Find the surface area generated by revolving one arch of the cycloid about the x-axis. Formula: $\displaystyle 2pi\int_{a}^{b}{y(t)}\sqrt{{x'(t)^2}+{y'(t)^2}} dt$ Thanks.
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Originally Posted by jffyx Find the surface area generated by revolving one arch of the cycloid about the x-axis. Formula: $\displaystyle 2pi\int_{a}^{b}{y(t)}\sqrt{{x'(t)^2}+{y'(t)^2}} dt$ Thanks. One arch is parametrised by $\displaystyle 0 \leq t \leq 2 \pi$. After doing the necessary differentiations, substituting and simplifying you get $\displaystyle S = 2 \sqrt{2} \pi \int_0^{2 \pi} (1 - \cos t)^{3/2} \, dt$. Your task now is to solve this integral.
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