Chapter 18 - Fundamental Theorems of Vector Analysis - 18.1 Green's Theorem - Exercises - Page 984: 14

$9 \pi$

The area of a given region can be computed as: $A=\dfrac{1}{2}\int_C xdy-y dx$ Here, we have $x=3 \cos \theta; y=3 \sin \theta$ Therefore, $A=\dfrac{1}{2}\int_C (3 \cos \theta)(3 \cos \theta) d \theta -(3 \sin \theta) (-3 \sin \theta) d \theta\\=\dfrac{1}{2}\int_0^{2 \pi} 9 d \theta\\=\dfrac{1}{2} (9) (2 \pi) \\=9 \pi$