integral { t from -1 to 1 } sqrt(1 - t^2) dt

Now let us find the area of the circle by adding up a bunch of rings "of width dx". The area of the circle between radius x and radius dx is essentially the width dx times the circumference, which at that radius is c x. So the area of the circle is

integral (x from 0 to R) (c x) dx = (1/2) c x^2 ] hi:R lo:0
= (1/2) c R^2 - (1/2) c 0^2
= (1/2) c R^2

But the area is k R^2 using that other constant k, so we see k = (1/2)c.

Just rename k to pi, so that the area is pi R^2 and the circumference is 2 pi R or pi times diameter.

To find the numerical value, just numerically estimate

pi = k = 2 integral { t from -1 to 1 } sqrt(1 - t^2) dt

Answer 3

all calculus describes is the area under a curve, thats the whole premise, Newton invented calculus in order to relate his laws and 50 years or so of work, its whole basis is on something infinitely small or uncalculably small, all he did was invent a system of solving anyone in theory can invent a mathematical subject if one has results and needs to invent some form to make them beleivable i.e. calculus, this in essence is formalism, to understand anything in life, find out why it exists in terms of who we are and how it can be applied philosophically and you will immediately began to enjoy learning it,

O ok, so divide the circle into 5 sements or 5 triangles now do this like 901 hehe times on another circle, then divide it to infinty, now for the first circle put all the 5 triangle up and down and you will see a square with rounded top edges, now do this to a circle with infinte number of triangles and you will see that it will see thatt the rectangle's top flattens out as you approach infinity, that is calculus, nothing more or less everything is just a play on geometry in every mathematical theory, that is why pi is a number without end because 3.14......every area of a circle will always be uncertain becaause pi is not exact in essence so inr eality all you can do is approach infinity to get something so accurate that exceeds the limit of for instance 10^45, which happens to be the amount of atoms in all the universe, any number beyond that implies philosophical exploration where one asks why even bother having a number beyond the range of existence,,,,, mathematicians thus cause they realize there inherent frustation start inventing different geomatrical theories because they love math and deep down inside hope that there therom may provide work for someone in 60 years looking for a mathematical description of an idea,, thats where string theory came into play, all they did was be imaginitive and they stumbled upon a theorem of some italian in the 60 who had no applicable use to it!,

Answer 4

They already answered before i have a chance to answer.

As to your second question, Pi is 3.1416... because:

This is a calculated value. It's just the ammount of times the diameter can go into the cercumference of the circle. It has been calculated through many trial tests. Try it at home with a string the size of the diameter and see how many times it goes into the cercumference.

Answer 5

equation of the cirle is:
y^2+x^2=R^2

which means that
y = sqrt(R^2-x^2)
as well as y = - sqrt(R^2-x^2)

so this is what you should be integrating (but I do not know how :) )

Answer 6

no i dont know how any of that works because i havent learned any of that yet but you can look up that kind of stuff on the internet in differnt homework-help websites

Answer 7

This link explains it perfectly, using both polar co-ordinates and "strips":

http://mathworld.wolfram.com/Circle.html

Answer 8

sqrt(R - x^2) dx
You have to use trig substitution.
Let x = sqrt(R)*sin(u), then dx = sqrt(R)*cos(u) du
sqrt(R - R(sinu)^2) sqrt(R)*cos(u) du
= sqrt(R)sqrt(1-(sinu)^2)*sqrt(R)*cos(u) du
= R*cos(u)*cos(u) du
= R*(cosu)^2 du
Use the trig identity
(cosu)^2 = (1 - cos(2u))/2
R*(cosu)^2 du
= R(1-cos(2u))/2 du
= R(u - (1/2)sin(2u))/2 + C
= R[(1/2)u - (1/4)sin(2u)] + C
Now,
x = sqrt(R)sin(u)
therefore,
sin(u) = x/sqrt(R)
cos(u) = sqrt(R-x^2)/sqrt(R)
sin(2u) = 2sin(u)cos(u)
sin(2u) = 2(x/sqrt(R))(sqrt(R-x^2)/sqrt(R))
= 2x*sqrt(R-x^2)/R
R[(1/2)u - (1/4)sin(2u)] + C
= R[(1/2)arcsin(x/sqrt(R)) - (1/4)2x*sqrt(R-x^2)/R]
= R[(1/2)arcsin(x/sqrt(R)) - (1/2)x*sqrt(R-x^2)/R]
= (1/2)sqrt(R)arcsin(x/sqrt(R)) - (1/2)sqrt(R-x^2)
Plug in your limits and you should have your answer.