What makes double integrals tricky is finding the bounds in non-rectangular regions. Here we go through what that means and practice a few examples.
What we're building to
- If you wish to perform an integral over a region of the
-plane that is not rectangular, you have to express each of the bounds of the inner integral as a function of the outer variable.or alternatively,
The trouble with non-rectangular regions
Consider the function
Its graph looks like this:
We will find the volume under a portion of this graph. Unlike the last article, this volume will not lie above a rectangular region on the
-plane. Instead, we will look for a volume whose base is a triangle. The triangle pictured below, to be specific.
This is a right isosceles triangle, one of whose legs connects the points
and on the -axis, while the other leg connects the points and . The volume above this triangle and below the graph of looks like this:
This is similar to the problem I showed in the last article, which introduced the double integral. And indeed, the way to solve it is similar.
- Find a formula for slices of area using an integral.
- Use a second integral to add those infinitely many slices of area into a volume.
What gets tricky now is the bounds. For example, consider the slices of this volume which represent constant
values. The following animation shows what these slices look like, as the constant -value varies back and forth between and .
The height of one of these slices changes based the height of the graph of
above its base. But the length of the base of the slice also changes. For example, when , the value of at the base can range from to , as in the vertical red stripe pictured below.
, the value of ranges from to :
This means when we set up an integral to find the area of one of these constant-
-value slices, the upper bound is written in terms of .
As far as our computations are concerned, it's perfectly fine to have one of the bounds written in terms of
. After all, we'll end up with an expression in terms of anyway. Go ahead and work out the integral for yourself:
From here, there is nothing new. Multiply this value by
to give it a little depth, and hence make it an infinitesimal volume. Then when we integrate it with respect to , the bounds are constants, and , since this is where the base of our triangle sits on the -axis.
The total volume is therefore
Integrating over a disk
Now let's try something a little harder: finding the volume under a graph bounded by the unit disk. The unit disk on the
-plane is all points such that
For example, the volume underneath the graph
bound by the unit disk looks like this
Once again, consider slices of this volume which correspond to constant
Think about what the base of each of these slices looks like on the
-plane. Each slice corresponds with some vertical stripe in the unit disk.
Using the pythagorean theorem, we can find the
-values which determine the top and bottom of this stripe as a function of the -value that the stripe represents.
We can now find the area of one of these constant-
-value slices by integrating with respect to . Again, where this is different from the case of rectangular regions is that the bounds are each a function of .
Concept check: Which of the following integrals represents the area of a constant-
-value slice of the volume we are looking for?
Work through it: This is a heavier computation than previous examples, but if you feel up to it, compute this integral to get a formula for the area of a constant-
-value slice, as a function of .
Area of a constant-
-values in the unit disk range from to , so to find the volume we are looking for, integrate the expression you just found with respect to between the value and . As before, you can imagine this as adding up many, many paper-thin volumes.
This turns out to be a tricky integral, but for pragmatism's sake we can solve it using any ol' computer algebra system or numerical integration tool, such as Wolfram Alpha.
Slice the other way: Shark fin region
Sometimes it's easier to consider constant-
-value slices, which involves cutting your region in -plane along horizontal stripes. For example, consider region of the -plane satisfying the following properties:
This region kind of looks like a shark's dorsal fin:
The upper right corner of the region is where the curve
meets the line . That point is .
Let's find the volume of a solid that has this region as its footprint, and whose height is determined by a relatively simple multivariable function:
Here's what the volume looks like:
This time, imagine cutting constant-
-value slices of this volume. This will give the area above a horizontal strip of our shark fin region, such as the one pictured below in red.
Concept check: If one of these horizontal stripes corresponds to a value
, what are the bounds on the -value of the stripe? That is, what are the -coordinates of the left and right ends of this line as a function of ?
Concept check: Which of the following integrals represents the slice of area above one of these stripes, and under the graph of
, as a function of ?
Concept check: Solve this integral to find the area of the constant
-value slices of our volume.
Area of constant-
Concept check: When we integrate this function of
to get the total volume, what bounds should we use?
Bring it on home: Solve this integral to find the volume of the region defined at the start of this section. (Feel free to use a calculator).
When you need to perform a double integral over a non-rectangular region, follow these steps.
- Start by cutting your region along slices that correspond with holding one of the variables constant. For example, holding
at some constant value will give a vertical stripe of your region.
- Find how to express the bounds of these stripes as a function of the other variable. For example, the top and bottom of a vertical stripe would be expressed as some function of
- When you set up your double integral, the inner integral will correspond to integrating along one of these stripes, and each of its bounds will be a function of the outer variable. If the inner integral corresponds to constant-
-values, the double integral as a whole might look like this:Alternatively, if you started with horizontal constant- -value slices, the double integral might look like this:
Want to join the conversation?
- The second bullet point of the summary...."the top and bottom of a vertical stripe would be expressed as some function of y" this should be "as some function of x" shouldn't it?(6 votes)
- Hello friends, In the third bullet of the summary where above the first integral the text states "Evaluates to some function of y" I think it should be x. Also in the third bullet of the summary where above the second integral the text states "Evaluates to some function of x" I think it should be y. If you have doubts about this comment, just contrast the summary [as of April 2019] with what is written in "What we're building to." I think the comment on the text made by Mr. Robles has been fixed. Wonderful explanation of the topic. Best wishes to all.(5 votes)
- At the end of the summary, it says "horizontal constant-y-value" slices and it shows an inner integral of dy. But in the shark-fin example, our slices were also horizontal, though with an inner integral of dx. What's up?(3 votes)
- You are right, what they really meant at the end of the summary was "vertical constant-x-value" (as indicated by the "dy")(4 votes)
- I am completely puzzled. I don't know what
"Using the pythagorean theorem, we can find the yyy-values which determine the top and bottom of this stripe as a function of the xxx-value that the stripe represents."
is referring to. Why do we use it here and how did we get the picture of the stripe?(3 votes)
- The arbitrary (any possible numbers) bounds on the integral ensure proportionality of the boundary of base area, which must be adhered to in the first place.
Thus, doing the integral yields the general formula for a constant x or y-slice.(2 votes)
- Regarding the first function in this article, f(x,y)=xy^2 it seems the order of integration matters in this problem. In the computation in the article we obtained the answer x^4/3. No problem, I see that. But then I tried to compute with the bounds in terms of y first (I changed the order of integration) and integrate with respect to x and obtained y^4/2: And no matter what I do, I can't reconcile this outcome. One way the answer is 32/15 and the other way it is 32/10(1 vote)
- maybe you made a mistake since I tried to calculate as well in terms of y and got the same answer: 64/30 = 32/15(1 vote)
- Hi, just wondering if there is a trick for knowing if a region is rectangular or not?(1 vote)