Edit 1:

Issues seem to emerge from within ImplicitRegion as well as Region Plot. Using $Surd[y^{2},{3}]$ or avoiding $y^{2/3}$ by using $y^{3}$ will result in a correct result of Integrate. Area and Integrate seem to return identical results.

In other case both the results of RegionPlot or Integrate[1,Element[{x,y},reg] of the following ImplicitRegion will be wrong, as verified by last result (In case no mistake was done):

$\text{reg7}=\text{ImplicitRegion}\left[\sin [x]\leq y\leq \cos [x],\left\{\left\{x,0,\frac{3 \pi }{2}\right\},y\right\}\right] $

The results and double checking them unfold as follows:

$\text{RegionPlot}[\text{reg7}]$ enter image description here

$\text{Integrate}[1,\{x,y\}\in \text{reg7}]$

$2 \left(\sqrt{2}-1\right)$

$\int_0^{\frac{3 \pi }{2}} \text{Abs}[ \cos [x]-\sin [x]] \, dx$

$2 \left(2 \sqrt{2}-1\right)$

Original Post:

I am trying to Integrate the area mentioned below using ImplicitRegion function, following this resource. This method worked in many other scenario. The plot was accurate, but the integration returned 1/10 which is not true as is verified by the traditional integration. I have tried integration over $-1\leq y\leq 0$ and $0\leq y\leq 1$ separately and also using Surd[x,3]^2 but things got much weirder. Can any one give a hint how I can improve or correct this?

$\text{reg}=\text{ImplicitRegion}\left[y\leq x\leq y^{\frac{2}{3}},\{x,\{y,-1,1\}\}\right]$

$\text{RegionPlot}[\text{reg}]$ enter image description here

$\text{Integrate}[1,y\in \text{reg}]$


$\int_{-1}^1 \left(y^{\frac{2}{3}}-y\right) \, dy$

$\frac{3}{5} \left(1+(-1)^{2/3}\right)$

  • 2
    $\begingroup$ You can format inline code and code blocks by selecting the code and clicking the {} button above the edit window. The edit window help button ? is useful for learning how to format your questions and answers. You may also find this meta Q&A helpful $\endgroup$ – Michael E2 Aug 14 '20 at 12:47
  • 2
    $\begingroup$ The region between Sin[x] and Cos[x] is Min[Sin[x], Cos[x]] <= y <= Max[Sin[x], Cos[x]] instead of Sin[x] <= y <= Cos[x]. I have updated my answer. $\endgroup$ – cvgmt Aug 15 '20 at 23:10

Updated answer.

The region between Sin[x] and Cos[x] is Min[Sin[x], Cos[x]] <= y <= Max[Sin[x], Cos[x]] instead of Sin[x] <= y <= Cos[x]

BTW, since $\max(f,g)-\min(f,g)=|f-g|$,that is why your integrate work.

$$ \int_a^b |f(x)-g(x)|\,\mathrm{d}x =\int_a^b \left [ \max(f(x),g(x))-\min(f(x),g(x))\right ]\,\mathrm{d}x$$

The set $$\{ (x,y) : \sin(x)\leq y\leq \cos(x) \} $$ is equivalent to $$\{ (x,y) : \sin(x) \leq y\; \text{and} \; y\leq \cos(x) \}$$ which implicitly satisfy a another condition say $\sin(x)\leq \cos(x)$

regionOne = 
      ImplicitRegion[Sin[x] <= y <= Cos[x], {{x, 0, 3/2 \[Pi]}, y}];
    regionTwo = 
       Min[Sin[x], Cos[x]] <= y <= 
        Max[Sin[x], Cos[x]], {{x, 0, 3/2 \[Pi]}, y}];
      RegionMeasure[regionone]}, {RegionPlot[regionTwo], 

enter image description here


Original answer

 reg = ImplicitRegion[y <= x <= CubeRoot[y]^2 && -1 <= y <= 1, {x, y}]
 DiscretizeRegion[reg, MaxCellMeasure -> {"Length" -> 0.012}]]
    Integrate[1, Element[{x, y}, reg]]
    Integrate[CubeRoot[y]^2 - y, {y, -1, 1}]

enter image description here


reg = ImplicitRegion[y^3 <= x^3 <= y^2 && -1 <= y <= 1, {x, y}]
Integrate[1, Element[{x, y}, reg]]
  • 2
    $\begingroup$ Or Integrate[Surd[y^2, 3] - y, {y, -1, 1}]. $\endgroup$ – user64494 Aug 14 '20 at 12:21
  • $\begingroup$ @cvgmt RegionPlot would render an incorrect plot in your first option (Same with using Surd[y^2,3] @user64494), even though Integrate result is true, which why I came across it.The second option involves a changing of the formula itself. I am still not able to figure out why I get, either 1/10 instead of 6/5,or a false plot or in other case the formula has to be manipulated. $\endgroup$ – pigeon Aug 15 '20 at 10:45
  • 1
    $\begingroup$ @pigeon It it the render problem of RegionPlot instead of CubeRoot or Surd $\endgroup$ – cvgmt Aug 15 '20 at 12:08
  • 2
    $\begingroup$ @pigeon you can also do Area[reg] $\endgroup$ – flinty Aug 15 '20 at 13:58
  • $\begingroup$ @flinty Despite seeing this many time, It just never crossed my mind due to the need to figure out how it works. It is just perfect in this scenario! Thanks $\endgroup$ – pigeon Aug 15 '20 at 16:57

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