Area of Loop of Folium of Descartes

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Theorem

Consider the folium of Descartes $F$, given in parametric form as:

$\begin {cases} x = \dfrac {3 a t} {1 + t^3} \\ y = \dfrac {3 a t^2} {1 + t^3} \end {cases}$


The area $\AA$ of the loop of $F$ is given as:

$\AA = \dfrac {3 a^2} 2$


Proof

From Behaviour of Parametric Equations for Folium of Descartes according to Parameter we have that the loop is traversed for $0 \le t < +\infty$.

We convert the parametric equation to polar form:

\(\ds r^2\) \(=\) \(\ds x^2 + y^2\)
\(\ds \) \(=\) \(\ds \dfrac {\paren {3 a t}^2} {\paren {1 + t^3}^2} + \dfrac {\paren {3 a t^2}^2} {\paren {1 + t^3}^2}\)
\(\ds \) \(=\) \(\ds \dfrac {\paren {3 a t}^2 \paren {1 + t^2} } {\paren {1 + t^3}^2}\)
\(\ds \tan \theta\) \(=\) \(\ds \dfrac y x\)
\(\ds \) \(=\) \(\ds \dfrac {3 a t^2} {1 + t^3} \dfrac {1 + t^3} {3 a t}\)
\(\ds \) \(=\) \(\ds t\)
\(\ds \leadsto \ \ \) \(\ds \theta\) \(=\) \(\ds \arctan t\)
\(\ds \leadsto \ \ \) \(\ds \dfrac {\d \theta} {\d t}\) \(=\) \(\ds \dfrac 1 {1 + t^2}\) Derivative of Arctangent Function


Then we have:

\(\ds \AA\) \(=\) \(\ds \dfrac 1 2 \int_{t \mathop = 0}^{t \mathop \to \infty} r^2 \rd \theta\) Area between Radii and Curve in Polar Coordinates
\(\ds \) \(=\) \(\ds \dfrac 1 2 \int_{t \mathop = 0}^{t \mathop \to \infty} \dfrac {\paren {3 a t}^2 \paren {1 + t^2} } {\paren {1 + t^3}^2} \dfrac 1 {1 + t^2} \rd t\) Integration by Substitution
\(\ds \) \(=\) \(\ds \dfrac {3 a^2} 2 \int_{t \mathop = 0}^{t \mathop \to \infty} \dfrac {3 t^2 \rd t} {\paren {1 + t^3}^2}\) Integration by Substitution


Then:

\(\ds u\) \(=\) \(\ds 1 + t^3\)
\(\ds \leadsto \ \ \) \(\ds \dfrac {\d u} {\d t}\) \(=\) \(\ds 3 t^2\)
\(\ds t\) \(=\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds u\) \(=\) \(\ds 1\)
\(\ds t\) \(\to\) \(\ds +\infty\)
\(\ds \leadsto \ \ \) \(\ds u\) \(\to\) \(\ds +\infty\)


which leads us to:

\(\ds \AA\) \(=\) \(\ds \dfrac {3 a^2} 2 \int_{t \mathop = 0}^{t \mathop \to \infty} \dfrac {3 t^2 \rd t} {\paren {1 + t^3}^2}\)
\(\ds \) \(=\) \(\ds \dfrac {3 a^2} 2 \int_{u \mathop = 1}^{u \mathop \to \infty} \dfrac {\d u} {u^2}\) Integration by Substitution
\(\ds \) \(=\) \(\ds \dfrac {3 a^2} 2 \intlimits {-\dfrac 1 u} 1 {+\infty}\) Primitive of Power
\(\ds \) \(=\) \(\ds \dfrac {3 a^2} 2 \paren {-0 - \paren {-1} }\) evaluating limits: $u \to +\infty \implies \dfrac 1 u \to 0$
\(\ds \) \(=\) \(\ds \dfrac {3 a^2} 2\)

Hence the result.

$\blacksquare$


Sources

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