Download An Introduction to Double Integrals Math Insight Suppose that you

An Introduction to Double Integrals
Math Insight
Suppose that you knew the hair density at each point on your head and you
wanted to calculate the total number of hairs on your head.
In other words, let (x,y) be a point that is x millimeters to the right
and y millimeters above some reference point, say your nose. We are assuming
that you already know the function f(x,y) that gives hair density in hairs per
square millimeter at point (x,y). The following might be a method to use your
knowledge of f(x,y) to estimate the number of hairs on your body.
1. Cut and flatten your skin so it lies in a plane.
(Although this has nothing to do with double integrals, brain mappers and
cartographers face similar problems. To map the brain or the surface of the
earth, one looks for ways to flatten these surfaces into a plane.)
2. Divide your skin into small rectangles of width Δx and height Δy.
3. Label each rectangle according to row i and column j. For rectangle ij, pick
some point in the rectangle, and call it (xij,yij). Since you know your hair density
function, you can look up the hair density of that point. It is simply f(xij,yij).
Label each rectangle by the number f(xij,yij).
4. If the hair density were constant in each rectangle, the number of hairs in
rectangle ij would be f(xij,yij) times the area of the rectangle. The area of the
rectangle is simply its width (Δx) times its height (Δy), i.e., the area is ΔxΔy.
Hence, the number of hairs in the rectangle is approximately f(xij,yij)ΔxΔy.
Source URL: http://mathinsight.org/double_integral_introduction
Saylor URL: http://www.saylor.org/courses/ma103/
Attributed to: [Duane Q. Nykamp]
www.saylor.org
Page 1 of 5
5. To estimate the total number of hairs on your head, you can add up the
(approximate) number of hairs in each rectangle. Using the above result, your
estimate for the total number of hairs is
where the sum is over all rectangles.
If in the above picture, each rectangle were 75 millimeters wide and 65
millimeters high, then the resulting estimate of the total number of hairs would
be (9+9+8+17+9+3+1+1+11+8+10+8+1+2+3+8+7+2+5+3)⋅75⋅65=609,375
6. The above result is only a rough estimate because it assumed that the hair
density was constant over each rectangle. You may also have noticed that
additional errors are introduced around the edges of your skin, where some
rectangles are only partly filled with skin. You can increase your accuracy by
decreasing the size of each rectangle (i.e., decreasing Δx and Δy). Of course, to
cover your whole head, you'll have to increase the number of rectangles as you
decrease their size.
7. To be really accurate, you should let the size of the rectangles go to zero (and
the number of rectangles go to infinity). In other words, you should take the limit
where Δx→0 and Δy→0.
As long as f(x,y) is a continuous function, this procedure will converge to a single
number, which would be the actual number of hairs on your head.
8. We can put this into math terms. When we took the limit as Δx→0 and Δy→0,
we ended up with the definite integral of the function f over your head. If we
denote by D the region of the plane that your skin occupied, then we write this
integral as
We refer to this integral as the double integral of f over D.
The sums of step 5 are the Riemann sums that approximate the integral. The
integral is the limit of the Riemann sums as the size of the rectangles goes to
zero. This is exactly the way you defined the integral in one-variable calculus.
Source URL: http://mathinsight.org/double_integral_introduction
Saylor URL: http://www.saylor.org/courses/ma103/
Attributed to: [Duane Q. Nykamp]
www.saylor.org
Page 2 of 5
You can read how we can interpret the double integral as volume1 underneath a
surface, just like you could interpret the regular one-variable integral as area
under a curve. In this case, we can also visualize the Riemann sum defining the
integral as the volume of many boxes, as illustrated in the below applet. (More
details on this volume interpretation and this applet can be viewed on this page2.)
Double integral Riemann sum. The volume of the small boxes illustrates a Riemann sum
approximating the volume under the graph of z=f(x,y) over the region D, i.e., the double
integral ∬DfdA for f(x,y)=cos2x+sin2y and D defined by 0≤x≤2 and 0≤y≤1. The volume of the
boxes is
where xi is the midpoint of the ith interval along the x-axis and yj is the midpoint of the jth interval
along the y-axis. The purple line of the cyan slider shows the volume estimated by the volume of
the boxes, and the blue line of the cyan slider shows the actual volume underneath the surface.
As Δx and Δy approach zero, the purple line approaches the blue line, illustrating how the
estimated volume approaches the actual volume.
I don't have good examples (other than the above hair-counting example) on
computing double integrals this way, as this is not how we typically compute
Source URL: http://mathinsight.org/double_integral_introduction
Saylor URL: http://www.saylor.org/courses/ma103/
Attributed to: [Duane Q. Nykamp]
www.saylor.org
Page 3 of 5
them. Instead, this page is about how we define a double integral. We have
better ways to compute double integrals (that is, unless you are a computer, in
which case chopping up the domain in pieces and computing a sum as an
approximation to an integral works pretty well).
You can also read examples of computing double integrals3 using the method in
which those of us who are not computers typically use, which is something called
an iterated integral4.
Source URL: http://mathinsight.org/double_integral_introduction
Saylor URL: http://www.saylor.org/courses/ma103/
Attributed to: [Duane Q. Nykamp]
www.saylor.org
Page 4 of 5
Notes and Links:
1. http://mathinsight.org/double_integral_volume
2. http://mathinsight.org/double_integral_volume
3. http://mathinsight.org/double_integral_examples
4. http://mathinsight.org/double_integral_iterated
Source URL: http://mathinsight.org/double_integral_introduction
Saylor URL: http://www.saylor.org/courses/ma103/
Attributed to: [Duane Q. Nykamp]
www.saylor.org
Page 5 of 5