# What does this double integral even represent?

I'm preparing for a calculus resit and I'm really confused by this double integral. I somewhat understand where the $$\int_{3y}^{12-3y}$$ comes from (just expressing $$y=\frac{1}{3}x$$ and $$y=-\frac{1}{3}(x-6)+2$$ in terms of x). I also get how to solve the integral (my answer was also 8), but now what this answer actually represents.

$$\int_0^2 \int_{3y}^{12-3y} (y)\ dx\ dy = 8$$

If you plot these two functions, the area enclosed by them and the x-axis is obviously 12 (just the area of two right-angled triangles, and I even checked it with integrals to make sure I wasn't going mad). I believe the double integral represents some sort of volume, but I don't understand what volume, because nowhere in this question a z-axis is mentioned.

I feel like I'm either missing something very fundamental, or this is just a really badly worded/confusing question. Either way, any help would be very much appreciated! :)  • The defintion of $y$ is not obvious. Both functions are named $y$. – Lord Commander Mar 8 at 21:46
• About the z axis - you integrate the function $z(x,y)=y$ over the 2d region of integration. – Koncopd Mar 8 at 21:47
• One interpretiation: If you have some region $A$ then $\frac{\iint y dA}{\iint dA}$ is the $y$ coordinate of the center of mass of that region and you can likewise do the same for $x$ to get the "center" point of the region (both integrals is over $A$ and the denominator is just the area of the region). – Winther Mar 8 at 21:54
• ... assuming unifrorm density. – NickD Mar 8 at 21:56
• Treating the integrand as $z(x,y)=y$ gives a geometrical interpretation where you are trying to find the volume of a pyramid whose base is the triangle $D$. Two of its sides are perpendicular to the $xy$-plane, and the third is given by $z(x,y)=y$. Its apex is over the point $(6,2)$, so the height is $2$. From this you can verify that the volume is $12\times 2/3=8$. – Elliot Yu Mar 8 at 22:33

Before we consider a double integral lets think about what a single integral represents. The usual interpretation of a single integral is the area under a curve. So $$\int_a^b f(x) \,\mathrm{d}x$$ represents the area between the $$x$$ axis (which is just $$y=0$$), the lines $$x=a$$, $$x=b$$ and $$y=f(x)$$. This often a valid and useful interpretation but it isn't the only interpretation. Another interpretation may arise with a more specific function. Say we have some string with length $$L$$ and we know that its mass per unit length at some point $$x$$ along the string is given by $$\rho(x)$$. Then the total mass of the string can be expressed as  $$m = \int_0^L \rho(x)\,\mathrm{d}x$$ To see why this is the case we think about what an integral is, really its just a sum of lots of infinitesimal bits (excuse the hand waving, I'm a physicist). So a discrete version of the integral above may be $$m \approx \sum_i \rho(x_i) \Delta x$$ where we approximate the integral as a sum and we split the string into finite lengths $$\Delta x$$ and we consider the density at the point $$x_i$$ within each length to be a good approximation of the density of the whole length.
Personally I think that this interpretation of the integral is better suited for generalisation to higher dimensions. We can similarly consider a function like $$\sigma(x, y)$$ which represents the density at a point $$(x, y)$$ on a plane. Then the integral $$\iint_S \sigma(x, y)\,\mathrm{d}x\mathrm{d}y$$ is just the mass of the surface, $$S$$, over which we are integrating.
There is still a valid interpretation of a double integral as a volume however. Imagine that the function $$f$$ represents the height above the $$x,y$$-plane. Then the integral [2, 3] $$\iint_S f(x, y)\,\mathrm{d}x\mathrm{d}y$$ represents the area between the $$x,y$$-plane and the function $$f$$ only above the volume in the $$x,y$$-plane that we call $$S$$. This interpretation can still hold even if we haven't got an explicitly constructed $$z$$-axis to think of the height as being along since this interpretation is just to aid in our thought process.
Notice how similar this is to the one-dimensional case with all dimensions shifted up by one. Single integrals become double integrals, intervals such as $$[a, b]$$ become regions of the plane, $$S$$, and areas, given by single integrals, become volumes, given by double integrals. The problem comes when we want to evaluate a triple integral and then with the "area under a curve" interpretation we have to consider the 4-dimensional hyper-volume under a volume and I don't know about you but that is not something I can picture.
So to answer your specific question, what does the given integral represent, the answer is: it depends. This is sort of a rubbish answer but it is true. If the function we are integrating, in this case simply $$y$$ is naturally thought of as a height then a volume is a valid interpretation (and probably the best interpretation in this case). If it can be thought of as a density then mass is probably a better interpretation. However both of these are just ways of thinking about the integral and really they are the same, we can view mass as simply a coordinate in some "mass-space" in a similar way to how we might consider momentum space or any other more abstract space.