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

This set of Differential and Integral Calculus Multiple Choice Questions & Answers (MCQs) focuses on “Rectification in Polar and Parametric Forms”.

1. Find the length of the curve given by the equation.
$$x^{\frac{2}{3}}+y^\frac{2}{3}=a^\frac{2}{3}$$
a) $$\frac{3a}{2}$$
b) $$\frac{-7a}{2}$$
c) $$\frac{-3a}{4}$$
d) $$\frac{-3a}{2}$$

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 $$\frac{-3a}{2}$$.

2. Find the length of one arc of the given cycloid.

x=a(θ-sinθ)
y=a(1+cosθ)


a) a
b) 4a
c) 8a
d) 2a

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.

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