sketch the region bounded by the graphs of f(y)=(y/Squareroot of(16-y^2)), g(y)=0, y=3 and find the area of the region.

Answer 1

x = y/√(16-y^2)
as seen on Wolfram
http://www.wolframalpha.com/input/?i=x+%3D+y%2F√%2816-y%5E2%29

taking horizontal slices
Area = ∫( y/(16-y^2)^(1/2) dy from 0 to 3
= - (16-y^2)^(1/2) | from 0 to 3
= -(16-9)^(1/2) - (-16^(1/2))
= -√7 + 4
= 4-√7 = appr 1.35