The area (in sq. units) of the region bounded by the curves $$x^2 + 2y - 1 = 0$$, $$y^2 + 4x - 4 = 0$$ and $$y^2 - 4x - 4 = 0$$ in the upper half plane is ___.

We need the area in the upper half plane bounded by the three curves: $$x^2 + 2y - 1 = 0$$ (i.e., $$y = \frac{1 - x^2}{2}$$, a downward-opening parabola with vertex $$(0, \frac{1}{2})$$), $$y^2 + 4x - 4 = 0$$ (i.e., $$x = 1 - \frac{y^2}{4}$$, a left-opening parabola with vertex $$(1, 0)$$), and $$y^2 - 4x - 4 = 0$$ (i.e., $$x = \frac{y^2}{4} - 1$$, a right-opening parabola with vertex $$(-1, 0)$$).

Finding intersection points in the upper half plane:

The two sideways parabolas intersect where $$4 - 4x = 4x + 4$$, giving $$8x = 0$$, so $$x = 0$$ and $$y = 2$$. The intersection point is $$(0, 2)$$.

The downward parabola meets the left-opening parabola: substituting $$y = \frac{1-x^2}{2}$$ into $$y^2 = 4 - 4x$$ gives $$\frac{(1-x^2)^2}{4} = 4 - 4x$$, so $$(1-x^2)^2 = 16 - 16x$$, which simplifies to $$x^4 - 2x^2 + 16x - 15 = 0$$. Testing $$x = 1$$: $$1 - 2 + 16 - 15 = 0$$. Factoring: $$(x-1)(x^3 + x^2 - x + 15) = 0$$. Testing $$x = -3$$ in the cubic: $$-27 + 9 + 3 + 15 = 0$$. So $$(x-1)(x+3)(x^2 - 2x + 5) = 0$$. The quadratic has negative discriminant, so real intersections are at $$x = 1$$ (giving $$y = 0$$) and $$x = -3$$ (outside the upper half plane). The relevant point is $$(1, 0)$$.

The downward parabola meets the right-opening parabola: substituting gives $$x^4 - 2x^2 - 16x - 15 = 0$$. Testing $$x = -1$$: $$1 - 2 + 16 - 15 = 0$$. And $$x = 3$$: $$81 - 18 - 48 - 15 = 0$$. So $$(x+1)(x-3)(x^2 + 2x + 5) = 0$$. The relevant intersection is $$(-1, 0)$$.

The bounded region in the upper half plane runs from $$(-1, 0)$$ to $$(0, 2)$$ along the right-opening parabola, from $$(0, 2)$$ to $$(1, 0)$$ along the left-opening parabola, and from $$(1, 0)$$ back to $$(-1, 0)$$ along the downward parabola.

We split the area integral at $$x = 0$$. From $$x = -1$$ to $$x = 0$$, the upper curve is $$y = 2\sqrt{x+1}$$ and the lower curve is $$y = \frac{1-x^2}{2}$$. From $$x = 0$$ to $$x = 1$$, the upper curve is $$y = 2\sqrt{1-x}$$ and the lower curve is $$y = \frac{1-x^2}{2}$$.

First integral: $$I_1 = \int_{-1}^{0}\left(2\sqrt{x+1} - \frac{1-x^2}{2}\right)dx$$.

$$\int_{-1}^{0} 2\sqrt{x+1}\,dx = 2 \cdot \frac{2}{3}(x+1)^{3/2}\Big|_{-1}^{0} = \frac{4}{3}(1 - 0) = \frac{4}{3}$$.

$$\int_{-1}^{0} \frac{1-x^2}{2}\,dx = \frac{1}{2}\left[x - \frac{x^3}{3}\right]_{-1}^{0} = \frac{1}{2}\left[0 - \left(-1 + \frac{1}{3}\right)\right] = \frac{1}{2} \cdot \frac{2}{3} = \frac{1}{3}$$.

So $$I_1 = \frac{4}{3} - \frac{1}{3} = 1$$.

Second integral: $$I_2 = \int_{0}^{1}\left(2\sqrt{1-x} - \frac{1-x^2}{2}\right)dx$$.

$$\int_{0}^{1} 2\sqrt{1-x}\,dx = 2 \cdot \left[-\frac{2}{3}(1-x)^{3/2}\right]_{0}^{1} = \frac{4}{3}(1 - 0) = \frac{4}{3}$$.

$$\int_{0}^{1} \frac{1-x^2}{2}\,dx = \frac{1}{2}\left[x - \frac{x^3}{3}\right]_{0}^{1} = \frac{1}{2} \cdot \frac{2}{3} = \frac{1}{3}$$.

So $$I_2 = \frac{4}{3} - \frac{1}{3} = 1$$.

The total area is $$I_1 + I_2 = 1 + 1 = 2$$ square units.