A decorative mirror is cut from a circular sheet of glass…

Step 1. The crossing points. $x^2 + 3x - 4 = 0$ gives $(x + 4)(x - 1) = 0$. The value $x = -4$ is outside the circle, so $x = 1$. Then $y = \pm\sqrt{3}$.

Step 2. The upper half. For $0 \le x \le 1$ the boundary is $y = \sqrt{3x}$. For $1 \le x \le 2$ the boundary is $y = \sqrt{4 - x^2}$.

Step 3. The integrals. The first piece is $\displaystyle\int_0^1 \sqrt{3x}\,dx = \dfrac{2\sqrt{3}}{3}$. The second is $\displaystyle\int_1^2 \sqrt{4 - x^2}\,dx = \left[\dfrac{x}{2}\sqrt{4 - x^2} + 2\sin^{-1}\dfrac{x}{2}\right]_1^2 = \pi - \dfrac{\sqrt{3}}{2} - \dfrac{\pi}{3} = \dfrac{2\pi}{3} - \dfrac{\sqrt{3}}{2}$.

Step 4. Double it. The area is $2\left(\dfrac{2\sqrt{3}}{3} + \dfrac{2\pi}{3} - \dfrac{\sqrt{3}}{2}\right) = \dfrac{\sqrt{3}}{3} + \dfrac{4\pi}{3} \approx 4.77$.