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Wednesday, April 17, 2013

 

Finding \(\pi r\) via integral

In general, the length of a differentiable function can be computed using the integral:
  \[ \begin{aligned} \displaystyle \int_a^b \sqrt{1+f'(x)^2} \, dx \\ \end{aligned} \]
To find the length of a semi-circle, we can start with:
  \[ \begin{aligned} x^2 + y^2 &= r^2 \\ \end{aligned} \]
which leads to:
  \[ \begin{aligned} & \displaystyle \int_{-r}^r \sqrt{1+\frac{x^2}{r^2-x^2}} \, dx \\ &= \displaystyle r \int_{-\pi/2}^{\pi/2} \cos\theta \sqrt{1+\frac{r^2\sin^2\theta}{r^2-r^2\sin^2\theta}} \, d\theta \\ &= \displaystyle r \int_{-\pi/2}^{\pi/2} \cos\theta \sec \theta \, d\theta \\ &= \pi r \end{aligned} \]
\(\Box\)


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