Area of a Circle A=\pi r^2 using integration
To find the area of the circle we simply integrate to find the area between x=0 and x=r and multiply by 4.
Area A=4\int^r_0 \sqrt{r^2-x^2}\,dx
Using the substitution x=r \sin\theta gives dx=r \cos\theta and the integral converts to
\displaystyle\begin{array}{ll} A&=\displaystyle 4\int^{\pi/2}_0 \sqrt{r^2-r^2\sin^2\theta}.r\cos\theta \,d\theta\\ &=\displaystyle 4\int^{\pi/2}_0 \sqrt{r^2(1-\sin^2\theta)}.r\cos\theta \,d\theta\\ &=\displaystyle4\int^{\pi/2}_0 \sqrt{r^2\cos^2\theta}.r\cos\theta \,d\theta\\ &=\displaystyle 4\int^{\pi/2}_0 r\cos\theta.r\cos\theta \,d\theta\\ &=\displaystyle 4\int^{\pi/2}_0 r^2\cos^2\theta \,d\theta\\ &=\displaystyle 4r^2\int^{\pi/2}_0 {1+\cos 2\theta\over 2}\, d\theta\\ &=\displaystyle 2r^2\left. \left( \theta+{\sin 2\theta\over 2}\right)\right|^{\pi/2}_0\\ &=\displaystyle 2r^2\left( {\pi\over 2} +{\sin\pi\over 2}-0- {\sin 0\over 2}\right)\\ &=\displaystyle\pi r^2 \end{array}
No comments:
Post a Comment