Useing a proper change of variables, compute the area enclosed by the hyperbola?

Question

Useing a proper change of variables, compute the area enclosed by the hyperbola y=1/x and y=2/x and the lines y=x and y=2x

Answer 1

Why change variables at all? The area is

integral from 1/sqrt(2) to 1 of (2x - 1/x) dx
plus
integral from 1 to sqrt(2) of (2/x - x) dx

which works out in closed form to

(ln 2) / 2

If you make the substitutions s = y/x and t = xy,
then you get:

s varies from 1 to 2
t varies from 1 to 2

st = y^2 so y = (st)^(1/2)
t/s = x^2 so x = (s^(-1/2)) (t^(1/2))

The determinant of the dx/ds dx/dt dy/ds dy/dt matrix is 2/s

So the enclosed area is the double integral s = 1 to 2, t = 1 to 2, of (2/s) ds dt.

Which is also (ln 2) / 2

Dan

Answer 2

Dan correctely calculated the area between the curves in the first quadrant. But hyperbolas come in pairs. There is an enclosed area in the third quadrant of equal area.