Revisions to Result of Integration over ImplicitPlot not as expected

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edited Aug 16, 2020 at 0:44

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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 implicitimplicitly satisfy a another condition say $\sin(x)\leq \cos(x)$

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

+++++++++++++++++++++++++++++++++++++++++++++++

Original answer

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

Or

reg = ImplicitRegion[y^3 <= x^3 <= y^2 && -1 <= y <= 1, {x, y}]
Integrate[1, Element[{x, y}, reg]]

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 implicit satisfy a another condition say $\sin(x)\leq \cos(x)$

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

+++++++++++++++++++++++++++++++++++++++++++++++

Original answer

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

Or

reg = ImplicitRegion[y^3 <= x^3 <= y^2 && -1 <= y <= 1, {x, y}]
Integrate[1, Element[{x, y}, reg]]

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

+++++++++++++++++++++++++++++++++++++++++++++++

Original answer

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

Or

reg = ImplicitRegion[y^3 <= x^3 <= y^2 && -1 <= y <= 1, {x, y}]
Integrate[1, Element[{x, y}, reg]]

edited body

Source Link

edited Aug 16, 2020 at 0:05

cvgmt

84.1k
6
97
179

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 $\max(f,g)-\min(f,g)=|f-g|$,Thatthat 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 implicit satisfy a another condition say $\sin(x)\leq \cos(x)$

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

+++++++++++++++++++++++++++++++++++++++++++++++

Original answer

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

Or

reg = ImplicitRegion[y^3 <= x^3 <= y^2 && -1 <= y <= 1, {x, y}]
Integrate[1, Element[{x, y}, reg]]

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 implicit satisfy a another condition say $\sin(x)\leq \cos(x)$

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

+++++++++++++++++++++++++++++++++++++++++++++++

Original answer

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

Or

reg = ImplicitRegion[y^3 <= x^3 <= y^2 && -1 <= y <= 1, {x, y}]
Integrate[1, Element[{x, y}, reg]]

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 implicit satisfy a another condition say $\sin(x)\leq \cos(x)$

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

+++++++++++++++++++++++++++++++++++++++++++++++

Original answer

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

Or

reg = ImplicitRegion[y^3 <= x^3 <= y^2 && -1 <= y <= 1, {x, y}]
Integrate[1, Element[{x, y}, reg]]

edited body

Source Link

edited Aug 15, 2020 at 23:55

cvgmt

84.1k
6
97
179

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 $$\{ (x,y) : \sin(x) \leq y\, \text{and} \, y\leq \cos(x) \}$$$$\{ (x,y) : \sin(x) \leq y\; \text{and} \; y\leq \cos(x) \}$$ which implicit satisfy a another condition say $\sin(x)\leq \cos(x)$

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

+++++++++++++++++++++++++++++++++++++++++++++++

Original answer

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

Or

reg = ImplicitRegion[y^3 <= x^3 <= y^2 && -1 <= y <= 1, {x, y}]
Integrate[1, Element[{x, y}, reg]]

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 implicit satisfy a another condition say $\sin(x)\leq \cos(x)$

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

+++++++++++++++++++++++++++++++++++++++++++++++

Original answer

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

Or

reg = ImplicitRegion[y^3 <= x^3 <= y^2 && -1 <= y <= 1, {x, y}]
Integrate[1, Element[{x, y}, reg]]

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 implicit satisfy a another condition say $\sin(x)\leq \cos(x)$

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

+++++++++++++++++++++++++++++++++++++++++++++++

Original answer

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

Or

reg = ImplicitRegion[y^3 <= x^3 <= y^2 && -1 <= y <= 1, {x, y}]
Integrate[1, Element[{x, y}, reg]]

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edited Aug 15, 2020 at 23:47

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edited Aug 15, 2020 at 23:34

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edited Aug 15, 2020 at 23:16

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edited Aug 15, 2020 at 23:07

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edited Aug 15, 2020 at 12:16

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edited Aug 15, 2020 at 12:04

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edited Aug 14, 2020 at 10:34

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created Aug 14, 2020 at 10:28

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