Integral Calculus Questions and Answers – Rectification in Polar and Parametric Forms

Answer: d
Explanation: We know that,
S=\(\int_{x1}^{x2}\sqrt{1+\frac{dy}{dx}^2}\)
\(y^\frac{2}{3}=a^\frac{2}{3}-x^\frac{2}{3}\)
Differentiating on both sides
\(\frac{2}{3} y^{\frac{2}{3}-1}= \frac{-2}{3} x^{\frac{2}{3}-1}\)
\(\frac{dy}{dx} = -\frac{y}{x}^{\frac{1}{3}}\)
\((\frac{dy}{dx})^2 = (\frac{y}{x})^{\frac{1}{3}}\)
\(1+(\frac{dy}{dx})^2=1+(\frac{y}{x})^\frac{2}{3}\)
Substituting from the original equation-
\(1+(\frac{dy}{dx})^2=(\frac{a}{x})\frac{2}{3}\)
\(\sqrt{1+\frac{dy^2}{dx}}=(\frac{a}{x})^{\frac{1}{3}}\)
\(S=\int_{a}^{0}(\frac{a}{x})^{\frac{1}{3}} dx \)
\(s=\frac{-3a}{2}\)
Thus, length of the given curve is \(\frac{-3a}{2}\).

x=a(θ-sinθ)
y=a(1+cosθ)

Answer: c
Explanation: We know that
\(s=\int_{\theta1}^{\theta2}\sqrt{(\frac{dx}{d\theta})^2+(\frac{dy}{d\theta})^2}\)
\(\frac{dx}{d\theta}=a(1-cos\theta)\)
\(\frac{dy}{d\theta}=a(-sin\theta)\)
\((\frac{dx}{d\theta})^2+(\frac{dy}{d\theta})^2=a^2(1-cos\theta)^2+a^2 sin^2\theta\)
\((\frac{dx}{d\theta})^2+(\frac{dy}{d\theta})^2=4a^2 sin^2\frac{\theta}{2}\)
\(s=\int_{0}^{2}\pi\sqrt{4a^2 sin^2\frac{\theta}{2}} d\theta\)
On solving the given integral, we get
s=8a
Thus length of one arc of the given cycloid is 8a.