$\frac{\mathrm{\θ}}{1\+r^2}$

$\frac{\mathrm{\θ}}{1\+r^2}$

$0$

$0$

$\mathrm{\theta }$

$\mathrm{\θ}$

$0$

$0$

$\frac{\mathrm{\pi }}{3}$

$\frac{1}{3}\mathrm{\π}$

$\mathrm{Student}\left[\mathrm{MultivariateCalculus}\right]\left[\mathrm{FunctionAverage}\right]\left(\,r\=..\,\mathrm{\θ}\=..\,\mathrm{coordinates}\=\mathrm{polar}\left[r\,\mathrm{\θ}\right]\,\mathrm{output}\=\mathrm{integral}\right)$

$\frac{{\∫}_0^{\frac{1}{3}\mathrm{\π}}{\∫}_0^{\mathrm{\θ}}\frac{\mathrm{\θ}r}{1\+r^2}\ⅆr\ⅆ\mathrm{\θ}}{{\∫}_0^{\frac{1}{3}\mathrm{\π}}{\∫}_0^{\mathrm{\θ}}r\ⅆr\ⅆ\mathrm{\θ}}$

$\mathrm{Student}\left[\mathrm{MultivariateCalculus}\right]\left[\mathrm{FunctionAverage}\right]\left(\,r\=..\,\mathrm{\θ}\=..\,\mathrm{coordinates}\=\mathrm{polar}\left[r\,\mathrm{\θ}\right]\right)$

$\frac{162\left(-\frac{1}{2}\ln \left(3\right)\+\frac{1}{4}\ln \left(9\+{\mathrm{\π}}^2\right)-\frac{1}{18}{\mathrm{\π}}^2\ln \left(3\right)\+\frac{1}{36}{\mathrm{\π}}^2\ln \left(9\+{\mathrm{\π}}^2\right)-\frac{1}{36}{\mathrm{\π}}^2\right)}{{\mathrm{\π}}^3}$