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I want to plot x against y in this relation: relation

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  • 3
    $\begingroup$ Please provide Mathematica code, this will increase your chance to get helpful answers! $\endgroup$ – Ulrich Neumann Jul 1 at 8:06
  • $\begingroup$ Your variable h is not defined, and you need to capitalize Exp (not exp). $\endgroup$ – Roman Jul 1 at 11:09
  • $\begingroup$ 1- it is type mistake, y is true (not h) 2- ok thanks I correct Exp. $\endgroup$ – R123 Jul 1 at 11:22
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You can try ContourPlot.

For example:

ContourPlot[Evaluate@Integrate[Sin[x  y  + x + (y - Pi/2) x  z], {z, 0, 2 Pi}], 
    {x, -Pi,  Pi}, {y, -Pi, Pi}, PlotRange -> Full, Exclusions -> None]

enter image description here

To show a single contour, say the contour where Integrate[Sin[x y + x + (y - Pi/2) x z], {z, 0, 2 Pi}] == Pi/2 you can use

ContourPlot[Evaluate[Integrate[Sin[x  y  + x + (y - Pi/2) x  z], {z, 0, 2 Pi}] == Pi/2], 
 {x, -Pi, Pi}, {y, -Pi, Pi}, Exclusions -> None, 
 ContourStyle -> Directive[Red, Thick], ContourShading -> None, 
 PlotRange -> Full]

enter image description here

Alternatively, you can use the option Contours -> {Pi/2}:

ContourPlot[Evaluate@Integrate[Sin[x  y  + x + (y - Pi/2) x  z], {z, 0, 2 Pi}], 
  {x, -Pi,  Pi}, {y, -Pi, Pi}, Exclusions -> None, ContourStyle -> Thick, 
 ContourShading -> None, PlotRange -> Full, Contours -> {Pi/2}]

same picture

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  • $\begingroup$ Thanks kglr. I will try it. but I have a question. What about the right side of this relation? The left side is equals to 2pi (=2pi). It seems that this is an integral equation. $\endgroup$ – R123 Jul 1 at 9:47
  • $\begingroup$ @Rojan, if you Integrate[f[x,y,z], {z, 0,2pi}]` is a function of (x,y). If you call that function g[x,y], you can use ContourPlot[ g[x,y] == level, {x,...}, {y...}] to get the locus of {x,y} points that satisfy g[x,y] == level. $\endgroup$ – kglr Jul 1 at 10:04

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