Area problem.

Frist of all, finding the points of intersection between the curves:

1) Between xy = 9 and x2 - y2 = 1:

Sub y = 9/x into x2 - y2 = 1:

x2 - 81/x2 = 1

x4 - 81 = x2

x4 - x2 - 81 = 0

x2 = (5√13 + 1)/2 since it must be positive

x = √[(5√13 + 1)/2] since first quadrant is considered

y2 = (5√13 - 1)/2

y = √[(5√13 - 1)/2]

2) Between xy = 9 and x2 - y2 = 4, by similar method, we get:

x = √(√85 + 2) and y = √(√85 - 2)

Now, rewriting the curves with x being the subject:

x = 9/y

x = √(1 + y2) for the first quadrant

x = √(4 + y2) for the first quadrant

Integrating along the y-axis, i.e. w.r.t. y, we should divide the integral into 2 parts:

1) From y = 0 to √(√85 - 2), between the curves x = √(4 + y2) and x = √(1 + y2)

2) From y = √(√85 - 2) to √[(5√13 - 1)/2], between the curves x = 9/y and x = √(1 + y2)

So the area is given by:
圖片參考：http://i1191.photobucket.com/albums/z467/robert1973/Sep11/Crazyint1.jpg

By change of variable
http://i1090.photobucket.com/albums/i376/Nelson_Yu/int-12.jpg