Double Integrals Limits of Integration

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I am just starting to do double integrals and came acorss an issue. I remembered from single integrals when we integrate from limits for say -1 to 1, we can double it and change integration limits to 0 to 1. Now, when is this the case? Basically, when can we not do this?
 

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  • #2
Hootenanny
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I am just starting to do double integrals and came acorss an issue. I remembered from single integrals when we integrate from limits for say -1 to 1, we can double it and change integration limits to 0 to 1. Now, when is this the case? Basically, when can we not do this?
This can be done in general if and only if integrand is even and you are integrating over a symmetric interval about the origin. That is, if [itex]f(x) = f(-x)[/itex] then
[tex]\int_{-a}^a f(x) \text{d}x = 2\int_0^a f(x) \text{d}x\;.[/tex]
For double integration over a rectangular region, you can simply applied the above rule separately to each integral. If [itex]f(x,y) = f(x,-y)[/itex], then
[tex]\int_{-a}^a \int_{-b}^b f(x,y) \text{d}y \text{d}x = 2\int_{-a}^a\int_0^b f(x,y) \text{d}y \text{d}x\;,[/tex]
and if [itex]f(x,y) = f(-x,-y)[/itex]
[tex]\int_{-a}^a \int_{-b}^b f(x,y) \text{d}y \text{d}x = 4\int_0^a\int_0^b f(x,y) \text{d}y \text{d}x\;.[/tex]

If you are integrating over a non-rectangular region in Cartesian coordinates, i.e. if the inner limits depend on the outer variable of integration, then you need to be a little more careful, but you can apply the rule.
 
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Hmm.. I see. Thanks alot. With the functions that arise in double integrals, how to know if the function is even?
 
  • #4
It does not need to be symmetrical around origo does it?

Basically if you have a function such that f(b-x)=f(b+x) then the function is symmetrical around b. And if the integration limits are b+a and b-a, you can use symmetry.

Just and example from the top of my head

[tex] \int_{0}^{2} \sqrt{(x-1)^2-1} dx \, = \, 2\int_{0}^{1} \sqrt{(x-1)^2-1} dx [/tex]

Also if a function is "odd" around a point b. Then the integral evaluates to zero. If it is not specified to find the area. If you are supposed to find the area, just halve one of the integration limits.

[tex] \Large \int_{{\frac{3\pi}{4}}}^{\frac{5\pi}{4}} \tan(x) dx [/tex]

This is an odd function symmetrical around a point b, which makes the integral zero. Same with functions of two variables.

To briefely sum it up

[tex]\int_{a}^{b} f(x) dx [/tex]

Is odd around [itex]c[/itex] if [itex]f(\frac{b+a}{2}-x) \, = \, -f(x) [/itex]
Is even around [itex]c[/itex] if [itex]f(\frac{b+a}{2}-x) \, = \, f(x) [/itex]
And the point it is symmetrical around is [itex]c=f(\frac{b+a}{2})[/itex]
 
  • #5
HallsofIvy
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In general, we do not talk about functions that are "odd" or "even" in two variables. Rather we talk about a function being "odd" or "even" in each variable.

f(x,y)= x^3y^3 is odd in both variables.
g(x,y)= x^2y^3 is even in x, odd in y.
h(x,y)= x^3y^2 is odd in x, even in y.
j(x,y)= x^2y^2 is even in both variables.

Of course, as is the case in one variable, the great majority of functions in two variables are neither even nor odd.

k(x,y)= x^2y^2+ x^3y^3 is neither even nor odd in either variable.
 

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