# Get Perimeter Region from Object

I can get the region of a disk $$\{x,y\}\in\{x^2+y^2 \leq 1\}$$ as follows:

Region[Disk[]] Now my question is: Is there also an automated way to get the perimeter region ?

$$\{x,y\}\in\{x^2+y^2=1\}$$

Many thanks !

• RegionBoundary[Region[Disk[]]]? – kglr Apr 6 at 6:18

Update: Composing RegionMember, RegionBoundary and FullSimplify:

ClearAll[boundaryRF]
boundaryRF = FullSimplify[RegionMember[RegionBoundary @ #, #2], #2 ∈ Reals] &;


Examples:

Grid[{#, boundaryRF[#, {x, y}]} & /@
{Disk[], Disk[{a, b}, r], Rectangle[], Triangle[]},
Dividers -> All] Grid[{#, boundaryRF[#, {x, y, z}]} & /@
{Ball[], Ball[{a, b, c}, r], Tetrahedron[], Cone[]},
Dividers -> All]


For Disk yes:

RegionBoundary[Disk[]]

Circle[{0, 0}]

RegionMember[%, {x, y}]

(x | y) ∈ Reals && x^2 + y^2 == 1


Works with symbolic parameters too:

RegionBoundary[Disk[{a, b}, r]]

Circle[{a, b}, r]

RegionMember[%, {x, y}]

(x | y) ∈ Reals && r > 0 && (-a + x)^2 + (-b + y)^2 == r^2


... and few other primitives:

RegionBoundary[Ball[]]

 Sphere[{0, 0, 0}]

RegionMember[%, {x, y, z}]

(x | y | z) ∈ Reals && x^2 + y^2 + z^2 == 1

RegionBoundary[Rectangle[]]

 Line[{{0, 0}, {1, 0}, {1, 1}, {0, 1}, {0, 0}}]

RegionMember[%, {x, y}] // FullSimplify

 (x | y) ∈ Reals && ((0 <= y <= 1 && (x == 0 || x == 1)) || (0 <= x <=
1 && (y == 0 || y == 1)))


Acknowledgement: Thank you Chip Hurst for reminding me that we can use RegionMember instead of

RegionRegionProperty[%, {x, y}, "FastDescription"][[1, -1]]


to get the region function.

• Very nice ! Thank you very much !! One question: when you type RegionRegionProperty[... what does the  stand for ? – james Apr 6 at 6:49
• @james, see tutorial/Contexts – kglr Apr 6 at 6:53
• Why not avoid the internal functions and use RegionMember[RegionBoundary[Disk[]], {x, y}]`? – Chip Hurst Apr 6 at 12:19
• @ChipHurst, yes off course!. It just did not occur to me. Perhaps you should post that as an answer? – kglr Apr 6 at 12:38
• I think you should update your post to avoid one thinking an internal function is needed. – Chip Hurst Apr 6 at 12:39