Answer to Question #204651 in Electrical Engineering for Kabir

Question

• Find the area of the surface of the solid formed by the revolution the curve, x=a(θ - sinθ) and Y=a(1- cosθ) about x-axis (y=0).

Answer 1

$A = 2\pi\int_{\alpha}^{\beta}x(θ) \sqrt{[x′(θ)]2+[y′)]2dθ}$

$Parameter\space θ \space varies\space from\space 0 \space to \space2π$

Derivatives:

$x ′ ( θ ) = ( θ − sin θ ) ′ = 1 − cos θ ,$

$y ′ ( θ ) = ( 1 − cos θ ) ′ = sin θ ,$

$[x′(θ)]^2+[y′(θ)]^2=(1−cosθ)^2+sin^2θ=1−2cosθ+cos^2 \theta+sin^2 \theta$

$=2-2cos \theta=4 sin ^2 \frac{\theta}{2}$

$A=2\pi\int_0^{2x} [(\theta-sin \theta).2sin \frac{\theta}{2}]d\theta=4 \pi[\int_0^{2x} \theta sin \frac{\theta }{2} d \theta- \int_0^{2x} sin \theta sin \frac{\theta}{2} d\theta]$

$I_2=\int_0^{2x} sin \theta sin \frac{\theta}{2}d \theta=\frac{4}{3} sin^3 \frac{\theta}{2} |_0^{2x}=0$

The area of the surface is ;

$=4x[I_1-I_2]$