Double Integrals: Interchangeability and Improper Integration Explained

  • Thread starter haha1234
  • Start date
  • Tags
    Integrals
In summary: I thought only about the integrals over a rectangle. I am sorry, I will do better!In summary, the order of integration can usually be interchanged, but there are restrictions described in Fubini's theorem. If the domain of integration is both x-simple and y-simple, the order can be changed. However, this may not hold true for infinite domains. A counterexample is given by the function f(x,y) = e^(-xy) - 2e^(-2xy) on the domain 1 ≤ x < ∞ and 0 ≤ y ≤ 1. The iterated integrals cannot be evaluated explicitly and Fubini's theorem fails in this case.
  • #1
haha1234
109
0
I started learning double integrals in this semester but I got some problems on it.
Is it matter if the order of integration interchange?
For example, V=∫f(x,y)dydx, can we rewrite the equation as V=∫f(x,y)dxdy?
If not, could you explain on it little bit:sorry:
Thanks!
 
Physics news on Phys.org
  • #2
In terms of appearance, ##dx\hspace{1mm}dy## and ##dy\hspace{1mm}dx## bear no difference in mathematical meaning. However, the order with which the integral calculation is carried out, in general, cannot be interchanged without changing the given form of the integration limits. For example, a question asks you to integrate a function ##f(x,y) = y## over an area bounded by ##x^2+y^2 \leq 9##, ##x>0##, and ##y>0##, which is the first quadrant of a circle with radius 3. The limits of the integral is not fixed for any value of ##x## and ##y##. In this case there are two ways to perform the integral, the first one is to integrate over a vertical stripe located at a fixed ##x## (which must lies within the said limiting area) and then integrate over ##x## from 0 to 3. The first integral over the stripe is obviously performed along ##y## direction from ##y=0## to ##y=\sqrt{9-x^2}##, this looks like
$$
\int_0^3 \int_0^\sqrt{9-x^2} ydy\hspace{1mm}dx
$$
In that integral, you will calculate the ##y## integration first followed by ##x## integration. However, you can also flip the order by integrating over ##x## first. But if you do this, you will have to change the limit as well. This means, you will now be calculating integral over a horizontal stripe of fixed ##y## first from ##x=0## to ##x=\sqrt{9-y^2} ##, and then integral over ##y## from 0 to 3.
$$
\int_0^3 y\int_0^\sqrt{9-y^2} dx\hspace{1mm}dy
$$
You should get identical answers from both paths. Note that, after you learn change of coordinate, you will see that this integral can be done in polar coordinates with fixed integral limits, which means the order of integration will not matter anymore.
 
  • #3
haha1234 said:
I started learning double integrals in this semester but I got some problems on it.
Is it matter if the order of integration interchange?
For example, V=∫f(x,y)dydx, can we rewrite the equation as V=∫f(x,y)dxdy?
If not, could you explain on it little bit:sorry:
Thanks!
Yes, usually you can switch the order of integration, but not always. The restrictions are described in Fubini's theorem (https://en.wikipedia.org/wiki/Fubini's_theorem).
 
  • #5
The iterated integrals need not be the same — even on a domain of integration that is "both x-simple and y-simple" — it if it permitted to be infinite in at least one direction.

Here is a standard counterexample: consider this function:

f(x,y) = e−xy − 2e−2xy

on the domain 1 ≤ x < ∞, 0 ≤ y ≤ 1.

Note that f(x,y) > 0 everywhere on this domain.

The two iterated integrals cannot be evaluated explicitly in terms of elementary functions, but using numerical approximation I get

∫ (∫ f(x,y) dy) dx ≈ 0.17048+

whereas

∫ (∫ f(x,y) dy) dx ≈ 0.52266+
, showing that Fubini's theorem can fail unless certain conditions are met — see Svein's link above to the Wikipedia article.
 
  • #6
zinq said:
f(x,y) = e−xy − 2e−2xy on the domain 1 ≤ x < ∞, 0 ≤ y ≤ 1.
Note that f(x,y) > 0 everywhere on this domain.
Sorry: f(x,0) = -1...
 
  • #7
Oops, did I say "Note that f(x,y) > 0 everywhere on this domain" ?

Wrong! Sorry about that.

Setting

f(x,y) = e−xy − 2e−2xy = 0​

to see where f(x,y) changes sign, first multiply by e2xy, which definitely is positive everywhere, to get:

exy = 2​

and so the expression changes sign along the locus of

xy = ln(2)​

where that passes through the domain 1 ≤ x < ∞, 0 ≤ y ≤ 1. In other words, the curve

S = {(x, ln(2)/x) | x ≥ 1}.​

(To prove this, it suffices to check that the critical points of the graph of z = f(x,y) do not intersect the curve S. This is left to the reader.

Numerical experiment suggests the minimum on the domain 1 ≤ x < ∞, 0 ≤ y ≤ 1 is -1 (the value taken at all points (x, 0) for x ≥ 1).

Tricky!
 
  • #8
In my answer I did not regard the improper integrals ...
 

What is a double integral?

A double integral is a type of integral that involves integrating over a two-dimensional region. It is denoted by the symbol ∫∫ and is used to find the volume under a surface in three-dimensional space.

What types of problems can be solved using double integrals?

Double integrals can be used to solve problems involving finding the area, volume, mass, and center of mass of two-dimensional shapes and three-dimensional objects. They are also used in physics and engineering to calculate various quantities such as electric fields and moments of inertia.

How do you set up a double integral?

To set up a double integral, you need to first determine the limits of integration for both the inner and outer integrals. This involves finding the range of values for both variables that define the region of integration. Once the limits are determined, the double integral is written as ∫∫f(x,y) dA, where f(x,y) is the function being integrated and dA represents the area element.

What are some common techniques for solving problems on double integrals?

Some common techniques for solving problems on double integrals include using Cartesian coordinates, polar coordinates, and changing the order of integration. Other techniques such as using symmetry and using the properties of integrals can also be helpful in simplifying the calculation process.

How do you evaluate a double integral?

Evaluating a double integral involves first setting up the integral and then using various integration rules and techniques to solve it. This can include using the Fundamental Theorem of Calculus, substitution, and partial fractions. In some cases, the integral may need to be solved numerically using techniques such as Simpson's rule or the trapezoidal rule.

Similar threads

  • Calculus
Replies
11
Views
2K
Replies
2
Views
1K
  • Calculus
Replies
24
Views
3K
Replies
4
Views
639
  • Calculus
Replies
5
Views
2K
Replies
2
Views
2K
Replies
2
Views
829
Replies
2
Views
178
Back
Top