How would you "find the area under the curve" of something that isn't quite a curve? The graph may be curved at places, but if it's not a function (the same x value has more than one corresponding y value) when how would you find the area of it? Is it possible? Just curious.

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    $\begingroup$ This is basically an ill-posed question. You would have to start from scratch and discuss what it means to do calculus on things that aren't functions. Unless of course you are just referring to finding the area of certain bounded regions? Like the area of the region bounded by $x^2+y^2=1$? $\endgroup$ Dec 2 '18 at 17:51
  • $\begingroup$ Do you mean something along the lines of integrating a multifunction? $\endgroup$
    – YiFan
    Dec 3 '18 at 7:47

Absolutely, this is called a curvilinear integral. It works when the curve is given by parametric equations. If the curve is closed, you can obtain its area by integrating one of $x\,dy$ or $-y\,dx$.

E.g. with a full circle,

$$x=r\cos t,y=r\sin t$$ and

$$A=\int_0^{2\pi}r\cos t\,d(r\sin t)=r^2\int_0^{2\pi}\cos^2t\,dt=\pi r^2.$$


Find a way to describe it with a function and integrate that. For example you might be trying to find the area between two curves. So you'd let $f(x)$ be the distance between the two curves and integrate that with respect to $x$.

BTW, integration (thought of geometrically) isn't necessarily about the area under a curve. Loosely speaking it's really about adding up infinitely many infinitely small areas. These might be slices under a curve, or they might be something else. For instance one way to derive the area of a circle is to use concentric slices: it's no accident that integrating $2\pi r$ gives $\pi r^2$. You could straighten them out and say they're slices of the area under a graph of $y=2\pi r$, but really theyre slices of the circle.

  • $\begingroup$ Breaking up into functions can be extremely ugly: "Consider an annulus centered at $(x, x)$ with inner radius $r$ and outer radius $R$, such that exactly 1/3 of its total area lies in the first quadrant of the plane. Solve for $x$ in terms of $R$ and $r$. A solution must exist by the intermediate value theorem." Now you have to deal with the inner and outer curves, from above and below, and it gets even worse if $x\sqrt{2} > r$ (which is rather hard to determine since $x$ is unknown and $r$ varies freely). $\endgroup$
    – Kevin
    Dec 2 '18 at 19:48
  • $\begingroup$ @Kevin That definitely sounds like a case where my second paragraph applies! Start by changing the coordinates so the annulus is centred on the origin I suppose, with the original axes being a couple of lines the required portion must lie between, then take concentric slices, get the arc length of a slice in terms of its radius, then integrate that? That would be my attempt (probably subdivided into slices which intersected one line, both or none) $\endgroup$
    – timtfj
    Dec 2 '18 at 20:37

Another, more encompassing way to look at integrals is not to look at functions, but introduce a class of objects which can be integrated, called (differential) forms. An example of such forms are $f(x) dx$ and $g(x) dx$. Already based on this little information, we can already derive a number of nice properties: they are linear (because integrals are), we know how they transform under a class of changes of variables (the chain rule): $f(y) dy = f(y(x)) y'(x) dx$ etc. In this way, we can already think of a one-dimensional integral not as an integral of a function (where the integral is $\int (\cdot) dx$ with a slot for a function) but as an integral of a form (where the integral is $\int (\cdot)$ with a slot for a form).

What's interesting about them is their generalization to multivariate calculus. In the simplest case, an integral $\int f(x,y) dx dy$ leads to the form $f(x,y)dx dy$, and we would think of differential forms as just a bit of notational hocus-pocus. Their real power comes from the way they deal with variable changes. In multivariate integration, a change of variables is accompanied by the Jacobian, which is a generalization of the derivative term $y'(x)$ in the one-dimensional case. Expressing $x,y$ as functions of a different set of variables $x(u,v),y(u,v)$, the integral transforms as $$ \int f(x,y)dxdy = \int f(x(u,v),y(u,v)) |J(u,v)| du dv, $$ where $$ |J(u,v)| = \left|\det\left(\left.\frac{\partial(x,y)}{\partial(u,v)}\right|_{x=x(u,v),y=y(u,v)}\right)\right|=\left|\frac{\partial u}{\partial x}\frac{\partial v}{\partial y}-\frac{\partial u}{\partial y}\frac{\partial v}{\partial x}\right| $$ is the absolute value of the determinant of the Jacobian matrix. This seems like a complicated prescription, and it is, but we can automate this process by making our forms a tiny bit more powerful: first, we introduce the wedge product (I'm restricting myself to two dimensions for the sake of simplicity) $dx\wedge dy$ with the property $$ dx\wedge dy = -dy\wedge dx. $$ This antisymmetry will end up leading to all the right signs for the evaluation of the determinant. Second, we make the $d$ a tiny bit more powerful, by defining its effect on an arbitrary function $$ d f(x,y) = \frac{\partial f}{\partial x}dx + \frac{\partial f}{\partial y}dy, $$ additionally requiring linearity in the obvious way. Third, we define that any integral of a differential form $f\,dx\wedge dy$ can be evaluated the usual way $\int f\,dx\wedge dy:=\int f\, dx\, dy$ (this defines the sign in particular).

How is this useful? Let's plug in our previous change of variables, and evaluate $f(u,v)du\wedge dv$ which we can think of as shorthand for $f(u(x,y),v(x,y))du(x,y)\wedge dv(x,y)$. Looking only at the $d$'s we get as a first step $$ du(x,y)\wedge dv(x,y)=\left(\frac{\partial u}{\partial x}dx + \frac{\partial u}{\partial y}dy\right)\wedge\left(\frac{\partial v}{\partial x}dx + \frac{\partial v}{\partial y}dy\right). $$ Distributing out the parentheses while taking into account that from antisymmetry $dx\wedge dx = -dx \wedge dx=0$, we get $$ du(x,y)\wedge dv(x,y)=\frac{\partial u}{\partial x}dx \wedge \frac{\partial v}{\partial y}dy + \frac{\partial u}{\partial y}dy\wedge \frac{\partial v}{\partial x}dx. $$ Collecting terms, taking into account that $dy\wedge dx = -dx\wedge dy$, we obtain $$du(x,y)\wedge dv(x,y)=\left(\frac{\partial u}{\partial x}\frac{\partial v}{\partial y}-\frac{\partial u}{\partial y}\frac{\partial v}{\partial x}\right) dx\wedge dy, $$ where we have recovered the determinant of the Jacobian. In other words, our forms automatically give us the right transformation of the integral (modulo a sign from the missing absolute value, but the meaning of a negative volume is situational anyway). This procedure straightforwardly generalizes to higher dimensions and to spaces that are more complex than $\mathbb{R}^n$, always effectuating the correct change of integral measure automatically, and we can write our form in terms of whatever coordinates are most useful.

Another way how this automatic handling of the change of variables makes life simpler are forms like $fdx + gdy$, taking us back to your original question. We can now leave evaluation as an exercise to the reader :-)

In this sense it is actually much more natural to think of integrals as operations on differential forms instead of functions.

  • $\begingroup$ Are you trying to drown the OP ? ;-) $\endgroup$
    – user65203
    Dec 3 '18 at 10:17
  • $\begingroup$ Lol at least I could refrain from introducing manifolds, tensors and the cohomology. $\endgroup$
    – tobi_s
    Dec 4 '18 at 2:41

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