Area Enclosed By Algebraic Curve

Calculus Level 3

The equation ( x 2 + y 2 ) 3 = 4 x 2 y 2 (x^2 + y^2)^3 = 4x^2 y^2 describes a curve in the Cartesian plane. What is the area of the region enclosed by this curve?

Hint : First convert Cartesian coordinates to polar .

π 4 \dfrac{\pi}{4} π 6 \dfrac{\pi}{6} π 2 \dfrac{\pi}{2} π 8 \dfrac{\pi}{8}

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Sameer Kailasa by Sameer Kailasa , 25, USA

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1 solution

Tom Engelsman
Jan 4, 2021

Let x = r cos ⁡ θ , y = r sin ⁡ θ x = r \cos \theta, y = r \sin \theta over 0 ≤ θ ≤ 2 π . 0 \le \theta \le 2\pi. Converting the above curve into polar coordinates yields;

( r 2 ) 3 = 4 ( r 2 cos ⁡ 2 θ ) ( r 2 sin ⁡ 2 θ ) (r^2)^3 = 4(r^2 \cos^{2} \theta)(r^2 \sin^{2} \theta) ;

or r 6 = 4 r 4 cos ⁡ 2 θ sin ⁡ 2 θ r^6 = 4r^4 \cos^{2} \theta \sin^{2} \theta ;

or r 2 = ( 2 cos ⁡ θ sin ⁡ θ ) 2 r^2 = (2\cos \theta \sin \theta)^2 ;

or r 2 = f 2 ( θ ) = sin ⁡ 2 2 θ . r^2 = f^{2}(\theta) = \sin^{2} 2\theta.

The enclosed area is now computed the following integral:

A = 1 2 ∫ 0 2 π f 2 ( θ ) d θ = 1 2 ∫ 0 2 π sin ⁡ 2 2 θ d θ = π 2 . A = \frac{1}{2} \int_{0}^{2\pi} f^{2}(\theta) d\theta = \frac{1}{2} \int_{0}^{2\pi} \sin^{2} 2\theta d\theta = \boxed{\frac{\pi}{2}}.

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