Answer:
[tex]\frac{10\pi}{3}[/tex]
Step-by-step explanation:
According to the information of the problem we have to compute the following integral.
[tex]{\displaystyle \int\limits \int} \frac{1}{3} + \sqrt{x^2 + y^2} \, dA[/tex]
Where the region of integration is
[tex]R = \Big\{ (r,\theta) : 0 \leq r \leq 2 , \,\,\,\, \frac{\pi}{2} \leq \theta \leq \frac{3\pi}{2} \Big\}[/tex]
If you plot, that is just a circle between [tex]\pi/2[/tex] and [tex]3\pi/2[/tex], which is just half of the circle on the negative part of the plane.
When you switch coordinates
[tex]{\displaystyle \int\limits \int} \frac{1}{3} + \sqrt{x^2 + y^2} \, dA = {\displaystyle \int\limits_{0}^{2} \int\limits_{\pi/2}^{3\pi/2}} \bigg(\frac{1}{3} + r \bigg)r \, d\theta\, dr = \frac{10\pi}{3}[/tex]
