I'm trying to find the area bounded by the curve $x^3=ay^4 – x^2y$.
First, I have an example of a solution to finding area of the following curve: $x^3 + y^3 = 3axy, a > 0$.
In the solution, they first express the equation parametrically as following. Let $t=\frac{y}{x}$. Then $$x = \frac{3at}{1 + t^3}, y = \frac{3at^2}{1 + t^3}.$$
And then they say that the curve describes a loop if the parameter $t$ ranges from $0$ to $+\infty$, which I don't understand why (like, how do you arrive at that). Here's the picture they've shown:
And I can also see this if I put the parametric graph in Desmos, let $t$ be between $0$ and some arbitrarily large number (the $+\infty$ part of the range described) and then slide $a$:
Now let's go back to my original equation. If I, in the same fashion, set the parameter $t=\frac{y}{x}$, the equation would be the following: $$x=\frac{1+t}{at^4}, y=\frac{1+t}{at^3}.$$
Next I've tried playing around with different ranges of $t$ in Desmos in order to obtain a loop similar to the one in the first gif, and I've managed to do that if I let $t$ be between an arbitrarily large negative number ($-\infty$) and $-1$:
But I really have no idea how to rigorously proof that in that range of the parameter $t$, the curve describes a loop.
All we need to do then is calculate the area by this formula:
$$S=\frac{1}{2}\int_{t_1}^{t_2} |y(t)x'(t)-x(t)y'(t)|dt,$$ where $t_1$ and $t_2$ are the bounds of the range for which the parameter makes the curve describe a loop. Still, the question is: how do you determine that range? Because if I put one of the bound of the range for my equation $0$, the the integral would diverge, but with $-1$ it doesn't.
Best Answer
There is some geometric intuition that we can apply here.
In the example you provide (which describes a curve called a folium), the parametrization $t = y/x$ has a geometric interpretation: it means $t$ is a parameter that describes the slope of the corresponding parametrized point $(x(t), y(t))$ relative to the origin.
In other words, for each $t$, the point $(x(t), y(t))$ lies on the line $y = tx$, by construction. So if the implicit equation is $x^3 + y^3 = 3axy$ and the parametrization yields $$(x(t), y(t)) = \left(\frac{3at}{1+t^3}, \frac{3at^2}{1+t^3}\right), \tag{1}$$ then we can see that when $a > 0$, $(x,y)$ will be located in the first quadrant whenever $0 < t < \infty$ since $x > 0$ and $y > 0$ for such $t$; moreover, the line $y = tx$ on this interval of $t$ sweeps across the first quadrant, and because of uniqueness, there is only one point being chosen to be on the curve for each such line; i.e., the curve is simple with no self-intersection on this interval.
Finally, the fact that $(x(0), y(0)) = (0,0)$ and $\lim_{t \to \infty} (x(t),y(t)) = (0,0)$ implies that the curve is closed.
Having established this, the question of the curve's behavior for $t < 0$ is of course easier to understand. When $-1 < t < 0$, the line $y = tx$ makes an angle with the positive $x$-axis between $-\pi/4$ and $0$. So on this $t$-interval, these lines will sweep out points on the curves in the second quadrant. Indeed, we can formally see this: when $-1 < t < 0$, then $1 + t^3 > 0$ and again, for $a > 0$, $x(t) < 0$ but $y(t) > 0$. As $t \to -1^+$, $x \to -\infty$ and $y \to +\infty$.
When $-\infty < t < -1$, the line $y = tx$ sweeps out angles from $-\pi/2$ to $-\pi/4$, and now we are in the fourth quadrant, since $1 + t^3 < 0$ hence $x(t) > 0$, $y(t) < 0$.
This also explains why the folium is reflected through the origin when $a < 0$.
For your curve, the parametrization is $$(x(t), y(t)) = \left(\frac{1+t}{a t^4}, \frac{1 + t}{a t^3} \right). \tag{2}$$ Since $a$ is just a scaling constant, it does the same thing as in the folium example, so for simplicity let us assume $a = 1$. Then as before, $t = y/x$ implies that $t$ is a slope parameter. However, unlike the folium, here the parametrization is undefined when $t = 0$.
When $t > 0$, it's clear that $(x(t), y(t))$ lies entirely in the first quadrant. Moreover, as $t \to 0^+$, we see that $(x(t), y(t)) \to (\infty, \infty)$, but because $y/x = t$, the curve actually approaches the positive $x$-axis. This seems counterintuitive, but for example, $t = 0.01$ corresponds to $(x,y) = (1.01 \times 10^8, 1.01 \times 10^6)$.
As $t \to \infty$, we have $(x(t), y(t)) \to 0$, so the curve approaches the origin as the slope of the point approaches vertical. So we know that there is no closed loop in the first quadrant, for $0 < t < \infty$.
There are no problematic cases when $-\infty < t < 0$ because the parametrization is smooth and well-defined on this interval. But since $t = -1$ corresponds to $(0,0)$, we see that the curve must form a closed loop for the interval $-\infty < t < -1$, hence this loop occurs in the second quadrant (as $-\infty < x(t) < 0$ and $0 < y(t) < \infty$). In fact, this loop has a maximum distance from the origin, corresponding to the solution of $\frac{d}{dt}\left[x(t)^2 + y(t)^2\right] = 0$, which is the unique real root of the cubic $$2t^3 + 3t^2 + 3t + 4 = 0,$$ or $$t = \frac{1}{2}\left( \sqrt[3]{-6 + \sqrt{37}} - \sqrt[3]{6 + \sqrt{37}} - 1\right) \approx -1.42944. \tag{3}$$
Therefore, for your curve, the appropriate interval of integration is $-\infty < t \le -1$.
Addendum: See the following animated image for your curve. The red ray sweeps out slopes from $-\infty < t < \infty$.