Iterated Integrals over Rectangles

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How To Compute Iterated Integrals

Now that we know what double integrals are, we can start to compute them. The key idea is to work with one variable at a time.

In order to integrate over a rectangle $[a,b] \times [c,d]$, we first integrate with respect to one variable (say, $y$) for each fixed value of $x$. That's an ordinary integral, which we can compute using the fundamental theorem of calculus. We then integrate the result over the other variable (in this case $x$), which we can also compute using the fundamental theorem of calculus. So a 2-dimensional double integral becomes two ordinary 1-dimensional integrals, one inside the other. We call this an iterated integral.

There are two ways to see the relation between double integrals and iterated integrals. In the bottom-up approach, we evaluate the sum $$\sum_{i=1}^m\sum_{j=1}^n f\left(x_{i}^*,y_{j}\right)^* \,\Delta x\,\Delta y=\sum_{i=1}^m \left(\sum_{j=1}^n f\left(x_{i}^*,y_{j}^*\right) \,\Delta x\,\right) \Delta y,$$ by first summing over all of the boxes with a fixed $i$ to get the contribution of a column (as indicated with the parentheses on the second sum), and then adding up the columns. (We could do this in the other order, by reversing the summations.)

$f\left(x_{i}^*,y_{j}^*\right)\, \Delta y\, \Delta x$ is the contribution of the volume of a tower over a single rectangle to our double integral, which approximates the volume under the surface $f(x,y) and over the rectangle.

$\displaystyle\left(\sum_{j=1}^n f\left(x_{i}^*,y_{j}^*\right) \,\Delta y\,\right) \Delta x$ is the sum of volumes of towers over all the rectangles in a single column, approximating the volume under the surface over the column. Here, $\Delta x$ is the width of the column. As $n \to \infty$, the sum over $n$ turns into an integral, and we get $$\displaystyle\left (\int_c^d f\left(x, y\right)\,dy \right )\, \Delta x.$$

Adding up the columns then gives $\displaystyle\sum_{i=1}^m \left (\int_c^d f\left(x_i^*, y\right)\, dy \right )\, \Delta x$. Taking a limit as $m \to \infty$ turns the sum into an iterated integral, giving us the actual volume over the large rectangle, under the surface:$$\int_a^b \left ( \int_c^d f(x,y)\, dy \right ) \,dx.$$

This approach is explained in the following video, and an example is worked out. (Video Fix? However, there is a small error. At the beginning it says that we're going to integrate over the rectangle $[0,1] \times [0,2]$, but for the rest of the video the region $R$ is actually the rectangle $[0,2] \times [0,1]$.)

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How To Compute Iterated Integrals