# Visualization of double integrals

So, we just started learning double integrals in class.

$$\iint f(x,y) \mathrm{d}x \mathrm{d}y$$

Our teacher said that performing this operation within certain limits of x and y yields the volume under that function confined within these limits.

She also mentioned that when $$f(x,y)=1$$ we're finding the 2D area, again confined within limits. However, she mentioned that this is only valid when $$f(x,y)=1$$, and not for any other constant value the function may represent.

Now, I'm confused.

(i) Although dimensionally the integral will represent xy, why does finding the double integral not mean finding the volume of the cuboid within the limits?

(ii) If it infact means finding the area, why is it only true for $$f(x,y)=1$$ and not some other constant value?

• – Hans Lundmark Apr 27 at 4:54
• @HansLundmark Thank you so much!!! So, according to this, if $f(x,y)=2$, I can interpret it to be the volume of a cuboid of height as $2$ units, or simply twice the area of a rectangle. Is this correct? – Smriti Sivakumar Apr 27 at 5:10
• Yes, that's correct. – Hans Lundmark Apr 27 at 6:36

## 2 Answers

There are often several ways one can interpret integrals. The viewpoint in single-variable integrals as "(signed) area under the graph" is too limited, because we often have many many many other reasonable interpretations. It all depends on the actual purpose at hand.

Suppose $$S\subset\Bbb{R}^2$$ is some region in the plane, and $$f:S\to \Bbb{R}$$ is a given function; for now assume all the nice regularity that is needed, I just want to get some basic ideas across for how one can interpret the integral $$\int_Sf\,dA$$

• Suppose $$f\geq 0$$ is a non-negative function. Then, we can imagine the graoh of $$f$$ as being a "surface lying over $$S$$". Then, the quantity $$\int_Sf\,dA$$ can be interpreted as the volume of the region bounded by the graph of $$f$$ and the region $$S$$. So, as a consequence, if we take $$f=1$$, we can interpret $$\int_S\,dA$$ as the volume of the solid with cross-section $$S$$, and height $$1$$. For example, if $$S$$ is the unit disk in the plane $$S=\{(x,y)\in\Bbb{R}^2\,:x^2+y^2\leq 1\}$$, then we can interpret $$\int_S\,dA$$ as the volume of the cylinder having $$S$$ as the cross section and having height $$1$$.

• A more direct interpretation of $$\int_S\,dA$$ is as the area of the planar region $$S$$, because recall that the integral is approximated by the sums $$\sum_{i=1}^n\text{area}(R_i)$$, where $$R_i\subset S$$ are rectangles, so atleast in nice cases, this approximation had better yield the area of $$S$$ in the limit. This should be reasonable geometrically because area times a height of $$1$$ should be the same thing as the volume under the given region. If you really want to you can think of this as integrating a constant unit-less $$1$$ vs integrating $$1$$ meter to give the distinction of interpretation between area vs volume.

• Another interpretation, again supposing $$f\geq 0$$, is to think of $$f$$ as describing the "surface mass density of $$S$$". So, imagine $$S$$ as a thin sheet of a material (like a very thin (not necessarily uniform) aluminum foil which you cut up into a certain shape), and suppose $$a\in S$$ is a point. Then, we can interpret $$f(a)$$ as the mass density at the point $$a$$. Recalling the relation "mass equals (surface)density times area", we thus interpret $$\int_Sf\,dA$$ as being the total mass of the thin sheet $$S$$.

• Now, if we allow $$f$$ to take on both positive and negative values, then a physical interpretation is to think of $$f$$ as the electric charge density, rather than the mass density (which must be non-negative). Then, reasoning as above, we think of $$\int_Sf\,dA$$ as being the net charge contained in $$S$$.

• If $$f$$ has some other special properties we can give it some other interpretation. For example, if $$M$$ is some two-dimensional surface, then often times we are able to parametrize the surface using a function $$g:S\to M$$ (for example, think of $$M$$ as the unit sphere, and $$S=(0,\pi)\times (0,2\pi)$$ as the domain for spherical coordinates parametrization). Then, the surgace area of $$M$$ can be calculated by the formula \begin{align} \text{area}(M)&=\int_S\underbrace{\left\lVert\frac{\partial g}{\partial u}\times \frac{\partial g}{\partial v}\right\rVert}_{:=f(u,v)}\,du\,dv \end{align} Don't worry about the exact formula for $$f(u,v)$$ here. My point is that sometimes, the thing you're integrating doesn't really have anything to do with the region $$S$$; rather you may be interested in the surface $$M$$, and $$S$$ and $$f$$ only comes up as an intermediate calculation. Concretely, we have (again don't worry about where exactly the formulas come from) \begin{align} \text{area of unit sphere}&=\int_{(0,\pi)\times (0,2\pi)}\sin\theta\,d\theta\,d\phi\\ &=\int_0^{\pi}\sin\theta\,d\theta\int_0^{2\pi}\,d\phi\\ &=4\pi \end{align} In this example, we have $$f(\theta,\phi)=\sin\theta$$ and $$S=(0,\pi)\times (0,2\pi)$$ is simply a rectangular region in the plane. Obviously, it is not very insightful to think of the integral $$\int_Sf\,dA(\theta,\phi)$$ as being the "mass of $$S$$" or as "the volume under the graph of $$f$$" or anything like that. The integral only comes in here as an intermediate calculation step to compute the surface area of the unit sphere

So you see, because there are so many possible interpretations of integrals, it is unwise to limit oneself to thinking only in terms of the "(signed) volume under the graph", or something like that. The integral is a very flexible tool, whose purpose is to "add up a lot of small things", and the things we're adding up then gives us the interpretation. I'm sure there are also many other possible ways of interpreting double integrals.

As a side remark: we can do the same thing with line integrals, triple integrals or really anything else. There are often many ways of viewing the same thing, and in some situations, one interpretation is more useful than another.

• Interesting points about the flat alumnium sheet and net charge, both of which are rather...flat. I was going to say that volume is unitless hence that interpretation and these have units, but that isn't really true. Hooray for science! – Jellyfish Apr 27 at 6:15

$$\int\int f(x,y)dxdy$$

will yield the volume under the function if $$f(x,y)\ge 0$$ inside the region. If $$f(x,y)\le0$$ inside the region, it will yield the volume of the function inside the region, but with a minus sign.

So if $$f(x,y)=1$$, the integral will yield the volume of a base with a height of $$1$$, which is the same as the area (because $$\text{surface area}=\text{volume}/1=\text{volume}$$. If $$f(x,y)=2$$, then the height of the region will be $$2$$, and you can find the area by dividing by the integral by $$2$$. It can also be seen mathematically by

$$\begin{split}\text{Area}&=\frac 1 2\int\int 2dxdy\\ &=\int \int 1dxdy\end{split}$$