By 苏剑林 | August 30, 2015
This article presents an attempt to achieve the conversion between the area of a closed curve and a line integral without relying on Green's Theorem. This approach is relatively easy to understand and generalize because it only utilizes the coordinate transformation of double integrals. As for whether this technique truly has practical value, I invite readers to comment. Suppose a simple closed curve on a plane is given by the following parametric equations:
\begin{equation}\left\{\begin{aligned}x = f(t)\\y = g(t)\end{aligned}\right.\end{equation}
where the parameter $t$ lies within some interval $[a, b]$, meaning $f(a)=f(b)$ and $g(a)=g(b)$. The problem now is to find the area of the region enclosed by this closed curve.
Green's Theorem
The usual approach to solve this is by using Green's Theorem:
\begin{equation}\iint\limits_D \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right)dxdy = \oint\limits_{\partial D} Pdx+Qdy \end{equation}
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 is simply the area within the region. There are many such $P$ and $Q$ that satisfy this, such as $Q=x$ and $P=0$, then:
\begin{equation}\iint\limits_D dxdy = \oint\limits_{\partial D} xdy=\int_a^b f(t)g'(t)dt\label{eq:xdy}\end{equation}
Thus, by using different $P$ and $Q$, one can construct various formulas for calculating the area. Among them, a formula that is often more convenient in many cases is:
\begin{equation}\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}\end{equation}
This is because the integral automatically becomes 0 for straight lines passing through the origin.
New Technique: Determinant Transformation
While utilizing Green's Theorem as described above is indeed convenient and impressive, for me, there are a few shortcomings:
- The foundation is Green's Theorem, which many people (including myself) consider to be relatively advanced content;
- How can this 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 very well; of course, one could consider the Stokes' Theorem in differential geometry, but that is even more advanced content.
In fact, I have been contemplating a more "natural" approach that does not rely on Green's Theorem. Hard work pays off, and I have found a solution that is quite satisfactory—more natural than using Green's Theorem—which I would like to share with everyone.
What do I mean by "natural"? My idea is to derive the result directly through integral transformation. Since calculating an area is essentially a double integral, if the double integral cannot be solved directly, why not consider a coordinate transformation? This line of thought is quite obvious, right? Okay, let's get to the point. Without loss of generality, assume the origin is located inside the closed curve, and this curve can "shrink" towards the origin proportionally, directly, and non-overlappingly. Expressed mathematically, this means the region can be represented by the following parametric equations:
\begin{equation}\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}\end{equation}
In this way, we can use $s$ and $t$ as the new coordinates for integration. The Jacobian determinant is (an absolute value might be needed, but this is a special case):
\begin{equation}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)]\end{equation}
So,
\begin{equation}\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}\end{equation}
We have once again arrived at formula $\eqref{eq:xdy-ydx}$! What was the process described above? We find that by assuming the closed curve can be transformed by shrinking non-overlappingly toward the origin in some manner, we can transform the coordinate system. By changing the method of shrinking, we can obtain different area formulas, such as:
\begin{equation}\left\{\begin{aligned}x = sf(t)\\y = g(t)\end{aligned}\right. ,\,t\in[a,b],s\in[0,1]\end{equation}
This represents a curve that can be compressed non-overlappingly onto the $y$-axis. Through integral transformation, we find the Jacobian determinant to be $J=f(t)g'(t)$, so:
\begin{equation}\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}\end{equation}
We have once again obtained formula $\eqref{eq:xdy}$! In other words, every formula corresponds to a specific method of shrinking. Of course, as can be derived from Green's Theorem, these formulas are all equivalent. However, without the help of Green's Theorem, these formulas are not strictly equivalent—they each have their own specific uses, because not all closed curves can shrink non-overlappingly to the origin using the same method. As for more generalized curves, such as when the origin is not within the curve, they can be handled through segmentation or translation, and the final result remains unchanged—in other words, the area of a shape is independent of its location.
Polar Coordinates
From the discussion above, it seems the results obtained by this new technique are no more numerous than those from Green's Theorem. Why labor over such a method? One reason is that the coordinate transformation method described above can be easily generalized to other coordinate systems, whereas applying Green's Theorem directly is not particularly convenient.
Let's consider polar coordinates. Suppose a simple closed curve is described by parametric equations in polar coordinates as follows:
\begin{equation}\left\{\begin{aligned}r = f(t)\\\theta = g(t)\end{aligned}\right.\end{equation}
and 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 within the closed curve, and assume this curve can shrink to the origin proportionally, directly, and non-overlappingly. Mathematically, this region is described by the following parametric equations:
\begin{equation}\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}\end{equation}
Comparing formula $\eqref{eq:r-s-t}$ with $\eqref{eq:x-y-t}$, we can see the difference between standard Cartesian coordinates and polar coordinates. The Jacobian of $\eqref{eq:r-s-t}$ is $J=f(t)g'(t)$, thus the area of the region is:
\begin{equation}\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}\end{equation}
The end of formula $\eqref{eq:r2ds}$ converts the area within a closed curve in polar coordinates into a line integral in polar coordinates, similar to formula $\eqref{eq:xdy}$ or $\eqref{eq:xdy-ydx}$. For curves that cannot shrink in this way, they can be treated by segmentation, and the final result remains $\eqref{eq:r2ds}$.
It might appear that the result of formula $\eqref{eq:r2ds}$ is trivial, as integrating over $dr$ first in $\iint\limits_{D}rdrd\theta$ yields a similar result. However, it is important to 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 $dr$ first in $\iint\limits_{D}rdrd\theta$ depends on the shape of the curve.
Spherical Coordinates
Finally, we conclude this article with the case in spherical coordinates. The sphere is an example of a two-dimensional surface. On a sphere, a closed curve can be determined by two parameters $\varphi$ and $\theta$:
\begin{equation}\left\{\begin{aligned}\varphi = f(t)\\ \theta = g(t)\end{aligned}\right.\end{equation}
Here $\varphi, \theta$ correspond to the 3D spherical coordinate transformation:
\begin{equation}\left\{\begin{aligned}&x = r\sin\varphi\cos\theta\\
&y = r\sin\varphi\sin\theta\\
&z = r\cos\varphi
\end{aligned}\right.\end{equation}
Following formula $\eqref{eq:r-s-t}$, we can obtain the following shrinking method (in spherical coordinates, the role of $\varphi$ is similar to $r$ in polar coordinates):
\begin{equation}\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}\end{equation}
Therefore,
\begin{equation}\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}\end{equation}
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