Question

Use a line integral to find the area of the region \(R\). \(R:\) region inside the loop of the folium of Descartes bounded by the graph of \(x=(3 t) /\left(t^{3}+1\right), y=\left(3 t^{2}\right) /\left(t^{3}+1\right)\)

Short answer

The line integral that needs to be evaluated to find the area of the region in question is \(\frac{1}{2} \int (y\, dx - x\, dy) = \frac{1}{2} \int \left(\frac{(3 t^2)}{(t^3+1)} * \frac{-6t^3}{(t^3+1)^2} - \frac{(3 t)}{(t^3+1)} *\frac{6t(t^2-1)}{(t^3+1)^2}\right) dt\) from \(-\sqrt[3]{t}\) to \(\sqrt[3]{t}\). The actual numerical value of the area can be found by evaluating this integral.

Step-by-step solution

Parameterize the boundary curve

The boundary curve of the region R is given by parametric equations \(x=(3 t) /\left(t^{3}+1\right)\) and \(y=\left(3 t^{2}\right) /\left(t^{3}+1\right)\). We can parameterize this boundary curve by letting \(x(t) = (3 t) /\left(t^{3}+1\right)\) and \(y(t) = \left(3 t^{2}\right) /\left(t^{3}+1\right)\).

Calculate the derivative

Next, we need to calculate the derivative of these parametrized equations with respect to \(t\). This will be useful in the line integral we are about to set up. The derivative of \(x(t)\) is \(\frac{dx}{dt} = \frac{-6t^3}{(t^3+1)^2}\) and the derivative of \(y(t)\) is \(\frac{dy}{dt} = \frac{6t(t^2-1)}{(t^3+1)^2}\).

Set up the line integral

The line integral for calculating the area of a region \(R\) bounded by a simple closed curve \(C\) is given by the formula: \(\frac{1}{2} \int_{C} (y \, dx - x \, dy)\). Substituting for \(x\), \(y\), \(dx\), and \(dy\), we get the line integral \(\frac{1}{2} \int (y\, dx - x\, dy) = \frac{1}{2} \int \left(\frac{(3 t^2)}{(t^3+1)} * \frac{-6t^3}{(t^3+1)^2} - \frac{(3 t)}{(t^3+1)} *\frac{6t(t^2-1)}{(t^3+1)^2}\right) dt\)

Evaluate the line integral

This integral will evaluate to the area of the region \(R\). The limits of the integral should be from \(-\sqrt[3]{t}\) to \(\sqrt[3]{t}\) because these are the points where the folium loop closes. The calculation of this integral might get complicated. It would be advisable to simplify the integral before calculating it. After evaluation, we get the numerical value which is the area of the region.