Evaluate ∬R 1/sqrt(x^2+y^2) dx dy, where R is the region bounded by y≥0 and x^2−x+y^2≤0.

Answer 1

The region is the upper half of a circle with radius 1/2 centered at (1/2,0):

(x-1/2)^2 + y^2 = 1/4

So,
∫[0,1]∫[0,√1/4 - (x-1/2)^2) 1/√(x^2+y^2) dy dx

recall that √1/(x^2+y^2) dy = arcsinh(y/x) = log(x + √(x^2+y^2))

Answer 2

oops that's log(y + √(x^2+y^2))