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Question: What is the area of the interior of the simple closed curve described by the equation

$\sqrt{(x-2)^2+(y-3)^2} + 2\sqrt{(x-3)^2+(y-1)^2} = 4$?

Comments: I came up with this specific problem myself in response to my earlier question, which I don't think was well posed, or at least it was not clear what I was after: to see how to find the area of the interior of a Jordan curve that is described by an implicit function. To see how this area looks like, I uploaded a picture from WolframAlpha:

enter image description here

As you can see, it is quite egg-like. We can generalize; the object could be called a weighted ellipse [edit: usually called a Cartesian oval] with an equation of the form

$\sqrt{(x-x_1)^2+(y-y_1)^2} + a \cdot \sqrt{(x-x_2)^2+(y-y_2)^2} = k$

where $(x_1,y_1)$ and $(x_2,y_2)$ are the Cartesian coordinates of the focal points of this weighted ellipse and $a$ is a weight (it equals $1$ in the case of an ordinary ellipse). As a bonus question, I would like to see how to find the area of this object.

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  • $\begingroup$ Without loss of generality we may assume $x_1=y_1=y_2=0$ and $x_2=1$. $\endgroup$ – apt1002 Nov 19 '13 at 1:51
  • $\begingroup$ That gives us the equation $\sqrt{x^2+y^2} + 2\sqrt{(x-1)^2+y^2} = 4$. This curve looks quite different from the one given, are you sure that the interior will have the same area? $\endgroup$ – Sid Nov 19 '13 at 2:03
  • $\begingroup$ I think you have failed to scale the "4". You need to divide it by $\sqrt5$. After computing the area, you will need to scale it back up by multiplying by 5. I'm afraid I still don't know how to compute the area, but this transformation certainly makes it easier. $\endgroup$ – apt1002 Nov 19 '13 at 3:32
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    $\begingroup$ All three terms in the defining equation of your curve are of a form that is preserved by rotations and enlargements. Rotations and enlargements can be used to move the two foci anywhere you want. The $\sqrt(5)$ is the distance between $(2, 3)$ and $(3, 1)$, i.e. the reciprocal of the enlargement factor. $\endgroup$ – apt1002 Nov 19 '13 at 3:55
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    $\begingroup$ This curve is called Cartesian oval first studied by René Descartes in 1637. I can't find a formula of its area online but I suspect it can be expressed in terms of elliptic integral. $\endgroup$ – achille hui Nov 19 '13 at 6:53
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This is an incomplete answer I have only covered the case where an oval is "large" in the sense it contains both of its foci in reasonable details.

To simplify the analysis, let start with $(x_1,y_1) = (-1,0), (x_2, y_2) = (1,0)$ and define $r, s, u, v$ such that

$$ \begin{cases}r^2 = (x+1)^2 + y^2\\s^2 = (x-1)^2 + y^2\end{cases} \quad\text{ and }\quad\begin{cases}u = \frac12 (r-s)\\v = \frac12(r+s)\end{cases} $$ We will look at the area of half of an oval in the upper half plane given by

$$\mathscr{O} = \big\{\; (x,y) \in \mathbb{R}^2 : y \ge 0,\, r + a s \le k \;\big\}$$

WOLOG, we can assume $a \ge 1$. Geometrically, there are three 3 possibilities:

  1. $k > 2 a$ - The oval is "large" in the sense it encloses both of its foci.
  2. $k = 2 a$ - The oval turns into a teardrop with a cusp at the focus $(-1,0)$.
  3. $2a > k > 2$ - The oval encloses one and only one of the focus $(1,0)$.

In the following analysis, we will concentrate on the $1^{st}$ case where $k > 2 a$.

Notice $u^2 + v^2 = \frac12 (r^2 + s^2) = x^2 + 1 + y^2$. On the upper half plane, we have $$ \begin{cases} x = \frac14 (r^2 - s^2) = uv\\ \\ y = \sqrt{u^2 + v^2 - u^2 v^2 -1 } = \sqrt{(1-u^2)(v^2-1)} \end{cases}$$

The area element becomes

$$dx \wedge dy = \frac{dx \wedge d y^2}{2y} = \frac{duv \wedge d(u^2 + v^2)}{2y} = \frac{(u dv + v du)\wedge(u du + v dv)}{\sqrt{(1-u^2)(v^2-1)}}\\ = \frac{(v^2 - u^2) du \wedge dv}{\sqrt{(1-u^2)(v^2-1)}} = \frac{( (2v^2-1)+(1-2u^2)) du\wedge dv}{2\sqrt{(1-u^2)(v^2-1)}} $$ Notice $$d u\sqrt{1-u^2} = \frac{1-2u^2}{\sqrt{1-u^2}}\quad\text{ and }\quad d v\sqrt{v^2-1} = \frac{2v^2-1}{\sqrt{v^2-1}} $$ We get $$dx \wedge dy = \frac12 d \left[\frac{u(1-u^2)dv - v(v^2-1)du}{\sqrt{(1-u^2)(v^2-1)}}\right]$$

When $k > 2a$, the half oval $\mathscr{O}$ get mapped to a quadrilateral $\mathscr{Q}$ in the $uv$-plane.

$$\mathscr{Q} = \big\{\; (u,v) \in \mathbb{R}^2 : -1 \le u \le 1; 1 \le v \le \alpha + \beta u \;\}$$ where $ \displaystyle \alpha = \frac{k}{a+1},\;\beta = \frac{a-1}{a+1}$. This gives us

$$2 \text{Area}(\mathscr{O}) = 2\int_{\mathscr{O}} dx \wedge dy = \int_{\partial\mathscr{Q}} \frac{u(1-u^2)dv - v(v^2-1)du}{\sqrt{(1-u^2)(v^2-1)}} $$ It is easy to check on $\partial\mathscr{Q}$, the horizontal and the two vertical edges contribute nothing.
As a result, the area of the full oval can be rewritten as

$$2 \text{Area}(\mathscr{O}) = \int_{-1}^1 \frac{(\alpha+\beta u)((\alpha+\beta u)^2-1) - \beta u(1-u^2)}{\sqrt{(1-u^2)((\alpha+\beta u)^2-1)}} du\tag{*1} $$ Abusing the notation, we will use $v$ as a shorthand for $\alpha + \beta u$ from now on.

Consider the quadratic equation $t = (\alpha+\beta t)(\alpha t + \beta)$. When $k > 2a$, one can verify above equation has two real roots $\eta$ and $\eta^{-1}$ for $t$. We will let $\eta$ be the unique root with $|\eta| < 1$. Define $\lambda, \mu$ and introduce a new variable $z$ such that

$$\begin{cases} \lambda &= \alpha + \beta \eta\\ \mu &= \alpha \eta + \beta \end{cases} \quad\text{ and }\quad u = \frac{z + \eta}{1 + \eta z}\;\;\iff\;\; z = \frac{u - \eta}{1 - \eta u} $$ One can show that $|\mu| < 1$ and we have $$ du = \frac{(1-\eta^2)}{(1+\eta z)^2}dz,\quad \begin{cases} u & = \frac{z+\eta}{1+\eta z }\\ 1 + u & = \frac{(1+\eta)(1+z)}{1+\eta z}\\ 1 - u & = \frac{(1-\eta)(1-z)}{1+\eta z} \end{cases} \quad\text{ and }\quad \begin{cases} v &= \frac{\lambda + \mu z}{1 + \eta z}\\ v + 1 &= \frac{(\lambda+1)(1 + \mu z)}{1+\eta z}\\ v - 1 &= \frac{(\lambda-1)(1 - \mu z)}{1+\eta z} \end{cases}$$ This implies

$$\begin{align} v(v^2 - 1) & = \frac{\lambda^2 - 1}{(1+\eta z)^3}(\lambda + \mu z)(1 - \mu^2z^2)\\ u(1-u^2) &= \frac{1 - \eta^2}{(1+\eta z)^3}( z + \eta )(1 - z^2)\\ \frac{du}{\sqrt{(1-u^2)(v^2-1)}} &= \sqrt{\frac{1-\eta^2}{\lambda^2-1}}\frac{dz}{\sqrt{(1-z^2)(1-\mu^2 z^2)}} \end{align}$$

Notice $\beta( 1 - \eta^2 ) = \mu - \lambda \eta = \mu ( 1 - \lambda^2)$, we can rewrite $(*1)$ as

$$2 \text{Area}(\mathscr{O}) = \sqrt{(1-\eta^2)(\lambda^2-1)} \int_{-1}^1 \frac{P(z) dz}{ (1+\eta z)^3 \sqrt{(1-z^2)(1-\mu^2 z^2)}}\tag{*2}$$

where $P(z) = (\lambda + \mu z)(1 -\mu^2 z^2) + \mu (z + \eta)(1-z^2)$.

By brute force, one can show that

$$\frac{P(z)}{(1+\eta z)^3} = A(\eta,\mu) + \left[ B(\eta,\mu) + C(\eta,\mu)\frac{\partial}{\partial \eta} + D(\eta,\mu)\frac{\partial^2}{\partial \eta^2}\right]\frac{1}{1+\eta z}$$

where $$\begin{cases} A(\eta,\mu) &= -\frac{\mu^3+\mu}{\eta^3},\\ B(\eta,\mu) &= \frac{\mu^4 + (\eta^4+1)\mu^2+\eta^4}{\eta^3\mu},\\ C(\eta,\mu) &= -\frac{{\mu}^{4}+\left( -2\,{\eta}^{4}+2\,{\eta}^{2}+1\right) \,{\mu}^{2}-2\,{\eta}^{4}}{{\eta}^{2}\,\mu}\\ D(\eta,\mu) &= \frac{{\mu}^{4}+\left( {\eta}^{4}-4\,{\eta}^{2}+1\right) \,{\mu}^{2}+{\eta}^{4}}{2\,\eta\,\mu} \end{cases}$$

Substitute this in $(*2)$, then in terms of following complete elliptic integrals of $1^{st}, 2^{nd}$ and $3^{rd}$ kind:

$$\begin{align} K(\mu) &= \int_0^1 \frac{dz}{\sqrt{(1-z^2)(1-\mu^2 z^2)}}\\ E(\mu) &= \int_0^1 \sqrt{\frac{1-\mu^2 z^2}{1-z^2}} dz\\ \Pi(\eta^2,\mu) &= \int_0^1 \frac{dz}{(1 - \eta^2 z^2)\sqrt{(1-z^2)(1-\mu^2 z^2)}} \end{align}$$

We have $$2\text{Area}(\mathscr{O}) = 2\sqrt{(1-\eta^2)(\lambda^2-1)} \left( A K(\mu) + \left[ B + C \frac{\partial}{\partial \eta} + D\frac{\partial^2}{\partial \eta^2}\right]\Pi(\eta^2,\mu) \right)$$ Please note that the partial derivatives of $\Pi(\eta^2,\mu)$ can be expressed in terms of the complete elliptic integrals itself.

$$\frac{\partial\Pi(n,\mu)}{\partial n} = \frac{1}{2n(\mu^2 - n)(n-1)}\left(n E(\mu) + (\mu^2 - n)K(\mu) + (n^2 - \mu^2)\Pi(n,\mu)\right)$$

So in principle, we can get rid of the partial derivatives above and express the area of the oval solely in terms of complete elliptic integrals directly.

Let us switch to the $3^{rd}$ case where $2 < k < 2a$. The oval now intersect the line segment joining the two foci and encloses one of the focus $(1,0)$.

Let $\displaystyle \sigma = \frac{1-\alpha}{\beta}$. On the $uv$-plane, the line $v = \alpha + \beta u$ intersect the line $v = 1$ at $( \sigma, 1 )$ before the line $u = -1$. The half oval $\mathscr{O}$ get mapped to a triangle $\mathscr{T}$:

$$\mathscr{T} = \big\{\; (u,v) \in \mathbb{R}^2 : u \le 1; 1 \le v \le \alpha+\beta u\;\big\}$$

Once again, we can express the area of half oval as a line integral over $\partial \mathscr{T}$ and the horizontal and vertical edges contributes nothing. As a result, the area of the full oval is given by

$$2\text{Area}(\mathscr{O}) = \int_\sigma^1 \frac{(\alpha+\beta u)((\alpha + \beta u)^2 - 1) - \beta u (1-u^2) }{\sqrt{(1-u^2)((\alpha + \beta u)^2-1)}}\tag{*1'} $$

Since the lower limit $\sigma$ now depends on $\alpha, \beta$. We need a different Mobius transform which map between $u$ and a new variable $\tilde{z}$ such that $$\begin{array}{rcr} u& &\tilde{z}\\ \frac{-1-\alpha}{\beta} & \longleftrightarrow & -\frac{1}{\tilde{\mu}}\\ -1 & \longleftrightarrow & \frac{1}{\tilde{\mu}}\\ \sigma = \frac{1-\alpha}{\beta} & \longleftrightarrow & -1\\ 1 & \longleftrightarrow & 1 \end{array}$$ to bring the integrand back to the canonical form. If one perform this substitution, one will obtain something very similar to $(*2)$ above. The derivation will be very complicated and I will stop here. However, it is sort of clear if one do the dirty work along this direction, ultimately we will be able to express the area of oval again in terms of complete elliptic integrals.

Update

Back to original problem. Notice the distance between $(2,3)$ and $(3,1)$ is $\sqrt{5}$. If we scale the oval by a factor $\frac{2}{\sqrt{5}}$, it will reduce to the case 3. we have covered before with following parameters:

$$( k, a ) = (\frac{8}{\sqrt{5}}, 2 )\quad\implies\quad(\alpha,\beta,\sigma) = ( \frac{8}{3\sqrt{5}}, \frac13, 3 -\frac{8}{\sqrt{5}})$$

The area of the original oval is given by the integral in $(*1')$ scaled by a factor $\left(\frac{\sqrt{5}}{2}\right)^2$:

$$\text{Area} = \frac{5}{4} \int_{3 -\frac{8}{\sqrt{5}}}^1 \frac{\left( {\left( \frac{u}{3}+\frac{8}{3\,\sqrt{5}}\right) }^{2}-1\right) \,\left( \frac{u}{3}+\frac{8}{3\,\sqrt{5}}\right) -\frac{u\,\left( 1-{u}^{2}\right) }{3}}{\sqrt{\left( {\left( \frac{u}{3}+\frac{8}{3\,\sqrt{5}}\right) }^{2}-1\right) \,\left( 1-{u}^{2}\right) }} du$$

Throwing this expression to WA gives us $$3.2040508691540993294644483966405124836692674716427904\cdots$$

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Minor progress: The integral in question is $$\int_{\gamma} y dx$$ where $\gamma$ runs around the Cartesian oval. To see this, break the integral up into $\gamma_1$ and $\gamma_2$ where $\gamma_1$ runs over the top of the oval and $\gamma_2$ over the bottom, both oriented from left to right. Then $$\int_{\gamma} y dx = \int_{\gamma_1} y dx - \int_{\gamma_2} y dx$$ and the second integral is the standard formula for the area between two curves.

The curve $\gamma$ is one of two connected components of the plane curve $$-711 - 184 x + 402 x^2 - 120 x^3 + 9 x^4 + 340 y + 80 x y - 12 x^2 y + 6 y^2 - 120 x y^2 + 18 x^2 y^2 - 12 y^3 + 9 y^4=0.$$

We can homogenize this to $$9 X^4 + 18 X^2 Y^2 + 9 Y^4 - 120 X^3 Z - 12 X^2 Y Z - 120 X Y^2 Z - 12 Y^3 Z + 402 X^2 Z^2 + 80 X Y Z^2 + 6 Y^2 Z^2 - 184 X Z^3 + 340 Y Z^3 - 711 Z^4=0$$

A degree $4$ curve usually has genus $3$. However, this curve has nodes at $(X:Y:Z) = (1:\pm i: 0)$, so it has genus $1$. If I cared enough about this problem, I would find a rational change of coordinates turning this into a cubic curve in Weierstrass form. The integral $y dx$ would then presumably be some sort of elliptic integral.

UPDATE:

Since distance is rotation symmetric, we can take any two points which are $\sqrt{5}$ apart and use them in place of $(2,3)$ and $(3,1)$. I choose the equation $$2 \sqrt{(x-r)^2 + y^2} + \sqrt{(x-r-\sqrt{5})^2 + y^2} = 4$$ where $r = (4-\sqrt{5})/3$. The reason for this choice is that it puts the point $(0,0)$ on the curve. Eliminating the radicals, this is $$128 x + (256 \sqrt{5} x)/3 + 16 x^2 - 96 \sqrt{5} x^2 - 48 x^3 + 24 \sqrt{5} x^3 + 9 x^4 - 128 y^2 - 32 \sqrt{5} y^2 - 48 x y^2 + 24 \sqrt{5} x y^2 + 18 x^2 y^2 + 9 y^4 \quad (\ast)$$ Again, we want to integrate $y dx$ around the connected component through $(0,0)$.

I make the substitution $u=x/(x^2+y^2)$, $v=y/(x^2+y^2)$. (Motivation: If we consider the homogenization of $(\ast)$, it is a degree $4$ equation passing through $(0:0:1)$ and with nodes at $(1:\pm i : 0 )$. Any conic through these points will intersect the curve at $3$ other points, since each node is a double intersection. A basis for the vector space of such conics is $XZ$, $YZ$, $X^2+Y^2$. So mapping the degree $4$ curve to $\mathbb{P}^2$ by $(X:Y:Z) \mapsto (XZ:YZ:X^2+Y^2)$ should have image a degree $3$ curve. Dehomogenizing, we want $(x,y) \mapsto (x/(x^2+y^2), y/(x^2+y^2))$.)

Conveniently, we can invert $u=x/(x^2+y^2)$, $v=y/(x^2+y^2)$ to give $x=u/(u^2+v^2)$, $y=v/(u^2+v^2)$. Plugging this into $(\ast)$ and clearing denominators, we now have the curve: $$(-384 - 96 \sqrt{5} + 384 u + 256 \sqrt{5} u) v^2 + (27 - 144 u + 72 \sqrt{5} u + 48 u^2 - 288 \sqrt{5} u^2 + 384 u^3 + 256 \sqrt{5} u^3)=0$$

Hooray! It's cubic, and almost in Weierstrass form. We now want to integrate, not $y dx$, but $$\frac{v}{u^2 + v^2} d \left( \frac{u}{u^2+v^2} \right) = \frac{v\left( (u^2+v^2 ) du - 2(u du + v dv)\right)}{(u^2+v^2)^3}.$$

Anyone want to turn this into a standard elliptic integral and put it out of its misery?

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Here's the answer using polar coordinates. It turns out that the implicit curve is quadratic in $r$ in polar coordinates which means $r$ can be solved for in terms of the angle $t$ using the quadratic formula. Finally, using $A=\frac{1}{2} \int_0^{2 \pi} r(t)^2 dt$ allows for the area to be computed.

For simplicity, take $x_1=y_1=0$ since we can place the origin anywhere. We start with $$a \sqrt{x^2+y^2}+\sqrt{(x-x_2)^2+(y-y_2)^2}=k.$$ Notice I swapped where the $a$ occurs. It will make some slightly algebra nicer. Substituting $x=r \cos t$ and $y= r \sin t$ gives $$\sqrt{(r\cos t-x_2)^2+(r \sin t-y_2)^2}=k-ar$$ and after squaring $$r^2 - r(2 x_1 \cos t +2 y_2 \sin t+1)+(x_1^2+y_2^2)= k^2-2kar + a^2r^2.$$

With dependencies on $t$, this function has the form $$r^2-r(A\cos t + B \sin t + C) + D = 0$$ where $A = \frac{2x_1}{1-a^2}$, $B = \frac{2y_2}{1-a^2}$, $C = \frac{1+2k}{1-a^2}$, $D = \frac{x_1^2+y_1^2-k^2}{1-a^2}$. Given our initial parameters, perhaps there are bounds on $A,B,C$ and $D$. I don't care. I'll take the above as the new definition of our curve and find the area in terms of the new parameters making assumptions as I go.

Using the quadratic formula yields $$ r = \frac{1}{2}\left[(A\cos t + B \sin t + C)\pm \sqrt{(A\cos t + B \sin t + C)^2-4D} \right].$$

We have the form $r(t)=p(t) \pm \sqrt{q(t)}$ for $r$. Assume parameters are chosen so that $q(t)>0$. If we $D<0$ we have one branch, otherwise two curves satisfy the original equation. Then the area of the region enclosed is given by

$$A = \frac{1}{2}\int_0^{2\pi} r(t)^2 dt = T_1 + T_2 + T_3,$$ with $T_1 = \frac{1}{2}\int_0^{2\pi} 2 p(t)^2 dt$, $T_3 = \pm\frac{1}{2}\int_0^{2\pi} p(t) \sqrt{q(t)} dt$, and $T_3 = \frac{1}{2} \int_0^{2\pi} q(t) dt$.

More explicitly, this is $$T_1 = \frac{1}{8} \int_0^{2\pi} (A \cos t+B \sin t+C)^2 = \frac{1}{8}(\frac{2 \pi A^2}{2} + \frac{2 \pi B^2 }{2} + 2\pi C) = \frac{\pi}{8}(A^2+B^2+C)$$

$$T_2 = \pm \frac{1}{4} \int_0^{2\pi} (A\cos t + B \sin t + C) \sqrt{(A\cos t + B \sin t + C)^2-4D} \; dt$$

$$T_3 = \frac{1}{8} \int_0^{2\pi} \left[(A\cos t + B \sin t + C)^2-4D \right] dt = \frac{\pi}{8}(A^2+B^2+C-4D)$$

As seen, the first and third terms are easily computable. The second requires work. Because $A \cos t + B \sin t = \alpha \sin(t+\delta)$ for $\alpha = \sqrt{A^2+B^2}$ and $\delta$ depending on $A$ and $B$, we can substitute this in the expression for $T_2$ and do a change of variables to get

$$T_2 = \pm\frac{1}{4} \int_0^{2\pi} (\sqrt{A^2+B^2} \sin t + C) \sqrt{(\sqrt{A^2+B^2} \sin t + C)^2-4D} \; dt.$$

I highly doubt the antiderivative of the above can be expressed in terms of elementary functions. Perhaps someone can work this into an expression involving elliptic integrals.

If I get to it, I'll work out the degenerate case of an ellipse. In that situation, my guess is that the $T_2$ term vanishes. I've almost certainly made an error or two--please let me know if I did. However, the overall approach is sound.

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  • $\begingroup$ Regarding the degenerate case of an ellipse, then we have that $a=1$ so $A,B$ etc are undefined. However, I suppose that the third equation simplifies (it must do so because we have a rather simple expression for the area of an ellipse given the coordinates of its foci and the necessary constant). I think you have a nice approach but I am surprised that the degree of $r$ lowers when we change the coordinate system. Could we find a coordinate system when it is linear? $\endgroup$ – Sid Nov 24 '13 at 21:29
  • $\begingroup$ I'm not sure I believe that $T_2$ must simplify. In general, finding the arbitrary area enclosed by a level curve should be hard. It's like solving an arbitrary ODE or finding the antiderivative in elementary terms of some random function. You have to use the tricks that work. Switching to polar coordinates just works in this case. What do you mean about a coordinate system when it is linear? $\endgroup$ – abnry Nov 25 '13 at 12:36
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Simple minded approach.
Take an area of $3500 \times 3500$ pixels (not really) and count the pixels inside the curve.

DEPRECATED

VERY LATE UPDATE (2021).

Take an area of $3^{13} \times 3^{13}$ pixels (not really) and count the pixels inside the curve. The following method is supposed to work for any convex closed (Jordan) curve that divides the plane.
Start with a grid of $3 \times 3$ squares and refine the grid three times each iteration.
Leave out a square as soon as all four vertices of it are guaranteed to be inside or outside the curve.
Continue (recursively) with the remaining squares, until the maximum grid refinement is reached.
Pictures say more that a thousand words:

enter image description here

Free accompanying (Delphi Pascal) source code available at this place:

Output after 13 iterations: $$ 3.204044 < \mbox{Area} < 3.204057 $$ Which is quite in agreement with (and inferior to) the above answer given by achille hui.

There is also a version available with binary instead of ternary refinement.
In addition, the software has been enhanced with Isoline Refinement. The meaning of the latter is best explained with a picture ($\color{red}{red}$ line):

enter image description here

With isoline refinement, a much better approximation of achille hui's result is found. Agreement is now within 12 digits. As a bonus, a value for the length of the perimeter is obtained as well:

Area = 3.20405086915288E+0000
Perimeter = 6.457419618 86746E+0000
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from geometry point, the equation $\sqrt{(x-2)^2+(y-3)^2} + 2\sqrt{(x-3)^2+(y-1)^2} = 4$ is equal to:

$\sqrt{x^2+y^2} + 2\sqrt{(x-\sqrt{5})^2+y^2} = 4$ or $\sqrt{(x-\sqrt{5})^2+y^2} + 2\sqrt{x^2+y^2} = 4$ which is symmetry to $x$ axis

so for first one:

$y = \dfrac{ \sqrt{-9 x^2+24 \sqrt{5} x-16 \sqrt{6 \sqrt{5} x+1}+20}}{3}$ for the half area.

but the integration is difficult. I have no idea except the numeric method.

edit: the area is:

$2\int_{x_1}^{x_2}ydx=3.2,x_1=2 (\sqrt{5}-2),x_2=\dfrac{2 (\sqrt{5}+2)}{3}$

and for general a,b,if it is not rational,it is complex. if it is rational, then we can simplify in same idea.

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    $\begingroup$ I think this answer would be better suited as a comment since you don't arrive at a solution. $\endgroup$ – Sid Nov 20 '13 at 17:15
  • $\begingroup$ Thanks for the update, but I'm not sure if I understand how you found this solution. Did you use a numeric method? $\endgroup$ – Sid Nov 21 '13 at 11:19
  • $\begingroup$ it is calculated by numberempire.com/definiteintegralcalculator.php. I had checked some webs and haven't find any possible formula for this. I am sure it is numeric method. elliptic integral seems had big difference for this. But I am not expert so you may wait for other answers. this is only for reference as it has many text so I didn't move to comments. if you insisted, I will move it. $\endgroup$ – chenbai Nov 22 '13 at 1:43
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    $\begingroup$ @Sid: Don't be silly. This answer wouldn't fit in a comment, would it? $\endgroup$ – TonyK Nov 22 '13 at 23:38
  • $\begingroup$ @chenbai That link you posted seems very useful. It would however be good to know the solution in the general case for which the numerical method does not suffice. I think I have an idea about how to proceed though, I will post it when I have meditated further on this. $\endgroup$ – Sid Nov 23 '13 at 10:21

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