Chapter 15: Multiple Integrals - Section 15.2 - Double Integrals over General Regions - Exercises 15.2 - Page 882: 31

$2 \pi$

Consider $\iint_{R} f(t, u) dA= \int_{-\pi/3}^{\pi/3} \int_{0}^{\sec t} 3 \cos t du dt= \int_{-\pi/3}^{\pi/3} [ 3u \cos t ]_{0}^{\sec t} dt$ or, $= \int_{-\pi/3}^{\pi/3} 3 \sec t \cos t dt$ or, $= \int_{-\pi/3}^{\pi/3} 3 dt$ Thus, we have $I=[3t]_{-\pi/3}^{\pi/3}=2 \pi$