Answer
$${\text{Surface area}} = \sqrt 2 - 1$$
Work Step by Step
$$\eqalign{
& f\left( {x,y} \right) = \ln \left| {\sec x} \right| \cr
& {\text{Find }}{f_x}\left( {x,y} \right){\text{ and }}{f_y}\left( {x,y} \right) \cr
& {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {\ln \left| {\sec x} \right|} \right] = \frac{{\sec x\tan x}}{{\sec x}} = \tan x \cr
& {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {\ln \left| {\sec x} \right|} \right] = 0 \cr
& {\text{The limits of the region }}R{\text{ are:}} \cr
& R = \left\{ {\left( {x,y} \right):0 \leqslant x \leqslant \frac{\pi }{4},{\text{ }}0 \leqslant y \leqslant \tan x} \right\} \cr
& \cr
& {\text{The area of the surface }}S{\text{ is given by}} \cr
& {\text{Surface area}} = \iint\limits_R {\sqrt {1 + {{\left[ {{f_x}\left( {x,y} \right)} \right]}^2} + {{\left[ {{f_y}\left( {x,y} \right)} \right]}^2}} dA} \cr
& = \int_0^{\pi /4} {\int_0^{\tan x} {\sqrt {1 + {{\left[ {\tan x} \right]}^2} + {{\left[ 0 \right]}^2}} dydx} } \cr
& = \int_0^{\pi /4} {\int_0^{\tan x} {\sqrt {1 + {{\tan }^2}x} dydx} } \cr
& = \int_0^{\pi /4} {\int_0^{\tan x} {\sqrt {{{\sec }^2}x} dydx} } \cr
& = \int_0^{\pi /4} {\int_0^{\tan x} {\sec xdydx} } \cr
& {\text{Integrate with respect to }}y \cr
& = \int_0^{\pi /4} {\left[ {y\sec x} \right]_0^{\tan x}} dx \cr
& = \int_0^{\pi /4} {\sec x\tan x} dx \cr
& {\text{Integrate}} \cr
& = \left[ {\sec x} \right]_0^{\pi /4} \cr
& {\text{Surface area}} = \sec \left( {\frac{\pi }{4}} \right) - \sec \left( 0 \right) \cr
& {\text{Surface area}} = \sqrt 2 - 1 \cr} $$
