Shows the graphics corresponding to an exercise in Stewart's {\it Calculus} (3rd Ed., 1995, Section 13.1, p.837, #4); approximates the volume under a surface using a double Riemann sum, then calculates the volume exactly using an iterated double integral (corresponding notebook for users of Mathematica who want the commands used to generate these graphics)

We start with a graph of the function . We have restricted the domain of this function to the rectangle [0,2]x[0,3].

Now we consider the volume of the object that lies between the surface and the xy-plane (the graphs show the solid from different viewpoints).

To calculate the Riemann sum we break the domain rectangle into smaller rectangles using the partition x=1, y=1, y=2 (look at the xy-plane). In each subrectangle we select a point (in this case, the middle of the subrectangle). We turn each of the subrectangles into a "box" by giving it a height, which is determined by the value of f when it is evaluated at the selected point within the subrectangle.

This animation shows the accumulation of six boxes used to approximate the volume under our surface.

We get a "double Riemann sum" by adding the volume of the boxes along one direction (perhaps along the y-axis) first to get a sum for an entire "row" -- that's one Riemann sum. Then we add the sums for the rows together -- this is another Riemann sum. And we get a "double Riemann sum" like: . As we take more and more boxes these sums are represented as integrals as we get a "double integral."

Another way to fill the volume of the region is with "slices." This animation shows the accumulation of slices parallel to the xz-plane.

Consider one "slice" in the animation:

The integral calculates the area of such a slice (think about how you learned to calculate the area under a curve). Note that in this integral y is acting like a constant. Also note that if we worked out this integral, our answer would be in terms of y. That happens because the area of the slice depends on how far along the y axis we are. So if we have infinitely many of these slices filling up the region under the surface, we can add up their areas -- using another integral -- to find the volume of that region: . This is called "an iterated double integral", in which the inner integral calculates the area of a slice and the outer integral adds up all those areas.

The next animation shows the accumulation of slices parallel to the yz-plane.

For this animation (with the slices going the other way), the iterated double integral would be: . This process of changing how we take our slices is called "reversing the order of integration." Warning: reversing the order of integration is not always as straightforward as it was here!

In this example the volume under the surface is 44. Using the double Riemann sum from above should give a good approximation of this number. Computing either of the double integrals should give the answer 44. Note that this example is simple enough to do by hand and certainly does not require the power of a computer or even a calculator.

This work is part of the Multivariable Calculus with Mathematica Project at the University of Oklahoma.