Is there a method or shortcut (done by hand) to determine the area bounded by the curves of mixed equations?

Question

While playing around with equations and graphs for Valentine's Day, I found the following problem to be interesting, which involves different types of equations.

Let the parametric equations (denoted by $f$) of a heart curve be $$ \begin{array}{rl} x &= 16\sin^3(t) \\ y &= 13\cos(t) - 5\cos(2t) - 2\cos(3t) - \cos(4t) \end{array} $$ for $0 \leq t \leq 2\pi$. Let $g(x) = x^2$ be the equation of a line passing through a heart. Determine the area of the region, bounded by these curves for $x \geq 0$ and $f > g$.

Firstly, I tried transforming the given parametric equations into $y(x)$ form and then, determine the area integral, which turns out to be messy.

Secondly, I even tried writing $y = x^2$ in parametric form. However, this doesn't seem to click since the parameters may be different.

I'm interested in knowing the best method to approach this problem.

Answer 1

If a parametric curve $(x(t),y(t))$ circles the origin in a counter-clockwise direction, then the area it encloses is given by $$ A = \frac{1}{2} \int \left( x \frac{dy}{dt} - y \frac{dx}{dt} \right) dt $$ (This is found by noting that the area of the long narrow wedge subtended by $\vec{r} = (x,y)$ and $\vec{r} + \Delta \vec{r} = (x + \Delta x, y + \Delta y)$ is approximately $\frac{1}{2}|\vec{r} \times \Delta \vec{r}|$.)

In principle, you could use this to solve your problem. Your heart-shaped curve and the parabola intersect at one point with $x> 0$; call this $(x_0, y_0)$. A line from $(0,0)$ to $(x_0, y_0)$ divides the area into two sub-areas. The area between the parabola and this diagonal can be found by standard Calculus 101 techniques. The area bounded by the diagonal, the heart-shaped curve, and the $y$-axis can be found by integrating the above parametric area equation from $t = t_0$ to 0, where $t_0$ is defined such that $(x(t_0), y(t_0)) = (x_0, y_0)$. This latter integrand is expressible in terms of cosines and sines, so although the answer might be ugly, the indefinite integral does exist in closed form.

The problem with this method is that the point of intersection $(x_0, y_0) \approx (3.222, 10.379)$ may be impossible to write down in closed form. (For the record, $t_0 \approx 0.626257...$.) Playing around with Mathematica, it appears to be expressible in terms of sines and inverse tangents of roots of a certain 6th-order polynomial. If a numerical answer is all that's required, though, then you could certainly get an answer to some number of decimal places using this technique.