 Hello and welcome to this screencast on section 11.1, iterated integrals. Recall that last section we defined a double integral of a continuous function f over a rectangle r as a limit of the double Riemann sum given here. While this definition tells us exactly what a double integral is, it's not very helpful for determining the value of a double integral. Fortunately, there is a way to view a double integral as an iterated integral, which helps us make computations in many cases. The iterated integral is closely related to ideas from when we studied solids of revolution from single variable calculus, where we sliced the solid, found the volumes of each of the slices, and used an integral to add up these volumes. Let's look at a picture of what we'll be doing in this section. Suppose we want to compute the total volume of this function shown here. We will make a choice and slice the function in the y direction. We'll add up all the different slices that we get to get an approximation for the total volume. This approximation is a Riemann sum, so as the number of slices increases without bound, the approximation will approach the total actual volume of the solid. Let's take a look at the calculus behind these pictures. Let f be a continuous function on the rectangular domain R. Let a be a function defined by the integral from c to d of the function f with respect to y. Note that the function a determines the value of the cross-sectional signed area in the y direction for fixed values of y. This area comes from the solid bounded between the surface defined by f and the xy plane over the rectangle r. We can integrate a with respect to x and in doing so we have made an approximation of this integral using the Riemann sum displayed here. Each term of this Riemann sum gives us the volume of a slice coming from an approximation of the total volume bounded by f and the xy plane over r. Thus what we have shown here is that the limit of this Riemann sum on the right side is equal to the exact volume bounded by f and the xy plane over r, which we know from the last section is given by the value of this double integral. But on the other hand this limit of the Riemann sum is also equal to the integral of the function a that we defined on the previous slide. But the function a is defined by an integral of f. So we can plug that in for a here in the parentheses. And what we have shown here is that a double integral can be evaluated by first evaluating an integral of f with respect to y from c to d and then evaluating an integral with respect to x from a to b. Thus a double integral can be evaluated by two single variable integrals. This is called an iterated integral. We have made a choice to integrate with respect to y first here but we could have also integrated with respect to x first using a similar procedure which shows that this double integral could have also been evaluated this way here. The result from the previous slide is known as Fubini's theorem which tells us that no matter which order we integrate we get the same value.