MHB The Origin of the \hat x Term in Evaluating Double Integrals over Triangles

[bugatti79]

Folks,

Self reading a book in which an equation is given as

I_{mn}\equiv\int_{\Delta} x^m y^n dx dy

where we are integrating an expression of the form x^m y^n over an arbirtrary triangle.

Is the above actually a double integral because of the dxdy term? Ie can this be written

I_{mn}\equiv\int_{\Delta} x^m y^n dx dy= \int \int_{D} x^m y^n dA where D is the triangle?

Thanks

 

[Ackbach]

Technically, the $dA$ differential is a double integral, and $dx \, dy$ differentials signify an iterated integral. I think many authors don't make a huge distinction between the two. The double integral is the more general concept - a particular iterated integral is coordinate dependent, usually.

 

[bugatti79]

Technically, the $dA$ differential is a double integral, and $dx \, dy$ differentials signify an iterated integral. I think many authors don't make a huge distinction between the two. The double integral is the more general concept - a particular iterated integral is coordinate dependent, usually.

Thanks for that.

I have found a nice link - Double integrals as iterated integrals - Math Insight

Cheers

 

[bugatti79]

Folks,

Self reading a book in which an equation is given as

I_{mn}\equiv\int_{\Delta} x^m y^n dx dy

where we are integrating an expression of the form x^m y^n over an arbirtrary triangle.

Is the above actually a double integral because of the dxdy term? Ie can this be written

I_{mn}\equiv\int_{\Delta} x^m y^n dx dy= \int \int_{D} x^m y^n dA where D is the triangle?

Thanks

In the book I am reading they evaluate the following integral to be

\int_{\Delta} x dx dy= A \hat x where

\displaystyle \hat x= \frac{1}{3} \Sigma_{i=1}^3 x_i and A=\int_{\Delta} dx dy=xy

Where does \hat x come from? I realize its to do with the 3 coordinates of the triangle...