Corrected: The area element doesn't have the $r^2$ in it, so the integral should be $\int_{0}^{\arctan \frac{3}{2}} \int_0^{\frac{2}{\cos \theta}} r\;dr\;d\theta$. Alpha says this is $3$.
Corrected: The area element doesn't have the $r^2$ in it, so the integral should be $\int_{0}^{\arctan \frac{3}{2}} \int_0^{\frac{2}{\cos \theta}} r\;dr\;d\theta$. Alpha says this is $3$.