The intersection angle is given by $\cos\theta =\frac37$ and the area integral is given by

$$\begin{align} &2 \int_0^{\arccos\frac37} \frac12(3-3\cos \theta)^2 d\theta + 2\int_{\arccos\frac37}^{\pi/2}\frac12 (4\cos\theta)^2d\theta \\ =& \>4\pi -\frac{39\sqrt{10}}7+\frac{11}2\arccos\frac37=1.151 \end{align}$$

The intersection angle is given by $\cos\theta =\frac37$ and the integral for the enclosed area is given by

$$\begin{align} &2 \int_0^{\arccos\frac37} \frac12(3-3\cos \theta)^2 d\theta + 2\int_{\arccos\frac37}^{\pi/2}\frac12 (4\cos\theta)^2d\theta \\ =& \>4\pi -\frac{39\sqrt{10}}7+\frac{11}2\arccos\frac37=1.151 \end{align}$$

Note that the limits in the two integrals above ensure that the area is enclosed inside both curves.