Posted by andy on Thursday, July 25, 2013 at 2:11pm.
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.

calculus  Steve, Thursday, July 25, 2013 at 2:53pm
The region is the upper half of a circle with radius 1/2 centered at (1/2,0):
(x1/2)^2 + y^2 = 1/4
So,
∫[0,1]∫[0,√1/4  (x1/2)^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))

calculus  Steve, Thursday, July 25, 2013 at 2:54pm
oops that's log(y + √(x^2+y^2))
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