Notes to a video lecture on http://www.unizor.com
Double Integrals -
Volume of a Solid
Consider the following problem.
Given a smooth function of two variables f(x, y) (we will always consider smooth functions in terms of continuity and sufficient differentiability), with non-negative values (that is,
In the following we will use the word "area" in a sense of a two-dimensional part of a plane and as a quantitative measure of this part of plane. The context would clarify which one it is used in every case.
Our task is to find the "volume of a solid" - the measure of a part of coordinate space bounded on the top by a surface representing this function, on the bottom - the X,Y-plane and on the four sides by planes
There is no ready to use formula for such a volume. We do know how the volume of a rectangular parallelepiped is defined, it is a product of its three dimensions - length multiplied by width, multiplied by height, but not of such a complicated figure as the one we consider now.
We really have to define what the volume of this figure is and then attempt to calculate it based on values of function
We did have a similar problem in Geometry with the volume of solids and approached it as a sequence of approximations of a complex figure with simple ones. Let's do the same now.
We will use certain intuitive considerations to define the volume of a solid and will prove that this definition is mathematically valid.
The process of approximation starts with dividing rectangle
If point (xi, yj) represents the coordinates of a "top right" corner of a small base rectangle and Δxi=xi−xi−1, Δyj=yj−yj−1are sides of this small rectangle, the volume of a constructed rectangular parallelepiped equals to
ΔVi,j = f(xi, yj)·Δxi·Δyj
The combined volume of all these parallelepipeds approximates the volume of our solid. With increasing number of small base rectangles and, correspondingly, decreasing their dimensions we assume that there is a limit of the combined volume of all these parallelepipeds.
It can be proven that under certain conditions of smoothness of our function
Now let's consider how we can summarize the volumes of all parallelepipeds constructed on the basis of small rectangles. There are two simple methods.
1. We can go from one point y=yj to another, varying index j from 1 to N and for each value of y=yj summarize the volumes of all parallelepipeds based on all values of x=xi, varying index i from 1 to M. So, for each index j we calculate the sum of all Vi,j, varying index i.
Then we summarize these sums by index j deriving with an approximate volume
VM·N = Σj∈[1,N] Σi∈[1,M] Vi,j =
= Σj Σi f(xi, yj)·Δxi·Δyj =
= Σj [Σi f(xi, yj)·Δxi]·Δyj
2. On the other hand, we can go from one point x=xi to another, varying index i from 1 to M and for each value of x=xisummarize the volumes of all parallelepipeds based on all values of y=yj, varying index j from 1 to N. So, for each index i we calculate the sum of all Vi,j, varying index j.
Then we summarize these sums by index i deriving with an approximate volume
VM·N = Σi∈[1,M] Σj∈[1,N] Vi,j =
= Σi Σj f(xi, yj)·Δxi·Δyj =
= Σi [Σj f(xi, yj)·Δyj]·Δxi
Both approaches to summarizing the volumes of small parallelepipeds should result in the same value of a combined volume.
Let's examine now what happens with both our formulas for total volume when the rectangles are getting smaller and smaller.
Recall the definition of a definite integral of function φ(t) on a segment p ≤ t ≤ q as a limit of sums of areas of rectangles obtained by dividing segment [p,q] into small pieces:
Σk φ(tk)·Δtk → ∫pq φ(t) dt
We can now say that
Σj f(xi, yj)·Δyj converges to
φ(xi) = lim Σj f(xi, yj)·Δyj =
= ∫cd f(xi, y) dy
under condition of maximum among Δyj converges to zero. Then we continue with the second summation getting
Σi [Σj f(xi, yj)·Δyj]·Δxi →
→ Σi [∫cd f(xi, y) dy]·Δxi →
→ ∫ab[∫cd f(xi, y) dy] dx
On the other hand, we can similarly state that
Σi f(xi, yj)·Δxi converges to
ψ(yj) = lim Σi f(xi, yj)·Δxi =
= ∫ab f(x, yj) dx
under condition of maximum among Δxi converges to zero. Then we continue with the second summation getting
Σj [Σi f(xi, yj)·Δxi]·Δyj →
→ Σj [∫ab f(x, yj) dx]·Δyj →
→ ∫cd[∫ab f(x, y) dx] dy
But these are supposed to be the same results since the difference is only in the order of summation. The sums are the same, therefore the limits are the same, therefore we can write this limit as a double integral:
∫cd∫ab f(x, y) dx dy
∫ab∫cd f(x, y) dy dx
Both are the same and constitute two integrations of function f(x, y) performed consecutively, either by x from a to b (keeping y as a constant) and then by y from c to d or, in reverse, first by y from c to d (keeping x as a constant) and then by x from a to b.