# Standard Cartesian Plane

I have posted before.

Is the **standard, ** used cartesian coordinate plane this: Where the vertical axis represents $f(x) = y$ and the horizontal axis represents $x$?

I was wondering because in evaluation of

$$I = \int_{-\infty}^{\infty} e^{-x^2} dx$$

We take:

$$I = \int_{-\infty}^{\infty} e^{-y^2} dy$$

Then we take $I^2$.

Are you then using $f(y) = e^{-y^2}$ or just changing $x \to y$ ?? So $y$ is the independent variable?

• The vertical axis is the set of points $(x,y) \in \mathbb{R}^2$ such that $x=0$. The set of points such that $y=f(x)$ represents a curve in the $xy$ plane. Would it make you feel better if we wrote $$I = \int_{- \infty}^{+ \infty} e^{-t^2} \, dt$$ and $$I = \int_{- \infty}^{+ \infty} e^{- w^2} \, dw?$$ We use $x$ and $y$ because of inertia in switching to polar coordinates. In the example I gave we can write $t = r \cos(\theta), w = r \sin(\theta)$, but for some reason some people aren't comfortable with that and think only $x$ and $y$ can be written in polar coordinates. Jan 6, 2015 at 18:01
• But the problem is, when you combine it, when it becomes an iterated integral, then do you have a $3D$ plot of the curve? It becomes $e^{-(t^2 + w^2)}$ but you cant have a $3D$ plot because the nor $w$ nor $t$ are the vertical axis in the 2D graph. Jan 6, 2015 at 18:04
• It's no longer a curve, it becomes a surface. Jan 6, 2015 at 18:07
• Exactly, how? Given: en.wikipedia.org/wiki/… Then one of the axes must be $y$ for example, and one of the axes must be $x$ for it to be a 3D system, and a surface.? Jan 6, 2015 at 18:08
• @MarkFantini, do you see? In the 3D plane there is a seperate $xy$ plane, so if we have $e^{-(x^2 + y^2)}$ there then the issue would be that either $y$ is a dependent variable for $x$ or either $x$ is a dependent variable for $y$. Jan 6, 2015 at 18:13

What you appear to be looking at is the usual way to evaluate $I=\int_{-\infty}^{\infty} f(x)\;dx$ where $f(x)=e^{-x^2}.$ We show that $$\begin{eqnarray} I^2 &=& \left(\int_{-\infty}^{\infty} f(x)\;dx\right) \left(\int_{-\infty}^{\infty} f(y)\;dy\right)\\ &=& \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} f(x) f(y)\;dx\;dy \\ &=& \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} g(x,y)\;dx\;dy. \end{eqnarray}$$ That is, instead of doing a single integration of the single-variable function $f$, we do a double integration of the two-variable function $g$, where $g(x,y)=e^{-(x^2+y^2)}.$ Then we change variables to polar coordinates in order to put the integral in a form that's much easier to evaluate.
While we can make a simple and reasonably complete graph of $y = f(x)$ in the $x,y$-plane consisting of one or more simple curves that pass the "vertical line" test (at least for reasonable functions $f$ such as the functions one uses for most practical purposes), we cannot graph a multivariable function $g(x,y)$ that way, because $g(x,y)$ assigns a value to every point in its domain, in this case every point in the plane. There are various other ways of plotting $g(x,y)$, but they involve different techniques such as three-dimensional visualization, contour lines, or other techniques to indicate that the function does not just have a single value for any given $x$, but rather has a value for every combination of $x$ and $y$.
I am reminded of this question, which is different from yours but also hinges on the confusion that can occur if one tries to apply principles of graphing single-variable functions to a problem concerning a multiple-variable function. Both single-variable and multiple-variable functions can be related to the same $x,y$-coordinate plane, but they relate to that plane in very different ways.
• I do not understand some things here.How does $f(x)f(y) = g(x,y)$ $x$ and $y$ are collinear points on the axis. Jan 8, 2015 at 13:21
• In $\int_{-\infty}^{\infty} f(x)\;dx$, the symbol $x$ is a name for the variable of integration. It is not a point, though sometimes it is helpful to relate values of $x$ to labeled points on the $x$-axis of a graph. The notation $g(x,y)$ in the equations is there merely to remind us that when we set up the double integral, the integrand is a function of two variables instead of just one. Jan 8, 2015 at 17:27
• Yes, $g$ is a new function. And yes, it is important that in these integrals, $x$ and $y$ are bound variables. Jan 8, 2015 at 18:24
• Okay, so how does $x^2+ y^2 = r^2$ when converting it into polar? Jan 9, 2015 at 12:36