Why does this integral lead to zero, while the other does not.

[FOIWATER]

why does this integral:

double integral (x+y) dA , A is the area bounded by the triangle with vertices (0,0) (0,1) (1,0)
equal 1/3

While this one, (x-y) dA , A is the area bounded by the same vertices
equal 0?

(not looking for the math, already worked them out, just looking for some logic behind why the second one would be zero).

Thanks!

 

[lurflurf]

well for the same domain

∫x dA=∫y dA=1/6

and integration is linear

 

[slider142]

Also, geometrically, it implies the function z = x - y has the same volume below the plane z = 0 restricted to that region as above it. Since the integral is geometrically measuring the signed volume between z = x - y and z = 0 restricted to that region, the two different signed volumes precisely cancel.

 

[Vargo]

Your integral is like the sum of the values of f(x,y) over the region. You are looking at a region that is symmetric with respect to reflection across the line y=x. In other words, for each (a,b) in your region, the reflected point (b,a) is also in the region.

For the function f(x,y)=y-x, notice that f(y,x) = -f(x,y). So when you add up the values of f over the region, they cancel each other out. By similar logic, the following functions would all yield 0 when integrated over the same region:
sin(x-y)
(x-y)^2sin(x-y)
5(y-x)+10(y-x)^7

 

[FOIWATER]

makes sense can't believe I didn't see it.

Thanks, guys