Find the length of curve r=1-cos (tita)

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find the length of curve r=1-cos (tita)

pls help to formulat an eqn for f(x,y)

thanx
 
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Hello teng,

the curve is given in polar coordinates.

\theta \, \mapsto \, (r(\theta)\cos(\theta),r(\theta)\sin(\theta))

with \theta_1\leq\theta\leq\theta_2

You can make use of the appropriate formula to calculate the length L.

L=\int_{\theta_1}^{\theta_2}\sqrt{(r'(\theta))^2+r^2(\theta)} \, d\theta

Regards,

nazzard
 
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Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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