 Hello and welcome to this screencast on section 11.3 Double Integrals Over General Regents. In this screencast, we will look at how to evaluate double integrals over general regions that are not just rectangles. First, let's take a look again at double integrals over rectangles. Let F be a continuous function and consider the rectangle R given here. As we saw last section with iterated integrals, we can evaluate this double integral of F over R by integrating first with respect to y, then x, or we can evaluate first with respect to x, then y. Note that on the left side, when we integrate first with respect to y, we are slicing the rectangle in the y direction. Each one of the red slices goes from y equals 2 to y equals 4, so the size of the slice in the y direction is constant. We take these slices for each x value from 0 all the way to 6. These slices are what gives us these constant values in the iterated integral. Similarly, on the right hand side, when we integrate with respect to x first, we are slicing the rectangle in the x direction. Each one of the red slices is constant length in the x direction from 0 to 6. We take these slices for each y value from 2 all the way to 4. Integrating over rectangles gives us these constant values in the limits of integrals because our slices always have constant length. We will see for general regions, this isn't always the case. For example, consider integrating a continuous function F over the triangular region D here. If we try and set this up as an iterated integral slicing in the y direction, we see that our slices in pink do not have the same length. As our x values vary from left to right, we see that the height of the pink slices changes, so we can describe the height of the slices as a function of x. First note that all these pink slices begin where y equals 0. The height of the slice is determined by the hypotenuse of the triangle shown here, which is a line with equation y equals 3 halves x. This means our slices in the y direction run from y equals 0 to y equals 3 halves x. This means when we first integrate with respect to y, we get an integral of F from 0 to 3 halves x with respect to y. Next, we see that we want to take these slices of varying height from x equals 0 to x equals 2. So our second integral that we evaluate with respect to x will have constant limits from 0 to 2. Evaluating this iterated integral will give us the volume bounded by the surface F and the xy plane over the triangular region D. Note that a similar procedure applies for evaluating this particular double integral by first slicing in the x direction. See your textbook for an example of how to carry this out. Before we end this screencast, there are a couple important items to note about the structure of iterated integrals over general regions. We'll use the one that we just did as an example here. The first important thing to note is that the limits on the outer integral must be constant. So that's the outer integral here. To see why this is the case, recall the interpretation of a double integral is the signed volume bounded by the function F and the xy plane over the region D. The volume is a real number or it's a constant. If these outer limits here were not constant, then the output of our iterated integral would not be constant. And this would contradict the fact that we should get a volume out of this. The second important thing to note is that the limits on the inner integral must be constants or they must be in terms of the only remaining variable. That is, if the inner integral is with respect to y as we have here, then its limits on this inner integral must be in terms of x or constant numbers. And this holds vice versa. If the inner integral was with respect to x, then its limits would need to be constants or in terms of y. To see why this second bullet point must be the case, as we noted above, if it were not the case, the output of the iterated integral would not be constant. Which would again contradict the fact that this iterated integral gives us a volume.