# Find the area of ​the plane region defined by $x^{2/3}+y^{2/3}≤ 1$ , $x≥0$ , $y≥0$ using a change of variables

Find the area of ​​the plane region defined by $$x^{2/3}+y^{2/3}\leq 1$$ , $$x\geq 0$$ , $$y\geq 0$$ using the change of variables $$x=r\cos^3 \theta$$ , $$y=r\sin^3 \theta$$

The teacher used polar coordinates to express this region, which looks like so:

And went and used $$\Omega=\{(r,\theta)\in\mathbb{R}^2:0 to describe the region and then used the Jacobian to find the area as so: $$A=\frac{1}{4}\iint_{\Omega}dA=\frac{1}{4}\int_0^1 \int_0^{2\pi} |J|drd\theta$$ and I'm not really interested in the resolution.

But rather I'm interested in why that region we expressed with polar coordinates would be the same exact region as one described by a circle? Isn't it? It shocks me a little bit since the region $$\Omega$$ I was integrating over doesn't actually represent the star-shaped region but instead it represents a circle, but I do get the area of the star-shaped region and not the one of the circle, which I'm guessing it is due to the Jacobian, but this stunned me, is there anything else I can look into to help me see it better?

• Where is that circle that you mention? Commented Apr 24, 2023 at 21:29
• @JoséCarlosSantos sorry, now should be edited, $0<r<1$, $0<\theta <2\pi$ describes a circle Commented Apr 24, 2023 at 21:30
• Yes, it describes a circle. You want to know why? What has that to do with the computation of that area? Commented Apr 24, 2023 at 21:33
• @JoséCarlosSantos I don't know if those questions are rhethorical or not, I'll take them as if they aren't, yes I would like to know why if both regions (star-shaped and circle) are described with the same $\Omega$, why the integral can give 2 different areas (and yeah I'm sure it has to do with the Jacobian but I can't grasp my head around it) Commented Apr 24, 2023 at 21:37
• They are not described the same way. The unit circle is$$\left\{(r\cos\theta,r\sin\theta)\mid r\in[0,1]\wedge\theta\in[0,2\pi]\right\},$$whereas your region is$$\left\{\left(r\cos^{\color{red}3}\theta,r\sin^{\color{red}3}\theta\right)\mid r\in[0,1]\wedge\theta\in[0,2\pi]\right\}.$$Don't you see a difference? Commented Apr 24, 2023 at 21:45

As already mentioned in the comments, the transformation $$x = r \cos^3 \theta, \quad y = r \sin^3 \theta \tag{1}$$ is not a cartesian-to-polar transformation. It should be obvious why: if $$x^2 + y^2 = 1$$ is a circle in the Cartesian plane, then in the $$(r,\theta)$$ coordinate system this relation becomes $$r^2 (\cos^6 \theta + \sin^6 \theta) = 1. \tag{2}$$ Application of additional trigonometric identities results in the equivalent expression $$\cos 4\theta = \frac{8-5r^2}{3r^2}. \tag{3}$$ This does not even remotely resemble a unit circle, which in polar coordinates would simply be $$r = 1$$.
What does end up being a circle in the $$(r,\theta)$$ system specified by $$(1)$$, is the equation $$x^{2/3} + y^{2/3} = 1. \tag{4}$$ That is, after all, precisely why we chose to use that transformation: it allows us to map the shape described by $$(4)$$ into a circle in the new coordinate system, which is then easily integrated. But as the Jacobian is nontrivial, this also means the differential area element in the new system is not constant with respect to its coordinates, just like how in the usual polar system, the differential area element is proportional to its distance from the origin; i.e. $$dA = dx \, dy = r \, dr \, d\theta.$$
Geometrically, another way to conceptualize the effect of such a transformation $$(1)$$, is that it changes the boundary of the curve but also the way the region in inside the curve is "weighted" when its area is calculated.