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To find the area bounded by the curves $x = 1$, $y = x - 2$, and $y^2 = x$:

1. Points of Intersection:

   • Intersection of $y = x - 2$ (or $x = y + 2$) and $y^2 = x$:
     $y^2 = y + 2 \implies y^2 - y - 2 = 0 \implies (y - 2)(y + 1) = 0$
     So, $y = -1$ and $y = 2$.
     For $y = -1$, $x = 1$, which gives the point $(1, -1)$.
     For $y = 2$, $x = 4$, which gives the point $(4, 2)$.
   • Intersection of $x = 1$ and $y^2 = x$:
     $y^2 = 1 \implies y = \pm 1$, giving points $(1, 1)$ and $(1, -1)$.
   • Intersection of $x = 1$ and $y = x - 2$:
     $y = 1 - 2 = -1$, giving the point $(1, -1)$.

2. Setting up the Area Integral:
   The bounded region lies between $y = -1$ and $y = 2$:

   • From $y = -1$ to $y = 1$, the region is bounded on the right by $x = y + 2$ and on the left by $x = 1$.
   • From $y = 1$ to $y = 2$, the region is bounded on the right by $x = y + 2$ and on the left by the parabola $x = y^2$.

   $\text{Area } A = \int_{-1}^{1} ((y + 2) - 1) \, dy + \int_{1}^{2} ((y + 2) - y^2) \, dy$
   $A = \int_{-1}^{1} (y + 1) \, dy + \int_{1}^{2} (y + 2 - y^2) \, dy$

3. Evaluating the Integrals:
   $A_1 = \left[ \frac{y^2}{2} + y \right]_{-1}^{1} = \left( \frac{1}{2} + 1 \right) - \left( \frac{1}{2} - 1 \right) = 2$
   $A_2 = \left[ \frac{y^2}{2} + 2y - \frac{y^3}{3} \right]_{1}^{2} = \left( \frac{4}{2} + 4 - \frac{8}{3} \right) - \left( \frac{1}{2} + 2 - \frac{1}{3} \right) = \left( 6 - \frac{8}{3} \right) - \left( \frac{13}{6} \right) = \frac{10}{3} - \frac{13}{6} = \frac{7}{6}$

   $\text{Total Area } A = 2 + \frac{7}{6} = \frac{19}{6} \text{ square units}$