The straight line x = y, that is, y = x, and the polar coordinate is θ = π/4.
The straight line y = 1, that is, rsinθ = 1, r = 1/sinθ.
The integration domain D is a triangle with O (0 0,0), A (1, 1) and B (√ 3, 1) as vertices.
In polar coordinates π/6 ≤ θ≤π/4, 1 ≤ r ≤ 1/sinθ,
Then the original integral I = ∫ < π/6, π/4 >; dθ∫& lt； 0， 1/sinθ& gt； f(rcosθ，rsinθ)rdr