# Area using $\iint dxdy$

I know about finding a volume under a curve using $$\iint f(x,y) dxdy$$.

But what $$\iint dxdy$$ describe? Should I consider $$f = 1$$? I thought if I need to find an area, not the volume shouldn't the $$\text{Area} = \iint 0 dx dy$$. So that the $$z$$-axis is always zero? But of course, the area will also be zero.

I'm super confused about this concept. Why should I care to find an Area under region? Isn't that volume is something we should consider and not just the area?

Yes, consider $$f(x,y)=1$$. The volume under this curve is $$1$$ (the height) times the area of the base, so the numerical value of volume coincides with the area: they're the same thing.

• But shouldn't f(x,y) = 0. so that z-axis (height) is always zero and not one? So that it have no 3rd dimension and we can just multiply the length and breadth (dx and dy) in x and y axis to compute the area? – Vignesh War Feb 4 at 14:10
• We are not multiplying length and breadth because we're not finding the area of a rectangle, but an arbitrary region. We're multiplying the area of the base and the height on top of the base. If $f=0$ then the volume is $0$. – YiFan Feb 4 at 14:19
• Sorry. I am little confused back then. Now I understood. Thanks for explaining. – Vignesh War Feb 4 at 14:23
• @VigneshWar glad to help! – YiFan Feb 4 at 22:07

If $$B$$ is a measurable and bounded subset of $$\mathbb R^2$$, then the area of $$B$$ is given by

$$\iint_B 1 d(x,y).$$

One of the most common ways to think of a double integral is to think of it as the volume under a surface created by a function of two variables $$z=f(x,y)$$ and bounded by a region $$R$$ (typically rectangular in shape) in the $$xy$$ plane. If the given function of two variables is $$f(x,y)=1$$, then what you're basically doing is you're finding the volume of a solid whose height is $$1$$ and the base area is the area of the given region $$R$$. The volume of a solid whose height is $$1$$ is numerically equal to its base area. That's why

$$\iint_{R}1\,dx\,dy=area\ of\ region\ R.$$

You can think of $$\int \int f(x,y)dxdy$$ as "weighted sum" of the region. Your $$f(x,y)$$ function tells you "how important" each point is. If your weight is $$1$$, then each point is worth one unit, so the integral just "adds up" the points, and you get the area. If $$f(x,y)$$ is the height, then the integral is adding up all those heights, so you get the volume.

In calculus, there is a distinction between things that are infinitesimally small, versus things that are zero. The height is not zero, it is an infinitesimal. So instead of taking $$\int \int 0dxdy$$, we should take $$\int \int dxdydz$$. This is then the volume of the region. We can factor out the $$dz$$ and consider this to be $$(\int \int dxdy)dz$$. The quantity within the parentheses is the area; if we were calculating the volume of a region, we could calculate it by integrating along the $$z$$-axis and finding the area of each cross section, i.e. $$dV =Adz$$.

Keep in mind that whenever you calculated the size of a geometric quantity, anything of a larger dimension will be "infinitely" large, while anything of a smaller dimension will be "infinitesimally" small. For instance, if you're calculating areas, then any volume will have an infinite area (it's the sum of infinitely many areas stacked on top of each other), and any line segment will have infinitesimal area. We need to distinguish between a line segment having no area, versus a line segment having no measure at all. When we measure a line segment with cm$$^2$$, we will get zero, but when we measure it with cm, we get a positive number. And if we want to find the area of a region by integrating a bunch of lengths, then we need to distinguish between these two measures; when we integrate a bunch of lengths, we're adding up a bunch of lengths that each have zero area individually, but together have positive area.

Similarly, by multiplying $$dxdy$$ by the height, and treating the height as being zero to get the area, you're measuring the region in three dimensions. And measuring area (a two-dimensional quantity) in three dimensional units results in zero.