SOLUTION The region \(R\) can be described as \[
R = \{(x, y) | y \ge 0, 1 \le x^2 + y^2 \le 4\}
\] It is the half-ring shown in Figure 1(b), and in polar
coordinates it is given by \(1 \le r \le 2, 0
\le \theta \le \pi\). Therefore, by Formula 2, \[
\iint_R (3x + 4y^2) dA = \int_0^\pi \int_1^2 (3r \cos \theta + 4r^2
\sin^2 \theta) r dr d\theta
\] \[
= \int_0^\pi \int_1^2 (3r^2 \cos \theta + 4r^3 \sin^2 \theta) dr d\theta
\]
\[
= \int_0^{\pi} \left[r^3 \cos \theta + r^4 \sin^2 \theta
\right]_{r=1}^{r=2} d\theta = \int_0^\pi (7 \cos \theta + 15 \sin^2
\theta) d\theta
\] \[
= \int_0^\pi \left[ 7 \cos \theta + \frac{15}{2}(1 - \cos 2\theta)
\right] d\theta
\] \[
= \left[ 7 \sin \theta + \frac{15\theta}{2} - \frac{15}{4}\sin 2\theta
\right]_0^\pi = \frac{15\pi}{2}
\]
