# Integration Problem – Finding the area of quarter of a circle by integrating…

Integrate:

$\int _{ 0 }^{ r }{ \sqrt { { r }^{ 2 }-{ x }^{ 2 } } } dx$

Say that:

$x=r\sin { \theta }$

So…

$x=r\sin { \theta } \\ \\ x=ru\\ \\ \frac { dx }{ du } =r\\ \\ u=\sin { \theta } \\ \\ \frac { du }{ d\theta } =\cos { \theta } \\ \\ \therefore \quad \frac { dx }{ d\theta } =r\cos { \theta } \\ \\ \therefore \quad dx=r\cos { \theta } d\theta$

Now…

$When\quad x=r,\\ \\ r\sin { \theta =r } \\ \\ \sin { \theta } =1\\ \\ \therefore \quad \theta =\frac { \pi }{ 2 } \\ \\ As\quad \left\{ 0\le \theta \le \frac { \pi }{ 2 } \right\} \\ \\ When\quad x=0,\\ \\ r\sin { \theta } =0\\ \\ \sin { \theta } =0\\ \\ \therefore \quad \theta =0$

So you now have to integrate:

$\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \sqrt { { r }^{ 2 }-{ { r }^{ 2 }\sin ^{ 2 }{ \theta } } } } \cdot r\cos { \theta } d\theta \\ \\ =\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \sqrt { { r }^{ 2 }\left( 1-\sin ^{ 2 }{ \theta } \right) } } \cdot r\cos { \theta } d\theta \\ \\ =\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \sqrt { { r }^{ 2 }\cos ^{ 2 }{ \theta } } } \cdot r\cos { \theta } d\theta \\ \\ =\int _{ 0 }^{ \frac { \pi }{ 2 } }{ { \left( { r }^{ 2 } \right) }^{ \frac { 1 }{ 2 } } } { \left( \cos ^{ 2 }{ \theta } \right) }^{ \frac { 1 }{ 2 } }\cdot r\cos { \theta } d\theta \\ \\ =\int _{ o }^{ \frac { \pi }{ 2 } }{ r\cos { \theta } } \cdot r\cos { \theta } d\theta \\ \\ =\int _{ 0 }^{ \frac { \pi }{ 2 } }{ { r }^{ 2 } } \cos ^{ 2 }{ \theta } d\theta$

But first, realise that:

$\cos { \left( \theta +\theta \right) } =\cos { \theta } \cos { \theta } -\sin { \theta } \sin { \theta } \\ \\ \cos { 2\theta } =\cos ^{ 2 }{ \theta } -\sin ^{ 2 }{ \theta } \\ \\ \cos { 2\theta } =\cos ^{ 2 }{ \theta } -\left( 1-\cos ^{ 2 }{ \theta } \right) \\ \\ \cos { 2\theta } =\cos ^{ 2 }{ \theta } -1+\cos ^{ 2 }{ \theta } \\ \\ \cos { 2\theta } =2\cos ^{ 2 }{ \theta } -1\\ \\ 2\cos ^{ 2 }{ \theta } =\cos { 2\theta } +1\\ \\ \cos ^{ 2 }{ \theta } =\frac { 1 }{ 2 } \cos { 2\theta } +\frac { 1 }{ 2 }$

So you now integrate:

$=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ { r }^{ 2 } } \left( \frac { 1 }{ 2 } \cos { 2\theta } +\frac { 1 }{ 2 } \right) d\theta \\ \\ =\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { 1 }{ 2 } } { r }^{ 2 }\cos { 2\theta } +\frac { 1 }{ 2 } { r }^{ 2 }\quad d\theta$

And:

$If\quad p=\frac { 1 }{ 4 } { r }^{ 2 }\sin { 2\theta } =\frac { 1 }{ 4 } { r }^{ 2 }u\\ \\ \frac { dp }{ du } =\frac { 1 }{ 4 } { r }^{ 2 }\\ \\ u=\sin { 2\theta } =\sin { q } \\ \\ \frac { du }{ dq } =\cos { q } =\cos { 2\theta } \\ \\ q=2\theta ,\quad \frac { dq }{ d\theta } =2\\ \\ \frac { du }{ d\theta } =2\cos { 2\theta } \\ \\ \therefore \quad \frac { dp }{ d\theta } =\frac { 1 }{ 4 } { r }^{ 2 }\cdot 2\cos { 2\theta } \\ \\ =\frac { 1 }{ 2 } { r }^{ 2 }\cos { 2\theta }$

Therefore:

Now, to get the area of a circle, you multiply the final result by 4.

$4\cdot \frac { 1 }{ 4 } \pi { r }^{ 2 }\\ \\ =\pi { r }^{ 2 }\\ \\ \therefore \quad A=\pi { r }^{ 2 }$