# Calculate area from RegionPlot directly

I have a region plot, and I would like to calculate the area:

With[{b = 2, h = 2}, RegionPlot[
{(x)^2 + (y - b)^2 > h^2 + b^2 &&
(x)^2 + (y - b - h)^2 > h^2 + b^2 &&
(x - h)^2 + (y - 2 b - h)^2 > h^2 + b^2 &&
(x - h - b)^2 + (y - 2 b - h)^2 > h^2 + b^2 &&
(x - 2 h - b)^2 + (y - b - h)^2 > h^2 + b^2 &&
(x - 2 h - b)^2 + (y - b)^2 > h^2 + b^2 &&
(x - h - b)^2 + (y)^2 > h^2 + b^2 &&
(x - h)^2 + (y)^2 > h^2 + b^2}, {x, 0, 2 h + b}, {y, 0, 2 b + h}]]


I have tried defining it as an ImplicitRegion but Area evaluates to Infinity. I am sure I am doing something wrong that is very simple, but I can't see what.

• Well, if you can extract the polygons, there's this. – J. M.'s technical difficulties Jun 30 '15 at 3:16
• For approximate area: Area @ DiscretizeGraphics @ g. Where g is your graphics. – RunnyKine Jun 30 '15 at 5:10

ImplicitRegion seems to work fine:

ir = With[{b = 2, h = 2},
ImplicitRegion[{(x)^2 + (y - b)^2 >
h^2 + b^2 && (x)^2 + (y - b - h)^2 >
h^2 + b^2 && (x - h)^2 + (y - 2 b - h)^2 >
h^2 + b^2 && (x - h - b)^2 + (y - 2 b - h)^2 >
h^2 + b^2 && (x - 2 h - b)^2 + (y - b - h)^2 >
h^2 + b^2 && (x - 2 h - b)^2 + (y - b)^2 >
h^2 + b^2 && (x - h - b)^2 + (y)^2 >
h^2 + b^2 && (x - h)^2 + (y)^2 > h^2 + b^2},
{{x, 0, 2 h + b}, {y, 0, 2 b + h}}]];

FullSimplify[Area[ir]]

(* (8 Pi)/3 - 4 (-7 + 2 Sqrt + Sqrt + 4 ArcCos[3/4]) *)


which agrees with the approximate area based on the plot per RunnyKine's comment

Area[DiscretizeGraphics[RegionPlot[ir]]]

(* 0.375737 *)


or the approximate area based on the discretized region

Area[DiscretizeRegion[ir]]

(* 0.374631 *)


The last value can be made more accurate by using a finer mesh.

Area[DiscretizeRegion[ir, PrecisionGoal -> 9]]

(* 0.374421 *)

• ah - thank you! I was using {x,y} at the end insterad of {{x, 0, 2 h + b}, {y, 0, 2 b + h}} - I didn't realise you needed the extra curly brackets around it & wondered why {x, 0, 2 h + b}, {y, 0, 2 b + h} didn't work!! – martin Jun 30 '15 at 6:44