Finding an area of parametric curve that lies above and below x axis.

Question

Find an area from $t \in <0, 2\pi>$ bounded by parametric curve:

$\begin{cases} x(t) = \cos^3{t} \\ y(t) = \sin^3{t}\end{cases}$

I believe the formula is: area $=\int^{2\pi}_{0}y(t)x'(t)dt$.

$x'(t)=-3\cos^2t\sin{t}$

$\Rightarrow -3\int^{2\pi}_{0} \sin^3t\cos^2t\sin{t}dt = \frac{1}{64}(-12t + 3\sin{(2t)} + 3\sin{(4t)} - \sin{(6t)}) = \frac{-3\pi}{16}$

I calculated the integral with wolfram as it's sort of complicated.

Area inside curve given by parametric equation

I found this thread but I don't understand why they're putting minus in front of the integral from the very beginning.

This is how the curve looks like:

My questions:

1. Why is there a minus in front of integral from the link I posted?

2. How should I figure out on my own that the graph looks like that (part of it lies below x axis and part above), and what should I do with that information? Is there any general rule to follow for all kinds of "simple" parametric curves?

3. Perhaps the most important: is there any online calculator that can calculate an area of parametric curves? At the moment I don't know how to check if my answers are correct.

Answer 1

I can give answers for questions $1$ and $2$:

1. There's a minus sign because, if $x(t)=\cos^3t$, by the chain rule, $\; x'(t)=3\cos^2t\times (-\sin t)=-3\cos^2\sin t$.
2. You can imagine the symmetries of the curve using the properties of the trigonometric functions:

– $x(t+\pi=-x(t)$, $\;y(t+\pi)=-y(t)$, so the curve has a symmetry w.r.t. the origin.

– $x(-t)=x(t)$, $\;y(-t)=-y(t)$ so the curve has a symmetry w.r.t. the $x$-axis.

– $x(\pi-t)=-x(t)$, $\;y(\pi -t)=y(t)$: the curve has a symmetry w.r.t. the $y$-axis.

– $x(\frac \pi 2-t)=y(t)$, $\;y(\frac \pi 2-t)=x(t)$: the curve has a symmetry w.r.t. the line $y=x$.

As the parameterisation has period $2\pi$, it shows the area is $$\int_0^{2\pi}\sin^3t(-3\cos^2t\sin t)\,\mathrm dt=4\int_0^{\frac\pi 2}\sin^3t(-3\cos^2t\sin t)\,\mathrm dt=-12\int_0^{\frac\pi 2}\sin^4t\cos^2t\,\mathrm dt.$$ Note the latter integral can easily computed by linearisation of the trigonometric monomial $\sin^4t\cos^2t$.