double integral: int(int(xy))dA over triangular region

[thebenji]

Evaluate the double integral:

I= Integral[Integral[xydA

over the triangular region with vertices (0,0) (1,0) and (0,1)

I tried it as if it were a square:

Integral[0 to 1]Integral[0 to 1] xy*dxdy

and wound up with 1/4

Then I divided it by two: 1/8

I knew that would most likely not work...but it was worth a shot.

How to do?

 

[galactus]

The equation of the line from (0,1) to (1,0) is y=1-x.


\(\displaystyle \L\\\int_{0}^{1}\int_{0}^{1-x}(xy)dydx=\frac{1}{24}\)

 

[thebenji]

But the answer is not 1/8.

The answer that I came up with and you came up with is, however, 1/8.

Perhaps the error is not ours.

 

[galactus]

Sorry, I had the wrong line equation. I fixed my original post.

The line from (1,0) to (0,1) is y=1-x