# Please, help to find the area enclosed by the curves: ContourPlot[{x + y^2 == 0, x + 3 y^2 - 2 == 0}, {x, -2, 2}, {y, -2, 2}]

Please, Help to find the area between two curves.

ContourPlot[{x + y^2 == 0, x + 3 y^2 - 2 == 0}, {x, -2, 2}, {y, -2, 2}]

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• You need to know how to integrate ... – Dr. belisarius Oct 6 '15 at 16:49

Area@ImplicitRegion[x + y^2 >= 0 && x + 3 y^2 - 2 <= 0, {x, y}]
(* 8/3 *)


or

rgn = DiscretizeRegion@ImplicitRegion[x + y^2 >= 0 && x + 3 y^2 - 2 <= 0, {x, y}];
Area[rgn]
(* 2.66664 *)


or

eqs = {x + 3 y^2 - 2 == 0, x + y^2 == 0};
xs = First@Solve[#, x] & /@ eqs;
pts = Solve[Reverse@eqs, {x, y}];
NIntegrate[Subtract @@ (x /. xs), Evaluate@{y, Sequence @@ (y /. pts)}]
(* 2.66667 *)


or do it by hand.

• Thank you very much! Its really good. – Bed Dhakal Oct 7 '15 at 1:42

The area you are looking for is enclosed by the curves

x == -y^2
x == -3 y^2 + 2


They intersect here:

Solve[-y^2 == -3 y^2 + 2, y]
(* {{y -> -1}, {y -> 1}} *)


Plotting the area of interest:

Plot[{-y^2, -3 y^2 + 2}, {y, -1, 1}, Frame -> True, Axes -> None,
Filling -> {2 -> {1}}] As you can see the area between the curves is simply the difference between areas under each individual curves:

Integrate[-3 y^2 + 2 - (-y^2), {y, -1, 1}]
(* 8/3 *)

• Thank you very much! It really worked for me. – Bed Dhakal Oct 7 '15 at 1:42