For this exercise we have to use the knowledge of curve and limits to calculate the integral section, in this way:
[tex]A= 0.3622[/tex]
So performing the limits of integration we have that the area will be:
[tex]A= \int\limits^a_b {\frac{1}{2} r^2} \, d\theta\\\\=\int\limits^{\pi}_{\pi/2} {\frac{1}{2} (e^{-\theta/6})} \, d\theta\\\\= \frac{-3}{2} e^{-\theta/3} = \frac{-3}{2} e^{-\pi/3}(e^{-\pi/6}-1)\\\\= 0.3622[/tex]
See more about integration at brainly.com/question/1470138
