Since it seems to have not been mentioned yet: yet another way to obtain an approximation of the area of your Lamé curve is to use the shoelace method for computing the area. Here's a Mathematica demonstration:

pts = First[Cases[ ParametricPlot[{Sqrt[Abs[Cos[t]]] Sign[Cos[t]], Sqrt[Abs[Sin[t]]] Sign[Sin[t]]}, {t, 0, 2 π}, Exclusions -> None, Method -> {MaxBend -> 1.}, PlotPoints -> 100] // Normal, Line[l_] :> l, ∞]]; PolygonSignedArea[pts_?MatrixQ] := Total[Det /@ Partition[pts, 2, 1, 1]]/2 PolygonSignedArea[pts] 3.7081447086368127 

The value thus obtained is pretty close to the results in the other answers.

Surprisingly, there is in fact an undocumented built-in function for computing the area of a polygon:

Graphics`Mesh`MeshInit[]; PolygonArea[pts] 3.708144708636812 

But, what you really should know is that Lamé curves have been well studied, and there is in fact a closed form expression for the area of a Lamé curve. Given the Cartesian equation

$$\left|\frac{x}{a}\right|^r+\left|\frac{y}{b}\right|^r=1$$

the formula for the area of a Lamé curve (formula 5 here) is

$$A=\frac{4^{1-\tfrac1{r}}ab\sqrt\pi\;\Gamma\left(1+\tfrac1{r}\right)}{\Gamma\left(\tfrac1{r}+\tfrac12\right)}$$

In particular, for the OP's specific case, $a=b=1$, and $r=4$. Thus,

With[{a = 1, b = 1, r = 4}, N[(4^(1 - 1/r) a b Sqrt[π] Gamma[1 + 1/r])/Gamma[1/r + 1/2], 20]] 3.7081493546027438369 

This is the same as the answer Daniel obtained through more general methods.

Since it seems to have not been mentioned yet: yet another way to obtain an approximation of the area of your Lamé curve is to use the shoelace method for computing the area. Here's a Mathematica demonstration:

pts = First[Cases[ ParametricPlot[{Sqrt[Abs[Cos[t]]] Sign[Cos[t]], Sqrt[Abs[Sin[t]]] Sign[Sin[t]]}, {t, 0, 2 π}, Exclusions -> None, Method -> {MaxBend -> 1.}, PlotPoints -> 100] // Normal, Line[l_] :> l, ∞]]; PolygonSignedArea[pts_?MatrixQ] := Total[Det /@ Partition[pts, 2, 1, 1]]/2 PolygonSignedArea[pts] 3.7081447086368127 

The value thus obtained is pretty close to the results in the other answers.

Surprisingly, there is in fact an undocumented built-in function for computing the area of a polygon:

(* $VersionNumber < 10. *) Graphics`Mesh`MeshInit[]; PolygonArea[pts] 3.708144708636812 (* $VersionNumber >= 10. *) Graphics`PolygonUtils`PolygonArea[pts] 3.708144708636812 

But, what you really should know is that Lamé curves have been well studied, and there is in fact a closed form expression for the area of a Lamé curve. Given the Cartesian equation

$$\left|\frac{x}{a}\right|^r+\left|\frac{y}{b}\right|^r=1$$

the formula for the area of a Lamé curve (formula 5 here) is

$$A=\frac{4^{1-\tfrac1{r}}ab\sqrt\pi\;\Gamma\left(1+\tfrac1{r}\right)}{\Gamma\left(\tfrac1{r}+\tfrac12\right)}$$

In particular, for the OP's specific case, $a=b=1$, and $r=4$. Thus,

With[{a = 1, b = 1, r = 4}, N[(4^(1 - 1/r) a b Sqrt[π] Gamma[1 + 1/r])/Gamma[1/r + 1/2], 20]] 3.7081493546027438369 

This is the same as the answer Daniel obtained through more general methods.

Since it seems to have not been mentioned yet: yet another way to obtain an approximation of the area of your Lamé curve is to use the shoelace method for computing the area. Here's a Mathematica demonstration:

pts = First[Cases[ ParametricPlot[{Sqrt[Abs[Cos[t]]] Sign[Cos[t]], Sqrt[Abs[Sin[t]]] Sign[Sin[t]]}, {t, 0, 2 π}, Exclusions -> None, Method -> {MaxBend -> 1.}, PlotPoints -> 100] // Normal, Line[l_] :> l, ∞]]; PolygonSignedArea[pts_?MatrixQ] := Total[Det /@ Partition[pts, 2, 1, {1, 1}]]/2 PolygonSignedArea[pts] 3.7081447086368127 

The value thus obtained is pretty close to the results in the other answers.

Surprisingly, there is in fact an undocumented built-in function for computing the area of a polygon:

Graphics`Mesh`MeshInit[]; PolygonArea[Polygon[pts]] 3.708144708636812 

But, what you really should know is that Lamé curves have been well studied, and there is in fact a closed form expression for the area of a Lamé curve. Given the Cartesian equation

$$\left|\frac{x}{a}\right|^r+\left|\frac{y}{b}\right|^r=1$$

the formula for the area of a Lamé curve (formula 5 here) is

$$A=\frac{4^{1-\tfrac1{r}}ab\sqrt\pi\;\Gamma\left(1+\tfrac1{r}\right)}{\Gamma\left(\tfrac1{r}+\tfrac12\right)}$$

In particular, for the OP's specific case, $a=b=1$, and $r=4$. Thus,

With[{a = 1, b = 1, r = 4}, N[(4^(1 - 1/r) a b Sqrt[π] Gamma[1 + 1/r])/Gamma[1/r + 1/2], 20]] 3.7081493546027438369 

This is the same as the answer Daniel obtained through more general methods.

Since it seems to have not been mentioned yet: yet another way to obtain an approximation of the area of your Lamé curve is to use the shoelace method for computing the area. Here's a Mathematica demonstration:

pts = First[Cases[ ParametricPlot[{Sqrt[Abs[Cos[t]]] Sign[Cos[t]], Sqrt[Abs[Sin[t]]] Sign[Sin[t]]}, {t, 0, 2 π}, Exclusions -> None, Method -> {MaxBend -> 1.}, PlotPoints -> 100] // Normal, Line[l_] :> l, ∞]]; PolygonSignedArea[pts_?MatrixQ] := Total[Det /@ Partition[pts, 2, 1, 1]]/2 PolygonSignedArea[pts] 3.7081447086368127 

The value thus obtained is pretty close to the results in the other answers.

Surprisingly, there is in fact an undocumented built-in function for computing the area of a polygon:

Graphics`Mesh`MeshInit[]; PolygonArea[pts] 3.708144708636812 

But, what you really should know is that Lamé curves have been well studied, and there is in fact a closed form expression for the area of a Lamé curve. Given the Cartesian equation

$$\left|\frac{x}{a}\right|^r+\left|\frac{y}{b}\right|^r=1$$

the formula for the area of a Lamé curve (formula 5 here) is

$$A=\frac{4^{1-\tfrac1{r}}ab\sqrt\pi\;\Gamma\left(1+\tfrac1{r}\right)}{\Gamma\left(\tfrac1{r}+\tfrac12\right)}$$

In particular, for the OP's specific case, $a=b=1$, and $r=4$. Thus,

With[{a = 1, b = 1, r = 4}, N[(4^(1 - 1/r) a b Sqrt[π] Gamma[1 + 1/r])/Gamma[1/r + 1/2], 20]] 3.7081493546027438369 

This is the same as the answer Daniel obtained through more general methods.