Answer
$\dfrac{7 \pi \sqrt {10}}{3}$
Work Step by Step
Here, we have $z=1; z=4/3$ and $z=\dfrac{\sqrt {x^2+y^2}}{3}$
Then $r_r=( \cos \theta) i+(\sin \theta) j+(\dfrac{1}{3})k$
and
$r_{\theta}=(-r \sin \theta) i+(r \cos \theta) j$
Also, $|r_r \times r_{\theta}|=\dfrac{r \sqrt {10}}{3}$
Thus, the area is given by $A=\int_0^{2 \pi} \int_3^{4} \dfrac{r \sqrt {10}}{3}dr d \theta$
or, $A=\int_0^{2 \pi} \dfrac{7\sqrt {10}}{6} d \theta=\dfrac{7 \pi \sqrt {10}}{3}$
