Answer
$$\frac{9}{2}\ln \left( {\sqrt 2 + 1} \right)$$
Work Step by Step
$$\eqalign{
& \int_0^{\pi /4} {\int_0^3 {r\sec \theta } } drd\theta \cr
& = \int_0^{\pi /4} {\left[ {\int_0^3 {r\sec \theta } dr} \right]} d\theta \cr
& {\text{solve the inner integral}}{\text{, treat }}\theta {\text{ as a constant}} \cr
& = \int_0^3 {r\sec \theta } dr \cr
& = \sec \theta \int_0^3 r dr \cr
& = \sec \theta \left[ {\frac{{{r^2}}}{2}} \right]_0^3 \cr
& {\text{evaluating the limits for }}r \cr
& = \sec \theta \left[ {\frac{{{3^2}}}{2} - \frac{{{0^2}}}{2}} \right] \cr
& = \sec \theta \left( {\frac{9}{2}} \right) \cr
& = \frac{9}{2}\sec \theta \cr
& \cr
& = \int_0^{\pi /4} {\left[ {\int_0^3 {r\sec \theta } dr} \right]} d\theta = \int_0^{\pi /4} {\frac{9}{2}\sec \theta } d\theta \cr
& = \frac{9}{2}\int_0^{\pi /4} {\sec \theta } d\theta \cr
& {\text{integrating}} \cr
& = \frac{9}{2}\left( {\ln \left| {\sec \theta + \tan \theta } \right|} \right)_0^{\pi /4} \cr
& = \frac{9}{2}\left( {\ln \left| {\sec \frac{\pi }{4} + \tan \frac{\pi }{4}} \right| - \ln \left| {\sec 0 + \tan 0} \right|} \right) \cr
& {\text{simplifying}} \cr
& = \frac{9}{2}\left( {\ln \left( {\sqrt 2 + 1} \right) - \ln \left| {1 + 0} \right|} \right) \cr
& = \frac{9}{2}\ln \left( {\sqrt 2 + 1} \right) \cr} $$
