Double Integrals: Notation Question

[Zurtex]

Just a question of notation here, my lecturer will wright an integral like this:

$$\int_3^6 dx \int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}y \, dy$$

But mean this:

$$\int_3^6 \int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}y \, dy \, dx$$

Is this standard notation? It seems rather odd to me.

 

[Galileo]

I encounter expressions where the $dx$ or $dy$ comes first often as well.
$$\int_a^b dx f(x)$$
seems to be quite customary.

To be consistent however, I'd use:
$$\int_3^6 dx \int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}dy \;y$$

 

[dextercioby]

Zurtex,it's highly reccomendable you use the first notation...There may be situations (like this one
$$\int_{0}^{3}\int_{-4}^{\pi}\int_{-5}^{9}\sin(xyz+x\sqrt{y}\sqrt[\frac{\sqrt{3}}{2}]{z}) dx \ dz \ dy$$

) in which you never know what integration to do first...

Daniel.

 

[TheDestroyer]

This integral is not possible, because the range for x is from 3 to 6, while you are integrating on a circle has the only x range (-2<x<2)

and about the notation, first is better i advice to use it always.

 

[dextercioby]

If I'm not mistaking,the square roots in "x" disappear afer integrating wrt "y"...So the integral is possible...

Daniel.

P.S.Nothing is wrong.

 

[TheDestroyer]

But it's not a logic integral on the area of that square, is it?

 

[dextercioby]

Who gives a rat's a$$ what that integral represent,as long as it is correct?
BTW,it's not an area at all...
$$S=\iint_{D} dx \ dy$$
is the area of a plain domain from R^{2}.
That integral is something else,as u may see...

Daniel.

 

[TheDestroyer]

Yes, Maybe I'm wrong, hehehe,

 

[Zurtex]

Zurtex,it's highly reccomendable you use the first notation...There may be situations (like this one
$$\int_{0}^{3}\int_{-4}^{\pi}\int_{-5}^{9}\sin(xyz+x\sqrt{y}\sqrt[\frac{\sqrt{3}}{2}]{z}) dx \ dz \ dy$$

) in which you never know what integration to do first...

Daniel.

But that makes more sense to me than the first, the first just seems confusing and looks like he is multiplying them.

 

[dextercioby]

For "nice" cases,the theorem of iteration can be applied...But in this case,there's no multiplication/iteration,just an elegant way to saying what integration is performed first...

Daniel.