# Prove that the area of half of the unit circle is pi/2.

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### 4 Answers

Prove that the area of half of the unit circle is `pi/2` :

Consider the unit circle; centered at the origin with radius 1. We compute the area above the x-axis and beneath the circle. This is the semicircle above the x-axis.

The area is given by `int_(-1)^1 sqrt(1-x^2)dx`

(The curve extends from -1 to 1; the coordinates of a point on the semicircle are (x,y) where `x^2+y^2=1==>y=sqrt(1-x^2)` )

So we evaluate `int_(-1)^1 sqrt(1-x^2)dx`

Let `x=sintheta,dx=costheta d theta` . The limits of integration become `-pi/2,pi/2` so we have an equivalent integral:

`` `int_(-pi/2)^(pi/2)sqrt(1-sin^2theta)costheta d theta`

`=int_(-pi/2)^(pi/2) cos^2theta d theta` Note that `cos^2theta=1/2cos(2theta)+1/2`

`=1/2 int_(-pi/2)^(pi/2)(cos(2theta)+1)d theta`

`=1/2[theta+sin(2theta)|_(-pi/2)^(pi/2)]`

`=1/2[pi/2-(-pi/2)]`

`=1/2[pi]` as required.

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You need to remember the equation that helps you to evaluate the area of the circle such that:

`A = pi*r^2`

r represents the radius of circle

You also need to remember that the radius of the unit circle is of 1 unit, hence, evaluating the area of the unit circle yields:

`A = pi*1^2 = pi`

You need to evaluate the area of half of unit circle,a, hence, you need to divide the area of the unit circle by `2` :

`a = pi/2`

**Hence, evaluating the area of half of unit circle yields `a = pi/2.` **

To find the area of half the unit circle, this is the same as finding the integral of

`int_{-1}^1sqrt{1-x^2}dx` now let `x=sin u` so `dx=cosu du`

`=2int_{x=0}^{x=1}sqrt{1-sin^2u}cosudu` simplify with pythagorean identity

`=2int_0^{pi/2}cosu cos u du`

`=2int_0^{pi/2}cos^2udu` now use identity `cos^2u={1+cos2u}/2`

`=int_0^{pi/2}(1+cos2u)du`

`=(u+1/2sin2u)_0^{pi/2}`

`=pi/2`

**The area of the half unit circle is `pi/2` .**

Unit circle means, radius = 1

area element of a circle is given by:

dA = r d(theta) dr

So to get the area we must integrate "dA" with "r" varying from 0 to 1 and (theta) from 0 to Pi for half circle.

So,

area of half circle (A) = Integrate [ r d(theta) dr ] (r from 0 to 1)

(theta from 0 to Pi)

= Integrate[r dr] 0 to 1

* Integrate[ d(theta) ] 0 to Pi

= (r^2 /2) 0 to 1

* (theta) 0 to Pi

= Pi/2