0
$\begingroup$

I have two regions like this

reg1 = ImplicitRegion[(-5. + x)^2 + (-3. + y)^2 <= 0.25 || (0.386909 + x Tan[25 \[Degree]] <= y <= 5.61309 - x Tan[25 \[Degree]] && 0 <= x <= 5.), {x, y}]; reg2 = ImplicitRegion[(-5. + x)^2 + (-3. + y)^2 <= 0.49 || (0.166233 + x Tan[25 \[Degree]] <= y <= 5.83377 - x Tan[25 \[Degree]] && 0 <= x <= 5.), {x, y}]; Show[RegionPlot[reg1, AspectRatio -> Automatic], RegionPlot[reg2, AspectRatio -> Automatic]]

enter image description here

I need to see and work with their difference

RegionPlot[RegionDifference[reg2,reg1], AspectRatio -> Automatic]

but it brings about an error

RegionPlot::invplotreg: {RegionDifference[ImplicitRegion[Plus[<<2>>]^2+Plus[<<2>>]^2<=0.49||(0.166233 +Times[<<2>>]<=y<=5.83377 +Times[<<3>>]&&0<=x<=5.),{x,y}],ImplicitRegion[Plus[<<2>>]^2+Plus[<<2>>]^2<=0.25||(0.386909 +Times[<<2>>]<=y<=5.61309 +Times[<<3>>]&&0<=x<=5.),{x,y}]]} is not a valid region to plot. >>
Improve this question
$\endgroup$
5
  • $\begingroup$ What are w3 and w2? Maybe you meant reg2 and reg1; in that case RegionPlot[RegionDifference[reg2, reg1], AspectRatio -> Automatic] works with no problems. $\endgroup$ Commented Sep 23, 2016 at 12:10
  • $\begingroup$ Yes, sorry, I meant reg2, reg1. But it still doesn't work for me. $\endgroup$ Commented Sep 23, 2016 at 12:51
  • $\begingroup$ Try quitting the kernel, maybe you have some old definitions. $\endgroup$ Commented Sep 23, 2016 at 12:57
  • $\begingroup$ works here too, maybe a version issue. $\endgroup$ Commented Sep 23, 2016 at 18:22
  • $\begingroup$ Works in version 11 on Windows 10 $\endgroup$ Commented Sep 23, 2016 at 18:57

1 Answer 1

Reset to default
1
$\begingroup$ 
RegionDifference @@ (DiscretizeRegion /@ {reg2, reg1})

enter image description here

Improve this answer
$\endgroup$

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.