Compute the area of the region $D=\{(x,y)\in\mathbb{R}:x^{2/3}+y^{2/3}\leq1\}.$

Question

Compute the area of the region $D=\{(x,y)\in\mathbb{R}:x^{2/3}+y^{2/3}\leq1\}.$

HINT: Set $x=\cos^3{t}$ and $y=\sin^3{t}.$

The problem is easy using Greens theorem and the integral simplifies neatly to

$$A=\frac{3}{16}\int_{t_1}^{t_2}(1-\cos{2t}) \ dt.$$

However, I'm not really sure how the bounds for $t$ should be in the integral. Both $x$ and $y$ go from $0$ to $1$, can I just plug this in my parameterization and obtain min/max for $t$? I get different values depending on if I use $y$ or $x$.

So my question is: how should one determin the bounds for $t?$

Answer 1

We need to describe one cycle thus, from the original parametrization for the region, we need $t\in[0,2\pi]$.

Answer 2

Consider the constraint : $X^2+Y^2 \le 1$, $X,Y$ real.

$X^2+Y^2 =1$ is a circle.

Parametrize:

$X= \cos t$, $Y= \sin t$, $t \in [0,2π).$

Note:

$P$: $[0,2π) \rightarrow S$,

$S:=${$(X,Y)| X^2+Y^2 =1$} ,with

$P(t) = (\cos t,\sin t)$, is bijective.

And now consider the bijective transformation:

$x=X^3$, and $y =Y^3 $, or in terms of $t$,

the composition of the two bijective transformations:

$x=\cos^3 t, y = \sin^3 t$ , where $t \in [0,2π)$.

Comments welcome.

P.S. The above starts from a circle and ends up with the given constraint in the coordinates $x,y $(the problem).

The above steps can be done in reverse order starting from x,y ending up with the circle .