Area Under Polar Curves

Area Under Polar Curves

In this tutorial we will learn how to find the area enclosed by a curve represented in the polar form (area under polar curves).

Let r = f(θ) be the equation of a curve y = f(x) in polar (r, θ) form. where f(θ) is a single valued continuous function of (θ), then the area of the sector enclosed by the curve r = f(θ) and the two radii vectors θ = θ1 and θ = θ2 (where θ2 > θ1), is given as

Area\left( A\right) =\dfrac{1}{2}\int ^{\theta _{2}}_{\theta _{1}}r^{2}d\theta

Now take a look at the solved examples given below to understand the method/formula of finding Area Under Polar Curves we discussed right above.

Example: Compute the area of the region bounded by the curve r = a(cos2θ)1/2.
Solution: Consider the fig. 1.1 given in the thumbnail. The radius vector will describe the one fourth of the required area if ‘θ’ varies between 0 and π/4. Now the required area will be given as

\begin{aligned}Area\left( A\right) =4\cdot \dfrac{1}{2}\int ^{\pi /4}_{0}r^{2}d\theta =2\int ^{\pi /4}_{0}a^{2}\cos2\theta d\theta \\ =2a^{2}\int ^{\dfrac{\pi }{4}}_{0}\cos 2\theta d\theta =a^{2}\left( \sin 2\theta \right) _{0}^{\dfrac{\pi }{4}}\\ =a^{2}\left( \sin \dfrac{\pi }{2}-\sin 0\right) =a^{2}\end{aligned}

Example: Find the area of the region enclosed by the cardioid (fig. 2.1 in the thumbnail) r = a(1+cosθ).
Solution: The required area will be given as

\begin{aligned}Area\left( A\right) =\dfrac{1}{2}\int ^{2\pi }_{0}r^{2}d\theta =\dfrac{1}{2}\int ^{2\pi }_{0}a^{2}\left( 1+\cos \theta \right) ^{2}d\theta \\ =\dfrac{a^{2}}{2}\int ^{2\pi }_{0}\left( 1+2\cos ^{2}\left( \dfrac{\theta }{2}\right) -1\right) ^{2}d\theta \\ =2a^{2}\int ^{2\pi }_{0}cos^{4}\left( \dfrac{\theta }{2}\right) d\theta =\dfrac{3\pi a^{2}}{2}\end{aligned}

Now attempt at least five questions based on this topic. So this is it from this tutorial. Hoping you people will attempt a few problems based on the topic discussed in this tutorial. In the next tutorial, we will discuss the topic “Minimum Value of The Area Under The Curve”.

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