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Step-by-step Solution
1. Rewrite and identify the curves
1) The curve given by
x^2 + 2y \;-\; 1 = 0
can be rewritten as
y \;=\;\frac{1 - x^2}{2}.
This is a downward-opening parabola with vertex at (0,\tfrac12) .
2) The curve
y^2 + 4x \;-\; 4 = 0
can be rearranged as
4x = 4 - y^2 \;\Longrightarrow\; x = 1 - \frac{y^2}{4}.
This is a left-opening parabola with vertex at (1,0) .
3) The curve
y^2 \;-\; 4x \;-\; 4 = 0
can be rearranged as
4x = y^2 \;-\; 4 \;\Longrightarrow\; x = \frac{y^2}{4} \;-\; 1.
This is a right-opening parabola with vertex at (-1,0) .
We seek the area of the region bounded by these three curves in the upper half-plane ( y \ge 0 ).
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2. Find intersection points
• All three curves meet the x-axis (i.e., y=0 ) at:
– From x^2 + 2y - 1 = 0 , putting y=0 gives x^2 = 1 , so x=\pm 1.
– From y^2 +4x -4=0 , putting y=0 gives x=1.
– From y^2 -4x -4=0 , putting y=0 gives x=-1.
Hence, they intersect the x-axis at (-1,0) and (1,0).
• The curves
y^2 +4x -4=0 and y^2 -4x -4=0
intersect where
1 - \frac{y^2}{4} \;=\;\frac{y^2}{4} -1 \;\Longrightarrow\; 2 = \frac{y^2}{2} \;\Longrightarrow\; y^2 = 4 \;\Longrightarrow\; y=2
\text{ (upper half-plane)}.
Substituting y=2 in either one gives x=0.
So they intersect at (0,2) .
Hence, the three intersection points relevant in the upper half-plane are:
• (-1,0), \quad (1,0), \quad (0,2).
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3. Describe the bounded region
By examining the shapes and intersections:
• At y=0 , the three curves meet pairwise at x=-1 and x=1.
• The left-opening and right-opening parabolas – namely x = 1-\tfrac{y^2}{4} and x = \tfrac{y^2}{4}-1 – intersect again at (0,2).
• The curve y = \tfrac{1 - x^2}{2} only extends up to y = \tfrac12 for real x (where 1 - x^2 \ge 0 ).
Altogether, the region in the upper half-plane can be viewed in two vertical “slices” in terms of y :
(a) From y=0 up to y=\tfrac{1}{2} , the region is constrained simultaneously by:
• y \;\ge\; \frac{1 - x^2}{2} (above the downward parabola),
• x \;\ge\; \frac{y^2}{4} -1 , and
• x \;\le\; 1 - \frac{y^2}{4}.
A careful check shows the actual x -domain for each fixed 0 \le y \le \tfrac12 is split into two symmetric parts about x=0, because y \ge (1 - x^2)/2 implies |x|\ge \sqrt{\,1-2y}\,.
(b) From y=\tfrac{1}{2} up to y=2 , the curve x^2 +2y-1=0 no longer intersects (no real x if 1-2y\tfrac12. So the region is bounded only by
x=\frac{y^2}{4}-1 \quad\text{and}\quad x=1-\frac{y^2}{4},
together with y\ge \tfrac12 and going up to the intersection at y=2.
For each \tfrac12 \le y \le 2, the width in x is:
\[
\Bigl(1-\tfrac{y^2}{4}\Bigr)
-\Bigl(\tfrac{y^2}{4}-1\Bigr)
=2 - \frac{y^2}{2}.
\]
Thus,
\[
A_{2}
= \int_{y=\tfrac12}^{y=2}
\Bigl(2 - \tfrac{y^2}{2}\Bigr)\,dy
= \int_{\tfrac12}^{2} 2\,dy
- \int_{\tfrac12}^{2} \frac{y^2}{2}\,dy.
\]
1) \displaystyle \int_{\tfrac12}^{2} 2\,dy
= 2\Bigl[y\Bigr]_{\tfrac12}^{2}
= 2\bigl(2 - \tfrac12\bigr)
= 2\times \tfrac{3}{2}
= 3.
2) \displaystyle \int_{\tfrac12}^{2} \frac{y^2}{2}\,dy
= \frac12\int_{\tfrac12}^{2}y^2\,dy
= \frac12\Bigl[\frac{y^3}{3}\Bigr]_{\tfrac12}^{2}
= \frac{1}{6}\Bigl(2^3 - \bigl(\tfrac12\bigr)^3\Bigr)
= \frac{1}{6}\bigl(8 - \tfrac{1}{8}\bigr)
= \frac{1}{6}\times \tfrac{63}{8}
= \frac{63}{48}
= \frac{21}{16}.
\]
Hence,
\[
A_{2}
= 3 - \frac{21}{16}
= \frac{48}{16} - \frac{21}{16}
= \frac{27}{16}.
\]
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5. Total area
Finally, the total area in the upper half-plane bounded by these three curves is
\[
A_{\text{total}}
= A_{1} + A_{2}
= \frac{5}{16} + \frac{27}{16}
= \frac{32}{16}
= 2.
\]
Hence, the required area is \boxed{2} square units.
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6. Diagram (reference)
