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:

final part

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 }

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