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When we study double integral many Calculus textbooks state that for a region $R$ in the plane $$\iint_R1\ dA= \text{area bounded by }R $$ But double integral actually give the volume of a solid. This happens to be numerically equivalent. That is we get an answer like v cubic metres which we say is v square metres. How do physicists view this? Is not this ugly mathematics?

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closed as off-topic by Tommi Brander, James S. Cook, Adam, Xander Henderson, Chris Cunningham May 8 '18 at 12:49

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If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ Note: It will be ideal to describe the reason for down-voting this, and most any other, question, so that the OP (or someone else) can either justify its phrasing, edit, or remove. $\endgroup$ – Benjamin Dickman Oct 22 '16 at 18:43
  • $\begingroup$ This question is based on an incorrect reading of its example. The integrand is the number 1, not the letter l = length. $\endgroup$ – Jasper Oct 24 '16 at 18:52
  • $\begingroup$ MathJax tip: We have \iint and \iiint just for this, and \text{} for text. $\endgroup$ – Simply Beautiful Art Nov 12 '16 at 15:17
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    $\begingroup$ This is not a question about math education. It's an unclear question about the OP's lack of understanding about the topic. $\endgroup$ – Ben Crowell Nov 13 '16 at 3:26
  • $\begingroup$ @Ben Crowel. Its true that there is a gap in my understanding. Does not one usually ask question when unclear about a topic? If very clear about a topic normally I answer others asking question about it. I don't understand this comment by you at all. $\endgroup$ – P Vanchinathan Nov 13 '16 at 3:32
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A double integral represents integrating over an area. If the integrand is a height (i.e. with units of length), the result will be a volume (i.e. units of length$^3$). If the integrand is a flux (e.g. units of power/length$^2$), the result will be in units of power. If the integrand is unitless, the result will be an area. In any case, the result will have the same units as the integrand times length$^2$.

Of course, you don't have to integrate over a geometrical area; it could be time, temperature, pressure, etc., and you'd have to adjust the units accordingly.

The easiest way to think about the correct units is to realize the units are simply the units of the integrand times the units of the differentials (dx, dV, dA, dT, etc), just like you would do when multiplying any other values. Keep in mind that the units of a differential are the same as the units of the base value. So dx, dy, dz has units of length, dA has units of length$^2$, etc.

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You are integrating a function $z=f(x,y)$ but the units of $z$ do not have to be for length. The units could be for mass density, in which case the units of the double integral would be for mass. Or $z$ could be unitless. In this case, the units of the double integral would be for area.

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    $\begingroup$ For clarfication, this example is correct if the mass density is in units of mass/area, not mass/volume $\endgroup$ – pwcnorthrop Oct 24 '16 at 17:44
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Integrals do not inherently have units associated with them. They are mathematical tools which summarize a process of taking a limit of a sum of lots of small things.

For instance, the integral $$\int_{-1}^1 \sqrt{1-x^2} d x $$ could represent the area of half of the unit circle, or (if the velocity of some particle was described as $v(t) = \sqrt{1-x^2}$) as the distance traveled by a particle, or as a mass, or as whatever else. You have to choose units which are sensible to the problem at hand.

In this particular case, there is a bit of geometric insight: that one can calculated the area of a plane region as the volume of cylinder of height 1 with that region as the base.

One need not pass through the volume interpretation however: you can directly view this double integral as summing the areas of lots of small rectangles comprising the region.

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As copied from the comments on an English Language Learners post about how mathematical expressions are read aloud:

I was taught to use the terms "integral" and "integration" for the various integral signs (such as "∫", "∬", "∭", "∮", "∯", and "∰"). Similarly, I was taught to use the d notation for full derivatives with respect to a variable, and the partial notation for partial derivatives with respect to a variable.

I often say "double integral" when performing either an integral of an integral (such as ∬ a dx dy ) or an area integral (such as ∫ a dA ) or a surface integral (such as ∯ a dA ). Similarly, I often say "triple integral" or "volume integral" when performing either an integral of an integral of an integral (such as ∭ a dx dy dz ) or a volume integral (such as ∫ a dV ).

I was taught to use the term "sum" for discrete sums, such as when the (capital) "sigma" notation ( "𝚺" ) is used. I use the term "series" to refer to the sequence generated by evaluating the sum for the various values of n. For example, "1, 2, 3, 4, …" is the sequence of integers. "1, 1/2, 1/4, 1/8, …" is a corresponding geometric sequence. "1, 3/2, 7/4, 15/8, …" is the corresponding series, where each item in the series is the sum of the first i elements of the geometric sequence.

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