Professor Dave again, let’s revisit integration.
We’ve pretty much wrapped up linear algebra, but before diving into more advanced topics, let’s use a few tutorials to go over some
things we haven’t covered yet. First off, let’s go back to integration, and see what
happens when we find double or triple integrals. We spent more than a dozen tutorials going
over integration in depth in the calculus series, and we learned that it is a useful
tool to find the area beneath a curve given by some function, f(x). Now let’s add a
dimension. We can think of double integrals as giving the volume beneath a surface given
by the function f(x, y). We are now integrating over two variables, x and y, so for each variable
we have an integration sign, and to indicate the variables involved, we have dx and dy.
Performing multiple integrations is not all that daunting on the surface, it’s simply
a matter of doing the innermost integral first, and then moving outwards. We can use parentheses
to more concretely illustrate this, doing the dx integral inside first, while treating
y as a constant, then integrating the result over y just the same as we would for any variable.
The order in which we do the integrations does not matter as long as we keep track of
which variable has which bounds. In this example we see that dx is associated with the x integrand,
which runs from a to b, while dy is associated with the y integrand, which runs from c to
d. If we wanted to take the y integral first, we would write it like this instead, with
the dy first, and the y integral on the inside. As long as we keep proper track of the variables
and their bounds, we can integrate in any order we want. In more complex problems, this
freedom lets us choose the order which will give the boundaries that are the simplest
to express. When we are looking for a volume within a
rectangular region in the xy plane, double integrals are extremely straightforward. For
example, let’s try to find the volume under the function f(x, y) = 1 + 4xy, where x ranges
from 0 to 1, and y ranges from 1 to 3. The volume will be given by the double integral
here. Let’s go ahead and integrate over x first. Integrating over x, we will treat
y as a constant and find the antiderivative with respect to x, which will be x + 2x2y,
which we will then evaluate from 0 to 1. Plugging in x = 1 we get 1 + 2y, and then we must subtract
the result when evaluated at x = 0, which is just zero. So after evaluating the first
integral, we are left with the integral of 1 + 2y with respect to y from 1 to 3. We treat
this the same as any integral over x, we just have a different variable to work with. The
antiderivative with respect to y will be y + y2 which we will now evaluate from 1 to
3. Plugging in y = 3 we get 3 + 9, then subtract the evaluation at y = 1 which is 1 + 1. We
get 12 - 2, which is 10. This is the result of our double integral, and the volume of
this section under the surface 1 + 4xy. What ends up making multiple integrals tricky
is when the integration domain depends on the variables we’re integrating over. For
example, when the region we are considering in the xy plane involves curves rather than
a rectangular area. In cases like this we will have to be more conscious of how we write
the bounds and the associated order of integration. What is likely to happen is that the bounds
for the first integration will include the other variable, while the bounds for the second
integration must be numbers, to get a final result that is also a number. This is an important
aspect to remember when thinking about the boundaries.
Let’s consider our surface to integrate to be given by the function f(x, y) = 3x2
- 3xy2, and we want to find the volume in the this shaded region between y = x2 and
y = 2x. It’s here that the challenge of establishing our boundaries from this information
starts to present itself. There are two ways to go about this problem, but first we must
take a look at the functions that determine our bounds. Because we have curves where x
and y are related to each other as our boundary, we must take that relation into account in
our integration. First, we must establish where exactly our boundary begins and ends.
We know we want this area between the functions y = 2x and y = x2, but we must determine the
start and end points ourselves. In this case they will be the points where the functions
intersect, or where 2x = x2. By simply moving 2x to the other side we see this is the same
as 0 = x2 - 2x, which factored, gives x times (x - 2). This means that our two points of
intersection are x = 0 and x = 2. We can then get the y coordinates of the two points by
plugging these values into either curve. Plugging in x = 0 into y = 2x gives y = 0, so our first
coordinate is (0, 0). Plugging in x = 2 into y = 2x gives y = 4, so our second coordinate
is (2, 4). Now is the point where we must choose how to express these bounds. We just
found the points of interest on the x and y axis. Our x coordinate can run from 0 to
2 and our y coordinate can run from 0 to 4. However, using both these bounds gives us
the rectangular area here, not the area between the curves. Instead of using both, we must
choose one, and then the curves will become the bounds for the other variable. If we choose
to run x from 0 to 2, then y must run from the lower curve to the higher curve, in this
case from x2 to 2x. We always must choose the lower values as our lower bounds and the
higher values as our upper bounds, otherwise we will be making an error in sign. So for
our volume calculation, we have the integration over x from 0 to 2, the integration over y
from x2 to 2x, of 3x2 - 3xy2 dy dx. Let’s step back and imagine we had chosen to let
y run from 0 to 4 instead. In this case we must now make our x bounds run from the lowest,
or leftmost curve. We are now wanting to start x at its lowest value relative to the curves
going to higher values. So y = 2x is now our lower bound for x, and y = x2 will be our
upper bound. But there’s a slight complication here, we want values of x to be our bounds
for x, but these curves are expressed as values of y. So for each curve we must express them
in the form of x equals some function of y. Both of these are relatively simple, for y
= 2x, by dividing by 2 we get x = y/2. For y = x2 we get x = root y. Keeping track of
which curve was the lower and upper bound, we want x to run from y/2 to root y. So expressing
our volume with these bounds, we end up with the integral over y from 0 to 4, and the integral
over x from y/2 to root y, 3x2 - 3xy2 dx dy. We now see that there are two ways to express
this same volume calculation. When establishing bounds it’s important to remember to keep
the last bound running between two set values, after which we must express the other integral
in terms of the curves of our boundary. There are often times where one choice of bounds
is more beneficial, or makes things easier to calculate. For example, roots often complicate
things as they are messier to deal with than whole integer exponents, and can cause trouble
dealing with negative values. Now let’s actually go ahead and do this
integration. To avoid the square root, we will choose to solve with the bounds where
x runs from 0 to 2 and y runs from x2 to 2x. For this we must integrate with respect to
y first. Taking the antiderivative with respect to y and treating x as a constant, we get
3x2y - xy3. Now don’t let the fact that our bounds involve x intimidate you, we still
treat them as if they were numbers, and evaluate our result from x2 to 2x. As such, we first
plug in y = 2x into our antiderivative and get 3x2(2x) - x(2x)3. Then we subtract from
that the other expression, where we are plugging in y = x2, so we get minus (3x2(x2) - x(x2)3).
Simplifying from there, we are left with the integral over x of 6x3 - 8x4 – the quantity
(3x4 - x7) or 6x3 - 11x4 + x7 dx where x runs from 0 to 2. We’re now in the home stretch
with just one last integral to evaluate. The antiderivative of this expression becomes
6x4/4 - 11x5/5 + x8/8 evaluated from 0 to 2. Plugging in 2 we get 6*(2)4/4 - 11*(2)5/5
+ (2)8/8. From this we must subtract the expression where we plug in x = 0, but this will simply
be 0. We can then plug this into a calculator and find the volume to be -14.4, or -72/5.
This negative value simply indicates that most of the surface 3x2 - 3xy2 in this area
is below the xy plane. And to be clear, we would have gotten this same answer if we went
through the calculation using the other way to express the bounds, we just decided that
this way was easier. As we can see, adding extra integrals is the
same as adding extra dimensions, and our overall process does not change. Three integrals can
be thought of as giving a “hypervolume” in 4-dimensions, 4 integrals would give a
hypervolume in 5 dimensions, and so on. Again, the most important thing to remember is the
order of integration, and just take things one step at a time. A rule of thumb is to
remember that each subsequent integral eliminates a variable. For example if we have a triple
integral of g(x, y, z) dx dy dz, where x runs from a to b, y runs from c to d, and z runs
from e to f. As we go through each integral we must be eliminating the variables x, y,
and z when we integrate each of them. Meanwhile the bounds can still depend on other variables.
When setting them up just keep in mind the order and what possible variables can still
go into the expression without reintroducing ones that have already been integrated out.
For this triple integral, if we integrate x first, its bounds a and b can depend on
either of the other two variables, y and z. If we integrate y next, the bounds c and d
can no longer depend on x but can still depend on z. And finally, the last variable z can
only vary between two constants as all the other variables have been integrated out.
And of course, the bounds cannot have the same variable that is being integrated.
There’s an incredible depth to multiple integrals that can’t be covered in a single
tutorial, but we’ve laid out the type of thinking needed to approach setting up bounds.
Multiple integrals are a key technique in a number of fields of math and science, so
getting down the basics for these is key. With that covered, let’s check comprehension.