This article primarily explores an attempt to convert the area of a closed curve into a line integral without relying on Green’s Theorem. This approach, which utilizes only the coordinate transformation of double integrals, is relatively easy to understand and generalize. As for whether this technique holds practical value, I welcome readers’ comments.
Suppose a simple closed curve on a plane is given by the following parametric equations: \left\{\begin{aligned} x &= f(t)\\ y &= g(t) \end{aligned}\right. where the parameter t lies within a certain interval [a,b], such that f(a)=f(b) and g(a)=g(b). The problem is to find the area of the region enclosed by this closed curve.
Green’s Theorem
The standard approach is to use Green’s Theorem: \iint\limits_D \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right)dxdy = \oint\limits_{\partial D} Pdx+Qdy It tells us that area integrals and line integrals can be transformed into one another. When \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}=1, the left side represents the area of the region. There are many pairs of (P, Q) that satisfy this, such as Q=x, P=0, which gives: \iint\limits_D dxdy = \oint\limits_{\partial D} xdy=\int_a^b f(t)g'(t)dt\label{eq:xdy} Thus, by choosing different P and Q, various formulas for calculating area can be constructed. One formula that is particularly convenient in many cases is: \iint\limits_D dxdy = \frac{1}{2}\oint\limits_{\partial D} -ydx+xdy=\frac{1}{2}\int_a^b [f(t)g'(t)-g(t)f'(t)]dt\label{eq:xdy-ydx} This is because the integral automatically vanishes for lines passing through the origin.
A New Trick: Determinant Transformation
While the use of Green’s Theorem is indeed convenient and impressive, for me, there are a few shortcomings:
1. It is based on Green’s Theorem, which many people (including myself) might perceive as a more advanced or "high-end" topic.
2. How can it be generalized to other coordinate systems (such as polar coordinates) or even other 2D surface coordinate systems (such as geometry on a sphere)? Green’s Theorem does not seem to handle this elegantly. Of course, one could consider Stokes’ Theorem from differential geometry, but that is an even deeper subject.
In fact, I have been contemplating a more "natural" approach that does not rely on Green’s Theorem. As the saying goes, "hard work pays off," and I have found a method that I find quite satisfactory and more intuitive than using Green’s Theorem. I would like to share it here.
What do I mean by "natural"? My idea is to derive the result directly through coordinate transformation. Calculating area is essentially a double integral. If the double integral cannot be solved directly, the obvious thought is to change the coordinate system. This logic is straightforward enough.
Now, to the main point. Without loss of generality, assume the origin lies within the closed curve and that the curve can be proportionally, radially, and without overlap "shrunk" to the origin. Mathematically, this means the region can be represented by the following parametric equations: \left\{\begin{aligned} x &= sf(t)\\ y &= sg(t) \end{aligned}\right. ,\,t\in[a,b],s\in[0,1]\label{eq:x-y-t} We can then use s and t as new coordinates for the integral. The Jacobian determinant is (taking the absolute value if necessary, though this is a specific case): J=\begin{vmatrix}\frac{\partial[sf(t)]}{\partial s} & \frac{\partial[sf(t)]}{\partial t}\\ \frac{\partial[sg(t)]}{\partial s} & \frac{\partial[sg(t)]}{\partial t}\end{vmatrix}=\begin{vmatrix}f(t) & sf'(t)\\ g(t) & sg'(t)\end{vmatrix}=s[f(t)g'(t)-g(t)f'(t)] Therefore: \begin{aligned} \iint\limits_D dxdy &= \int_a^b \int_0^1 s[f(t)g'(t)-g(t)f'(t)]dsdt \\ &= \frac{1}{2}\int_a^b [f(t)g'(t)-g(t)f'(t)]dt \end{aligned} We have arrived at formula [eq:xdy-ydx] once again!
What is the essence of this process? We assumed that the closed curve could shrink to the origin in a certain way without overlap, allowing for a coordinate transformation. By changing the method of shrinking, we can obtain different area formulas. For example: \left\{\begin{aligned} x &= sf(t)\\ y &= g(t) \end{aligned}\right. ,\,t\in[a,b],s\in[0,1] This represents the curve being compressed onto the y-axis without overlap. Through coordinate transformation, we find the Jacobian determinant to be J=f(t)g'(t), thus: \begin{aligned} \iint\limits_D dxdy &= \int_a^b \int_0^1 f(t)g'(t)dsdt \\ &= \int_a^b f(t)g'(t)dt \end{aligned} We have arrived at formula [eq:xdy] again!
In other words, every formula corresponds to a specific way of shrinking. From Green’s Theorem, we know these formulas are equivalent. However, if we do not rely on Green’s Theorem, these formulas are not necessarily equivalent in derivation but rather have their own specific uses, because not all closed curves can be shrunk to the origin without overlap using the same method. For more general curves, such as those where the origin is not inside, one can handle them through partitioning or translation. The final result remains unchanged—in other words, the area of a shape is independent of its position.
Polar Coordinates
From the discussion above, it might seem that this new trick doesn’t yield more results than Green’s Theorem. Why bother developing such a method? One reason is that the coordinate transformation method can be easily generalized to other coordinate systems, whereas applying Green’s Theorem directly is not always convenient.
Let’s consider polar coordinates. Suppose a simple closed curve is described by the parametric equations: \left\{\begin{aligned} r &= f(t)\\ \theta &= g(t) \end{aligned}\right. We want to find the area within the curve.
Naturally, we can consider how the curve shrinks to the origin. Again, without loss of generality, assume the origin is inside the closed curve and that the curve can be proportionally, radially, and without overlap "shrunk" to the origin. Mathematically, the region is described by: \left\{\begin{aligned} r &= sf(t)\\ \theta &= g(t) \end{aligned}\right. ,\,t\in[a,b],s\in[0,1]\label{eq:r-s-t} Comparing [eq:r-s-t] with [eq:x-y-t], one can see the difference between standard Cartesian coordinates and polar coordinates.
The Jacobian determinant of [eq:r-s-t] is J=f(t)g'(t), so the area of the region is: \begin{aligned} \iint\limits_{D}rdrd\theta&=\int_a^b\int_0^1 sf(t)\cdot f(t)g'(t)dsdt\\ &=\frac{1}{2}\int_a^b f^2 (t)g'(t)dt\\ &=\frac{1}{2}\oint\limits_{\partial D}r^2 d\theta\end{aligned}\label{eq:r2ds} The final part of [eq:r2ds] transforms the area within a closed curve in polar coordinates into a line integral in polar coordinates, similar to formulas [eq:xdy] or [eq:xdy-ydx]. For curves that cannot be shrunk in this manner, they can be partitioned, and the final result remains [eq:r2ds].
It might seem that the result in [eq:r2ds] is trivial, as integrating \iint\limits_{D}rdrd\theta with respect to dr first yields a similar result. However, note that the final \frac{1}{2}\oint\limits_{\partial D}r^2 d\theta is a line integral that works for any curve, whereas the feasibility of integrating \iint\limits_{D}rdrd\theta with respect to dr first depends on the shape of the curve.
Spherical Coordinates
Finally, we conclude this article with the case of spherical coordinates. A sphere is an example of a two-dimensional surface. On a sphere, a closed curve can be determined by two parameters \varphi and \theta: \left\{\begin{aligned} \varphi &= f(t)\\ \theta &= g(t) \end{aligned}\right. Here \varphi and \theta correspond to the 3D spherical coordinate transformation: \left\{\begin{aligned} x &= r\sin\varphi\cos\theta\\ y &= r\sin\varphi\sin\theta\\ z &= r\cos\varphi \end{aligned}\right. Following the logic of [eq:r-s-t], we can obtain the following shrinking method (in spherical coordinates, the role of \varphi is analogous to r in polar coordinates): \left\{\begin{aligned} \varphi &= sf(t)\\ \theta &= g(t) \end{aligned}\right. ,\,t\in[a,b],s\in[0,1]\label{eq:t-v-t} Thus: \begin{aligned} \iint\limits_{D}\sin \varphi d\varphi d\theta&=\int_a^b\int_0^1 \sin[sf(t)]\cdot f(t)g'(t)dsdt\\ &=-\int_a^b \left.\cos[sf(t)]\right|_0^1 \cdot g'(t)dt\\ &=\int_a^b [1 - \cos f(t)] \cdot g'(t)dt\\ &=\oint\limits_{\partial D} (1 - \cos \varphi) d\theta\end{aligned}\label{eq:sinds}
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