Answer
$\dfrac{\pi}{2}(2 \ln2-1)$
Work Step by Step
Our aim is to integrate the triple integral as follows:
$\int^{\ln2}_0 \int^{\sqrt{(\ln2)^2-y^2}}e^{\sqrt{x^2+y^2}} \space dx \space dy =\int^{\pi/2}_0 \int^{\ln2}_0 re^r dr d\theta $
or, $=\int^{\pi/2}_0(2 \ln2- [e^r]_0^{\ln (2)})d\theta $
or, $=\dfrac{\pi}{2}(2 \ln2-1)$
