Answer to Question #206145 in Calculus for KABIR

Question

1. 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

Ans:-

$x=a(\theta-Sin\theta)\ \ \ \ \ \ \ \ \ y=a(1-Cos\theta)\\ Y=0 \Rightarrow 1-Cos\theta=0 \Rightarrow Cos\theta=1 \Rightarrow \theta=0,2\pi\\$

$x=a(\theta-Sin\theta) \Rightarrow \ \dfrac{dx}{d\theta}=a(1-Cos\theta)$

$y=a(1-Cos\theta) \Rightarrow \dfrac{dy}{d\theta}=aSin\theta\\$

$\sqrt{(\dfrac{dy}{d\theta})^2+(\dfrac{dx}{d\theta})^2}= \sqrt{a^2(1-Cos\theta)^2+a^2Sin^2\theta}$

$\sqrt{a^2+a^2Cos^2\theta-2a^2Cos\theta+a^2Sin^2\theta}\\$

$\sqrt{2a^2-2aCos\theta} \ \ \Rightarrow \sqrt2a\sqrt{1-Cos\theta}=2aSin\dfrac{\theta}{2}$

Surface Area=$\int ^{2\pi} _0 2\pi a(1-Cos\theta) \times 2aSin\dfrac{\theta}{2} \ d\theta$

$\int ^{2\pi} _0 2\pi a \times 2Sin^2\dfrac{\theta}{2}\times 2aSin\dfrac{\theta}{2} \ d\theta$

$16\pi a^2 \int ^{2\pi} _0 Sin^3\dfrac{\theta}{2}dt$

$16\pi a^2 \int ^{\pi} _0 2 \times Sin^3x\ dt\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ [Take \dfrac{\theta}{2}=x]$

$32 \pi a^2I_3=32\pi a^2 \times \dfrac{2}{3}\\$

$\dfrac{64\pi a^2}{3} sq.units$