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Just started learning about double integrals literally $10$ minutes ago. I have a fairly good grip on the Riemann integral and so far it seems very similar, but we are just working with volumes instead of areas.

My question is this: we approximate the volume under the surface of a function by subdividing the total volume into little blocks with volume themselves. To do this we use the notation,

$$V \approx \sum_{i=1}^{n} \sum_{j=1}^{m}f(x_i,y_j)\Delta A.$$ I know this is essentially adding up all the smaller volumes of the rectangular prisms, but I'm not quite sure I get exactly what the double summation means. Does it mean we sum up all the possible combinations of $i's$ and $j's$ from $1$ to $n$ and $m$?

I'm guessing this is a little trivial but I've never actually used this double sum before so I just want to make sure exactly what it means and how I would calculate it. Thanks.

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You've essentially got it; there's only a little subtlety. To calculate the sum, first fix $i$ as $1$ and then calculate

$$\sum_{j=1}^{m}f(x_i,y_j)\Delta A$$

Then fix $i$ as $2$ and calculate it again. And so on. Then add up the results.

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  • $\begingroup$ Okay thanks so in total you would be computing $n\times m$ sums? I presume you do the same to calculate it the opposite way, (fix a $j$ then so the sum over $i$). $\endgroup$
    – Chris L
    Commented May 30, 2015 at 13:24
  • $\begingroup$ Yes. Your suspicion is also correct: if the double sum were written using the dummy variable $j$ first (instead of $i$), then you'd fix $j$ temporarily and sum over $i$. $\endgroup$
    – Ken
    Commented May 30, 2015 at 13:35

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