How to find the bounds of a triple integral
When a triple integral is applied to function it gives us a hypervolume.
This hypervolume comes from the space that is bounded by a given three dimensional region
and the surface of the function.
If however we apply a triple integral to a function that is always equal to 1, then it
simply gives us the volume of that three dimensional region.
In either case, you want to recgnize that the bounds of the triple integral are describing
some three dimensional space.
This means if we can fully describe the surfaces that make up that region, then we can quickly
and easily build the bounds for our triple integral.
To do this we can use the method of collapsing.
In this method we choose a variable, or direction to walk through the region.
When then descibe the surfaces that we must enter and exit as we move through it.
These surfaces are described by functions, and they become the bounds of our integral.
Specifically these bounds will be on the integral associated with our direction of movement.
Finally the region collapses onto the remaining axis so that we can repeat the process for
the remaining variables.
Let's see how this works when our region is being described using rectangular coordinates
Here we have a region that is being bounded by three surfaces.
The top describes the plane z= y + 1 The bottom is the parabolic cylinder z = x^2 + 1
and the side is the plane y = 1
We can begin by choosing to move through this region in the z-direction.
As you can see we'll have to enter in the bottom of the parabolic cylinder, and we'll
have to exit out the plane. So the bottom bound will be x^2 + 1 and the
top bound will be y + 1. Now let's collapse our region onto the remaing
variables of x and y.
Now our region is much simpler since it only has 2 dimensions.
Like before we want to choose a direction to move through this region so we can build
out next set of bounds. Let's choose the y-direction.
In this direction we would enter along the bottom surface given by the parabola y = x^2
and we would exit out the line y = 1. So the bottom bound is x^2 and the top bound is 1
Now collapses in the y-direction, leaving
us with just a thin region on the x-axis.
Since there is only one variable left, we can now only move in the x-direction.
Now we'll be entering at the point x = -1 and exiting at the point x = 1.
This gives us our final bounds.
Its important to note that the order in which we choose our variables does not matter as
long as our function is continuous within side the region.
To see this, let's reset our region, and come up with some brand new bounds.
We'll begin by moving through the region in the y-direction.
This will force us to enter the slanted plane of z = y + 1 and exit out the plane y = 1.
These will become our bottom and top bounds. Since we are moving in the y-direction, we
must describe both of these surfaces in terms of y.
So the bottom bound will be y = z - 1 and the top bound will be y = 1.
We can now collapse our region on the the remaining variables of x and z.
Let's now move through the collapsed region in the x-direction.
As you can see we'll enter in the left of a parabola and exit out the right of a parabola.
This parabola is given by the equation z = x^2 + 1
This may seem like a problem, but remember
we are moving in the x-direction so we want to describe these surfaces in terms of x.
Subtracting 1 from both sides and taking the square root, we see that we get two surfaces
from our one equation. x = -sqrt(z - 1) and x = sqrt(z - 1)
These will become the bottom and top bounds for our integral.
We now collapse in the x-direction.
This leaves us with a thin line segment along the z-axis.
As we move through this region, again in the z-direction, we would enter at z = 1 and exit at z=2.
These are the remaining bounds of our triple integral
Keep in mind that even if the order of our variables does not matter, we should still
pick the order with care so that our bounds are nice and simple.
This is especially true, if you need to evaluate this integral, once the bounds are set.
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