I've been studying Multivariable Calculus for a while; but I still don't quite know the difference between $f(x,y) = x^2 + y^2$ and $x^2 + y^2 = 9$. I know that the former graphs a paraboloid, while the latter a cylinder. But what's the difference (or 'theoretical difference' if you will) between having a function and having an equation.

Do we assume that the function maps to z-axis because of convention? Doesn't the equation map it's point to the z-axis as well?

I sometimes feel like this fact was skipped or even 'ignored' by the teachers.

Thanks in advance

  • $\begingroup$ A function is a mathematical object. An equation is a mathematical statement of the form $a=b$, which can be true or false. In practice, we often define functions via an equation, by taking the function to be that function that makes the equation true. $\endgroup$ Commented Jun 23, 2022 at 17:15
  • $\begingroup$ In practice mathematicians "abuse notation" a bit by saying things like "Consider the function $f(x,y) = x^2 + y^2$ to mean "Consider the function $f$ defined by the equation $f(x,y) = x^2 + y^2$, even though technically $f(x,y) = x^2 + y^2$ is an equation, not a function, but this is generally considered OK shorthand. $\endgroup$ Commented Jun 23, 2022 at 17:17
  • $\begingroup$ As for $x^2 + y^2 = 9$, this implicitly defines a set $\{(x,y) \in \mathbb{R}^2 | x^2 + y^2 = 9\}$. This is what you get if you consider the points $x,y$ with $f(x,y) = 9$. It might help to think of it as a contour corresponding to the height of $9$ on a topographic map of the plane where the elevation of a point $(x,y)$ is $x^2 + y^2$. $\endgroup$ Commented Jun 23, 2022 at 17:23

7 Answers 7


It depends on the context what you mean with $$f(x,y) = x^2+y^2 \tag 1$$

  • What it could be is the definition of a function $f:\Bbb R^2\to\Bbb R$, or over any other domains where concepts like squaring and adding makes sense, like $\Bbb Z$, $\Bbb Z/n\Bbb Z$, some field $K$ just to name a few. Only from the question one could infer that it's actually the definition of a function.

  • Only in some specific context $(1)$ represents a paraboloid, like: Let $P$ be the set of all points in $\Bbb R^3$ such that $$P=\{(x,y,f(x,y))\in\Bbb R^3\}$$ then $P$ is a paraboloid. In this context you are talking about the "plot of" some function defined by $(1)$ rather than about a function definition.

  • The function $f$ could be defined completely different, like: Let $f(x,y) = x-y$. Determine all solutions of $(1)$. Or: draw all points in the $x$-$y$-plane that satisfy $(1)$, which turns out to be a circle.

Similarly for the second equation: $$x^2+y^2 = 9\tag 2$$

Only from the supplied context one can infer that $(2)$ is used like

  • The definition of an algebraic variety, more specifically a quadratic surface: Let $C=\{(x,y,z)\in\Bbb R^3\mid x^2+y^2 = 9\}$. Then $C$ is a cylinder — or a circular pipe that extends to infinity, as is has neither top nor bottom.

  • Without that context, it could just as well be a circle in $\Bbb R^2$.

  • You can read it as a Diophantine equation.

  • You can understand it as implicit function definition.

So the bottom line is: What counts is the context.


Function notation

A function like $f(x,y) = x^2+y^2$ can be thought of as a machine that takes the input $(x,y)$ (a pair of real numbers) and returns the real number $x^2 + y^2$. Sometimes this is notated as $f:\mathbb R^2\to \mathbb R$ to emphasize the domain $(\mathbb R^2)$ of the function and the codomain $(\mathbb R)$.

Defining sets

Equations like $x^2+y^2=9$ are properly speaking sets of points. In your example, the proper way to describe the mathematical object "$x^2+y^2=9$" is as a set $\{(x,y,z): x^2+y^2=9\}$, which we read as "the set of all triples $(x,y,z)$ of real numbers such that $x^2+y^2=9$." The equation $x^2+y^2=9$ in this context is often referred to as the defining equation for this particular set of points, and the set itself can indeed be visualized as a cylinder in space. Notice how it is necessary to specify the context. The same equation makes sense in the plane as well, and the set $\{(x,y):x^2+y^2=9\}$ has the same defining equation, but it is a different collection of points (a circle) because it's in the plane.

Implicit functions

Some sets of points like $\{(x,y):x^2+y^2=9\}$ implicitly define functions, since we can solve for $x$ or $y$ as a function of the other variable, as in $y = \sqrt{1-x^2}$ for the portion of the circle lying in $\{(x,y):y>0\}$. This equation $y=\sqrt{1-x^2}$ implicitly defines a function $f\colon(-1,1)\to\mathbb R$ by $f(x) = \sqrt{1-x^2}$. Here is a good example of where the function notation "$f\colon A\to B$" helps keep track of important data about the implicit function.

A takeaway

It's more useful to think of functions "dynamically" as rules or machines that take inputs to outputs, and sets of points "statically" with a defining equation (or multiple defining equations) that defines the set of points. Not all sets of points have defining equations. For example, some sets of points are defined by inequalities, such as $\{(x,y,z):z> 0\}$. This set of points is visualized as the "upper half of space."

If you continue studying math, you will sooner or later encounter the subject of linear algebra, or real analysis. In either subject, you will get a treatment of the more precise definitions of functions and sets of points, since these precise definitions are fundamental to building up a properly rigorous mathematical environment.

  • $\begingroup$ Hi! I already sat for Lineal Algebra ("everything" up until Linear Transformations). However, even then a lot of the concepts were skipped over. Maybe it's because I'm attending an Engineering University, and Engineers just skip said explanations in favor of... something. $\endgroup$ Commented Jun 23, 2022 at 20:11
  • $\begingroup$ Can’t a function also be viewed as a set? $\endgroup$ Commented Jun 24, 2022 at 1:35
  • $\begingroup$ @ToddWilcox: That's absolutely right, since a function can be identified with its graph, which is a set of points. I think for the purpose of helping distinguish functions and "equations" as in this question however, the right perspective is to view a function as a rule for assigning values to inputs. $\endgroup$
    – Alex Ortiz
    Commented Jun 24, 2022 at 14:55

This is actually more general than just "calculus", but unfortunately the way most maths is taught mixes up its different historical stages of development instead of just presenting a clean logical picture from the get-go based on the most up-to-date state of the field.

In a fully modern understanding, an equation is a statement: it is an assertion that something is true - namely, that two things are equal. "$=$" is a relation, and when you write

$$a = b$$

you are asserting that that relation holds true for the objects named "$a$" and "$b$". In this regard, such an equation is kind of like a mathematical "sentence", in the same way that this post is composed mostly of English sentences. Each one asserts something to be true - for example, the last sentence you read just asserted that "it is true that this post is mostly composed of English sentences". (Note of course that asserting something is true, and its being true, are two different thing.) Relations are like mathematics' "verbs".

A function is a specific type of mathematical object, which can be thought of as describing or defining some kind of "property" of objects of a particular kind, which can have different values depending on exactly which object we are looking at. For example, if we had a set of cars, all of which we notionally understand as having a single color (i.e. a "red car", regardless of whether 100% of every surface is red-colored), then we can create a "color function" which, when you give it a car as its input argument, becomes a symbol standing for the color of that car. That is, if we write it as $\mathrm{Color}(\cdot)$, and the car is $A$, then the combined symbol $\mathrm{Color}(A)$ has the same meaning as whatever the color is that our car has, e.g. if $A$ is a red car, then the expression $\mathrm{Color}(A)$ means the same thing as "red".

And thus we can write an equation

$$\mathrm{Color}(A) = \mathrm{red}.$$

When you wrote

$$f(x, y) = x^2 + y^2$$

you did write an equation, not a function. The function is just


Yes, not $f(x, y)$. Just $f$. $f(x, y)$ is the value of the function when arguments $x$ and $y$ are put in. What is the function? Well, that's the thing - you see, we can use such an equation to define a function implicitly, by saying that $f$ is the function which makes that equation true, i.e. it is the function that takes, when you give it values $x$ and $y$, the value $x^2 + y^2$. Because there is only one value for $x^2 + y^2$, the definition is unambiguous, and thus this equation serves "double duty" to define the function.

If you want to make it clear what you are doing is defining the function, you can use the assignment or walrus operator $:=$, and write

$$f(x, y) := x^2 + y^2$$

which means you are saying you are thus making $f(x, y)$ equal to $x^2 + y^2$ within the given context (or "scope", which is implicit, not explicit).

Now about "graphing". It is true that the graph of $f$, deefined aboev, is a paraboloid. But what do we mean by "graph"? There's actually two, though not unrelated, meanings at play here.

When we have a function $f$, its "graph" is the set of all ordered pairs $(X, f(X))$ (note here the use of capital $X$, a function with two arguments can also be considered a function of one argument consuming an ordered pair as its argument). In your case, this would be equivalent to ordered triples $(x, y, f(x, y))$, so it describes a surface in 3D space. That surface is, as you point out, a paraboloid.

When we have an equation, e.g.

$$x^2 + y^2 = 9$$

what we have is not "technically" called a graph but rather a solution set: it is the set of possible variable assignments for $x$ and $y$ that make the equation true. Here, that set is the same as the set of all Cartesian coordinates of the points on the circle of radius 3 centered at the origin. The "graph" is actually the set of coordinates, so it's an interpretation of this equation, not the equation "itself".

So what's the difference? A function is an object, an equation is a statement, but the two bear a variety of relationships that have to be elucidated carefully in order to make things clear.

  • $\begingroup$ Thank you so much for the answer; it was very eye opening $\endgroup$ Commented Jun 26, 2022 at 21:58
  1. The function specified as $$g(x,y) := x^2 + y^2$$ has maximal domain $\mathbb R^2,$ in which every input/point $(x,y)$ has a corresponding value $x^2 + y^2$ in $\mathbb R.$

    This function might output the temperature variation on an infinite plane.

  2. Now, not every point $(x,y)\in\mathbb R^2$ satisfies the conditional equation (where $g$ is defined as above) $$g(x,y) = 9;\tag1$$ its solution set—the collection of all points that satisfy it—$\{(x,y)\in\mathbb R^2\mid x^2+y^2=9\}$ traces out a circle on the $x$-$y$ plane.

  3. In 3D Euclidean space, points are triples $(x,y,z)\in\mathbb R^3,$ and the conditional equation $(1)$ is now satisfied by the cylinder $\{(x,y,z)\in\mathbb R^3\mid x^2+y^2=9\}$ instead.

  4. With $g$ defined as above, setting $z:=g(x,y)$ so that $z$ explicitly depends on $x$ and $y:$ $$z=x^2+y^2.$$ This equation is satisfied by the paraboloid $\{(x,y,x^2+y^2)\mid (x,y)\in\mathbb R^2\}$.

  5. Notice that while every $(x,y)$ combination can be found on the paraboloid, only those that satisfy $x^2+y^2=9$ can be found on the cylinder.

  6. Consider the part of the paraboloid that lies on the plane $z=9,$ and reset the latter as the new $x$-$y$ plane: this gives the circle $(1).$


What's the difference (or 'theoretical difference' if you will) between having a function and having an equation?

  • If you have a function you have a "transformation" (variables assigned to other one). If you have an equation you have a "thing to be solved" (unknown to be determined).

  • A function is a specific type of correspondence between two sets. An equation is an assertion of equality between two mathematical expressions involving at least one unknown.

  • Functions have graphs, which are specific type of sets. Equations have solutions, which are the values of the unknowns that make the equality true.

  • In some cases, the graph of a function has a geometric representation. In some cases, the set of solutions of an equation has a geometric representation as well. However, the fact that there are geometric representations associated with functions and equations does not imply that they are the same thing (actually, at least theoretically, any subset of $\mathbb R^3$ has a geometric representation regardless of whether it's the graph of a function, the set of solutions of an equation, or anything else).


I'm guessing you're coming from a multivariable calculus perspective. The idea you're circling is that of a smooth manifold. A manifold somehow makes precise our notion of a surface. Let's stick with the idea of a 2-dimensional manifold embedded in $\mathbb{R}^3$ for this discussion, though you should know that generalizations are possible.

There are at least 3 equivalent ways to describe a 2-dimensional manifold embedded in $\mathbb{R}^3$:

  1. It is locally the image of a function $\vec{x} : A \subset \mathbb{R}^2 \to \mathbb{R}^3$. For example, $\vec{x}(\theta , \phi) = (\sin\theta\cos\phi, \sin\theta \sin\phi, \cos\theta)$ has a subset of the unit sphere as its image.

  2. It is locally the graph of a function $f: A \subset \mathbb{R}^2 \to \mathbb{R}$. Formally, what this means is that we build a function like the one in 1 out of $f$ whose image is the manifold. For example, we may take $f(x,y) = \sqrt{1 - x^2 - y^2}$. We form the function $F(x,y) = (x, y, f(x,y)) = (x, y, \sqrt{1 - x^2 - y^2})$. We say the image of $F$ is the graph of $f$, and it too has a subset of the unit sphere as its image.

  3. It is locally the solution set (in more formal language "the pre-image of $0$") of a function $g : \mathbb{R}^3 \to \mathbb{R}$. For example, if $g(x,y) = x^2 + y^2 + z^3 - 1$, then the solutions to $g(x,y, z) = 0$ are a subset of (in fact the whole) unit sphere.

So there you have it. Three different ways to talk about 2-dimensional manifolds embedded in $\mathbb{R}^3$. I think the reason you are confused is because multivariable calculus often freely passes between these three different notions as needed without pausing to recognize that they are equivalent.

If you want to know more about these ideas, I can recommend Shifrin's book on differential geometry to get your feet wet.

If you want technical details, I can recommend Edwards Advanced Calculus of Several Variables, which goes into detail on the machinery that proves these descriptions are equivalent. Ultimately it comes down to something called the implicit function theorem.


We often see an expression like $x^2+y^2=9$ as a level surface to $f(x,y)=x^2+y^2$, i.e. $f(x,y)=9$.

For example, we might want to find the tangent plane to a point on the surface defined by $x^2+y^2=9$.

We then define $f(x,y)=x^2+y^2$. When $f(x,y)=9$ which is a constant, we know the derivative in every direction is zero which gives us that $\nabla f\cdot\hat u=0$ for every direction $\hat u$ from which we conclude that the gradient is normal to the surface. That normal and the point give us the tangent plane.


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