Answer to Question #115710 in Calculus for Navjot

Question #115710
Find the volume of solid obtained by revolving one arch of cycloid x=a(theta+sin theta) and y=a(1-cos theta)
1
Expert's answer
2020-05-14T17:42:20-0400

The line of cycloids is set parametrically. When it comes to the first arch, then there is a limitation:

0<0<2π0<0<2\pi and then the first arch is drawn when the parameter value changes within: 0t2π0\le t \le 2\pi. We use the formula for finding the body volume formed by a line given parametrically: V=πabf2(t)x(t)dtV=\pi \int_a^bf^2(t)x\prime(t)dt. Substitute ve into the formula and get: V=π02π(a(1cosθ)(a(θ+sinθ)dθ=π02π(a(1cosθ)(θ+cosθ)dθ=π02π(aθacos2θ+acosθaxcosθ)dθ=aπ02π(12cos2θ+12)dθaπ02πθcosθdθ+aπ02πcosθdθ+aπ02πθdθ=π2(2π1)aV=\pi\int^{2\pi}_0(a(1-\cos{\theta})(a(\theta+\sin{\theta})^\prime d\theta=\pi\int^{2\pi}_0(a(1-\cos{\theta})(\theta+\cos{\theta})d\theta=\pi\int^{2\pi}_0(a\theta-a\cos^2{\theta}+a\cos{\theta}- ax\cos{\theta)}d\theta=-a\pi\int^{2\pi}_0(\frac{1}{2}\cos{2\theta}+\frac{1}{2})d\theta-a\pi\int^{2\pi}_0\theta\cos{\theta}d\theta+a\pi\int^{2\pi}_0\cos{\theta}d\theta+a\pi\int^{2\pi}_0\theta d\theta=\pi^2(2\pi-1)a


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