#### Buzz Bloom

Gold Member

- 1,908

- 313

I am a bit confused by problem (12. a).

I am describing my confusion in a "SPOILER" box because I am not sure if it might unintentionally help someone solve the problem.

The area where f(x,y) ≠ 0 is a parallelogram bounded by the following four line segments connecting pairs of (x,y) pints.

In the TOP area, f(x,y) = -1.

In the BOTTOM area, f(x,y) = +1.

(Along the line L f(x,y) = -1 but this does not influence the value of the integration.)

Therefore the integral

∫

What confused me is: Why it is necessary to bring Fubini's theorem into the discussion?

If Fubini's theorem must be discussed, then I offer the following.

Since the order of the two integration variables does not change the zero result, this demonstrates that

I am describing my confusion in a "SPOILER" box because I am not sure if it might unintentionally help someone solve the problem.

[(0,0), (1,0)]

[(1,0), (1,3)]

[(1,3), (2,0)]

[(2,0), (0,0)]

This parallelogram is divided into two regions, TOP and BOTTOM. The dividing line L separating TOP and BOTTOM is[(1,0), (1,3)]

[(1,3), (2,0)]

[(2,0), (0,0)]

L = [(0,1), (1,2)].

The area of TOP is the same as the area of BOTTOM. Both areas are 1 square unit.In the TOP area, f(x,y) = -1.

In the BOTTOM area, f(x,y) = +1.

(Along the line L f(x,y) = -1 but this does not influence the value of the integration.)

Therefore the integral

∫

_{R}∫

_{R}f(x,y) dx dy = ∫

_{R}∫

_{R}f(x,y) dy dx = 0.

What confused me is: Why it is necessary to bring Fubini's theorem into the discussion?

If Fubini's theorem must be discussed, then I offer the following.

Since the order of the two integration variables does not change the zero result, this demonstrates that

__the order does not matter__, which is what Fubini's theorem proves.

Last edited: