setting up double integrals

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This is a recap of how we set up double integrals, using the example of finding the volume under z = 2 y between the cylinders x2 + y2 = 1 and x2 + y2 = 4. First, recall that our steps are the following:

To set up a double integral over a region, double int over R of f(x,y) dA, we go through the three steps
  1. Sketch the region,
  2. Consider one variable as fixed at successive values across the region, and determine the range of values it can assume when doing this -- this gives the outer limits in the double integral, which must be constants,
  3. At each of the possible values for the outer variable, determine the corresponding values for the other, inner variable.
Doing the third step requires us thinking about what, geometrically, fixing one variable means: if we fix x or y, this means that we're restricting ourselves to a vertical or horizontal line at some x or y value. If we fix r in polar coordinates, it means we're restricted a circle of that radius centered at the origin, and if we fix theta in polar coordinates, we're restricted to a line through the origin at an angle theta from the x-axis.

An Example

Let's consider this for the example given above, recapped here for your viewing pleasure:
Find the volume under z = 2 y between the cylinders x2 + y2 = 1 and x2 + y2 = 4.

We're going to approach this two ways, setting it up both in rectangular and polar coordinates. We'll illustrate the steps given above with a figure, too.

illustration of region of integration
figure 1: the region of integration for the example

Solution:

  1. First, sketch the region. This is shown in the figure to the right.
  2. Consider one variable as fixed at successive values across the region, and determine the range of values it can assume. We want to do this twice, once for rectangular and once for polar coordinates:
    Rectangular: let's fix x. Then, from the region shown, we can see that we must have
    0 <= x <= 2.
    Polar: in this case, let's fix theta. Then the region is defined in the first quadrant, which means that we must have
    0 <= theta <= pi/2.
    So, at this point, we know that we have (for rectangular and polar coordinates, respectively)
    int_0^2 int_?^? 2y dy dx
    or
    int_0^{pi/2} int_?^? 2(r sin(theta)) r dr dtheta
    (note in the second we've converted y into r sin(theta), and converted the dA into r dr dtheta).
  3. At each of the possible values for the outer variable, determine the corresponding values for the other, inner variable.
This last step we'll try to animate in a useful manner. Recall that we're thinking about moving through the values of the outer variable and finding the range of possible values for the inner variable at each. This is shown in the following figure. The first two frames of the animation are the region (which is 2D) and the volume (which is 3D). Click through the first couple of frames after those, noting the following:
  1. Every other frame shows us picking a different value of the outer variable (x or theta) and drawing an arrow showing the corresponding range of the inner variable (y or r).
  2. Then, the following frame draws in the actual slice of the volume (sort of; it really needs some more width to be a volume slice) that we're adding up for that value of the outer variable.
  3. For the rectangular coordinates, note that the limits on the inner variable (y) change at x=1, from sqrt(1-x2) <= x <= sqrt(4-x2) to 0 <= x <= sqrt(4-x2).
  4. Finally, notice how the beginning and ending arrows tell us the limits on the outer variable, x or theta.

image window
Select coordinates: rectangular  |  polar
Then step through or play animation:
Step through animation: backward one frame forward one frame
Play: play animation   speed: frames/second   Stop: stop animation

Be sure that you can see for both the rectangular and polar coordinate formulation how the arrows at successive values of the outer variables tell you the range on the inner variables, and how we get from there to the final integrals:

int_0^1 int_sqrt(1-x^2)^sqrt(4-x^2) 2y dy dx + int_1^2 int_0^sqrt(4-x^2) 2y dy dx
and
int_0^(pi/2) int_1^2 2 r^2 sin(t) dr dt

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double integrals
Last modified: Tue Mar 2 14:04:03 EST 2004
Comments to:glarose@umich.edu
©2004 Gavin LaRose, UM Math Dept.