Finding the area enclosed by an implicit function [closed]

How can I solve area of following plot by using double integral? closed as off-topic by Mark McClure, happy fish, bbgodfrey, Michael E2, yohbsMay 10 '17 at 3:54

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• "This question cannot be answered without additional information. Questions on problems in code must describe the specific problem and include valid code to reproduce it. Any data used for programming examples should be embedded in the question or code to generate the (fake) data must be included." – bbgodfrey, Michael E2, yohbs
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• Next time: Want an area?, show a region, this is a curve. Want an area using a specific method?, where are you stuck in implementation then? Write a formula at least. This is your third image-only question, they are not welcomed here, though answered sometimes, because people can't copy code/formulas from an image. – Kuba May 8 '17 at 8:28

You can convert to polar coordinates:

r[t_] := 2 Cos[t]^2
Integrate[r[t]^2/2, {t, 0, 2 Pi}]


yields $3\pi/2\approx 4.71239$

or you can use Green's Theorem (with $\vec{F}=\{-y/2,x/2\}$):=

Integrate[{-r[t] Sin[t], r[t] Cos[t]} .D[{r[t] Cos[t], r[t] Sin[t]}, t]],{t,0,2Pi}]/2


also yielding $3\pi/2$

or approximate using ImplicitRegion:

reg = ImplicitRegion[(x^2 + y^2)^3 <= 4 x^4, {{x, -2, 2}, {y, -2, 2}}]
RegionMeasure[DiscretizeRegion[reg, MaxCellMeasure -> {"Length" -> 0.01}]]


yields: 4.71238

See Kuba comment below for shorter ImplicitRegion solution:

Area@ImplicitRegion[(x^2. + y^2)^3. <= 4. x^4., {x, y}]


yields: 4.71239

• +1 Finite precision numbers will shorten ImplicitRegion solution: Area @ ImplicitRegion[(x^2. + y^2)^3. <= 4. x^4., {x, y}] – Kuba May 8 '17 at 8:14