Understand value of two dimensional integral with a delta function I calculated the following integral with maple (and analytically): $$\int_0^\infty \mathrm dy_1\int_0^\infty\mathrm dy_2  \delta\left(y_1+y_2-2\right)=2$$
Can anyone explain this results 'geometrically'? 
I am expecting this integral to be the line between the point (0,2) and (2,0) in 2 dimensions. But the length of this line is $\sqrt{2^2+2^2}=\sqrt{2} \cdot 2$.
 A: The delta function defines a line on which the integral is non-zero.  The output is 1 within the bounds of the integral.  For the first one, you get $1$ when $y_1+y_2=2$ and zero otherwise.  For the first quadrant, this is only valid when $y_1$ is between 0 and 2.  So the $y_1$ integral here becomes $$\int_0^2 1dy_1=2$$
The fact that it is non-zero along line does not change the fact that the second integral evaluated is only along 1 coordinate. 
A: In addition to Paul's answer, you may interpret the discrepancy you observe "geometrically" by thinking of a delta function as being approximated by characteristic function of $[-\varepsilon/2, \varepsilon/2]$ times $1/\varepsilon$ (i.e. function which is $1/\varepsilon$  on the interval and $0$ off it). Then the function you are integrating will be supported on a strip around the segment $y_1+y_2=2$ whose vertical slices are of length $\varepsilon$ so its thickness (in the direction perpendicular to the direction of the segment) is $\varepsilon/\sqrt{2}$. Thus the function's integral will be (approximately) $\varepsilon/\sqrt{2}\times \text{ length }\times 1/\varepsilon=\text{ length }/\sqrt{2}$, as you observe.
