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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! :)

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  • $\begingroup$ The defintion of $y$ is not obvious. Both functions are named $y$. $\endgroup$ – Lord Commander Mar 8 at 21:46
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    $\begingroup$ About the z axis - you integrate the function $z(x,y)=y$ over the 2d region of integration. $\endgroup$ – Koncopd Mar 8 at 21:47
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    $\begingroup$ 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). $\endgroup$ – Winther Mar 8 at 21:54
  • $\begingroup$ ... assuming unifrorm density. $\endgroup$ – NickD Mar 8 at 21:56
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    $\begingroup$ 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 $12\times 2/3=8$. $\endgroup$ – Elliot Yu Mar 8 at 22:33
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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 [1] $$m = \int_0^L \rho(x)\,\mathrm{d}x $$ 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.

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  • $\begingroup$ Damn, that is one elaborate answer! Thank you so much, really appreciate it! :) $\endgroup$ – Sam van der Kris Mar 9 at 12:51

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