The Area Enclosed by a Closed Curve: A New Trick
This post is mainly an experiment: an attempt to convert between the area enclosed by a closed curve and a line integral without going through Green's theorem. Because this approach relies solely on a change of variables in a double integral, it is relatively easy to understand and easy to generalize. As for whether this trick has any real practical value, I'll leave that to the readers to judge.
Suppose a simple closed curve in the plane is given by the parametric equations:
$$\begin{equation}\left\{\begin{aligned}x = f(t)\\y = g(t)\end{aligned}\right.\end{equation}$$
where the parameter $t$ lies in some interval $[a,b]$, i.e., $f(a)=f(b),g(a)=g(b)$. The question now is: find the area of the region enclosed by this closed curve. more
Green's Theorem
The usual approach is to use 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 a surface integral and a line integral can be converted into each other. When $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}=1$, the left-hand side computes the area of the region. There are many choices of $P,Q$ satisfying this, e.g. $Q=x,P=0$, giving
$$\begin{equation}\iint\limits_D dxdy = \oint\limits_{\partial D} xdy=\int_a^b f(t)g'(t)dt\label{eq:xdy}\end{equation}$$
As we can see, by choosing different $P,Q$'s, we can construct all sorts of formulas for computing area. One that turns out to be particularly convenient in many situations is
$$\begin{equation}\iint\limits_D dxdy = \frac{1}{2}\oint\limits_{\partial D} -ydx+xdy=\frac{1}{2}\int_a^b [f(t)g'(t)-g(t)f'(t)]dt\label{eq:xdy-ydx}\end{equation}$$
because its integral along any line through the origin automatically vanishes.
A New Trick: The Determinant Transformation
Using Green's theorem this way is indeed convenient and practical—impressively so. However, to me it leaves something to be desired:
1. It is built on Green's theorem, which most people (myself included) tend to regard as somewhat advanced material.
2. How do we generalize it to other coordinate systems (such as polar coordinates)? Or even to other two-dimensional surface coordinate systems (such as the geometry on a sphere)? Green's theorem doesn't seem to handle this very gracefully. Of course, one could bring in Stokes' theorem from differential geometry, but that's going even deeper.
In fact, I have long been mulling over a more "natural" approach that doesn't rely on Green's theorem. As they say, effort pays off—I eventually found an approach that I find fairly satisfying, one that feels more natural than using Green's theorem, and I'd like to share it here.
What do I mean by more "natural"? My idea is to derive the result directly via a change of variables. Computing an area is, at its core, a double integral; when a double integral can't be evaluated directly, the natural move is to change coordinates—simple enough, right? Alright, let's get to it. Without loss of generality, suppose the origin lies inside the closed curve, and suppose the curve can be "shrunk" to the origin proportionally, radially, and without overlap. Expressed mathematically, this means the region can be described by the following parametric equations:
$$\begin{equation}\left\{\begin{aligned}x = sf(t)\\y = sg(t)\end{aligned}\right. ,\,t\in[a,b],s\in[0,1]\label{eq:x-y-t}\end{equation}$$
This lets us use $s,t$ as new coordinates for the integral, with Jacobian determinant (an absolute value might be needed in general, but here it's a special case):
$$\begin{equation}J=\begin{vmatrix}\frac{\partial[sf(t)]}{\partial s} & \frac{\partial[sf(t)]}{\partial t}\\ \frac{\partial[sg(t)]}{\partial s} & \frac{\partial[sg(t)]}{\partial t}\end{vmatrix}=\begin{vmatrix}f(t) & sf'(t)\\ g(t) & sg'(t)\end{vmatrix}=s[f(t)g'(t)-g(t)f'(t)]\end{equation}$$
Therefore
$$\begin{equation}\begin{aligned}\iint\limits_D dxdy = &\int_a^b \int_0^1 s[f(t)g'(t)-g(t)f'(t)]dsdt \\ = &\frac{1}{2}\int_a^b [f(t)g'(t)-g(t)f'(t)]dt\end{aligned}\end{equation}$$
And we arrive once again at formula $\eqref{eq:xdy-ydx}$!
What kind of process have we just gone through? We see that we assumed the closed curve could be "shrunk" to the origin, without overlap, in some particular way, and then changed coordinates accordingly. Using a different shrinking scheme yields a different area formula, e.g.,
$$\begin{equation}\left\{\begin{aligned}x = sf(t)\\y = g(t)\end{aligned}\right. ,\,t\in[a,b],s\in[0,1]\end{equation}$$
meaning the curve can be compressed without overlap onto the $y$-axis. In that case, the change of variables gives Jacobian determinant $J=f(t)g'(t)$, so
$$\begin{equation}\begin{aligned}\iint\limits_D dxdy = &\int_a^b \int_0^1 f(t)g'(t)dsdt \\ = &\int_a^b f(t)g'(t)dt \end{aligned}\end{equation}$$
and once again we arrive at formula $\eqref{eq:xdy}$!
In other words, each formula corresponds to a particular way of shrinking the curve. Of course, from Green's theorem we know that all these formulas are equivalent. But if we don't invoke Green's theorem, then these formulas are not automatically equivalent—each has its own domain of applicability, because not every closed curve can be shrunk to the origin without overlap using the same scheme. As for more general curves, e.g., ones where the origin does not lie inside the curve, we can handle them by splitting into pieces or by translating the curve; the final result is unchanged—in other words, the area of a shape doesn't depend on where the shape is located.
Polar Coordinates
From the discussion above, it might seem like the results obtained with this new trick are no more powerful than what Green's theorem already gives us—so why bother going to all this trouble to develop such a method? One reason is that the change-of-variables approach above generalizes fairly easily to other coordinate systems, whereas working directly from Green's theorem is not as convenient.
Let's consider polar coordinates. Suppose a simple closed curve is described by the polar parametric equation
$$\begin{equation}\left\{\begin{aligned}r = f(t)\\\theta = g(t)\end{aligned}\right.\end{equation}$$
and we want to find the area enclosed by the curve.
Naturally, we can think about how the curve shrinks to the origin. Again without loss of generality, suppose the origin lies inside the closed curve, and again suppose the curve can be shrunk to the origin proportionally, radially, and without overlap. Expressed mathematically, the 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 this with formula $\eqref{eq:x-y-t}$, we can see the difference between ordinary Cartesian coordinates and polar coordinates.
The Jacobian determinant of $\eqref{eq:r-s-t}$ is $J=f(t)g'(t)$, so 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 final step in $\eqref{eq:r2ds}$ converts the area enclosed by the closed curve in polar coordinates into a line integral in polar coordinates, which is the same as formula $\eqref{eq:xdy}$ or $\eqref{eq:xdy-ydx}$. For curves that cannot be shrunk in this manner, we can split them into segments; the final result is still given by $\eqref{eq:r2ds}$.
It might look as though the result in $\eqref{eq:r2ds}$ is trivial, since integrating $dr$ over $\iint\limits_{D}rdrd\theta$ first directly gives roughly the same result. But note that the final expression $\frac{1}{2}\oint\limits_{\partial D}r^2 d\theta$ is a line integral, valid for any curve, whereas directly integrating $dr$ over $\iint\limits_{D}rdrd\theta$ first is only feasible for curves of a certain shape.
Spherical Coordinates
Finally, let's close this post with the case of spherical coordinates. A sphere is an example of a two-dimensional surface; on a sphere, a closed curve can be specified using the two parameters $\varphi,\theta$:
$$\begin{equation}\left\{\begin{aligned}\varphi = f(t)\\ \theta = g(t)\end{aligned}\right.\end{equation}$$
Here $\varphi,\theta$ corresponds to the usual three-dimensional 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 the pattern of $\eqref{eq:r-s-t}$, we can construct the following shrinking scheme (in spherical coordinates, $\varphi$ plays a role analogous to $r$ in polar coordinates):
$$\begin{equation}\left\{\begin{aligned}&\varphi = sf(t)\\ &\theta = g(t)\end{aligned}\right. ,\,t\in[a,b],s\in[0,1]\label{eq:t-v-t}\end{equation}$$
and therefore
$$\begin{equation}\begin{aligned}\iint\limits_{D}\sin \varphi d\varphi d\theta=&\int_a^b\int_0^1 \sin[sf(t)]\cdot f(t)g'(t)dsdt\\ =&-\int_a^b \left.\cos[sf(t)]\right|_0^1 \cdot g'(t)dt\\ =&\int_a^b [1 - \cos f(t)] \cdot g'(t)dt\\ =&\oint\limits_{\partial D} (1 - \cos \varphi) d\theta\end{aligned}\label{eq:sinds}\end{equation}$$
Translated automatically with claude-sonnet-5; all equations are reproduced verbatim from the source. Copyright remains with the original author.