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The graph of $y = x^2$ is a parabola opening upwards with its vertex at the origin (0,0).
The region is bounded by the parabola $y=x^2$, the horizontal line $y=9$, and the vertical line $x=0$. We need to find the point of intersection between $y=x^2$ and $y=9$.
Set $x^2 = 9$. This gives $x = \pm 3$. Since we are bounded by $x=0$, we consider the positive value $x=3$. So the intersection point is (3,9).
We want to find the area between $y=9$ and $y=x^2$ from $x=0$ to $x=3$. The area is given by the integral: $$A = \int_{0}^{3} (9 - x^2) \, dx$$
$$A = \int_{0}^{3} (9 - x^2) \, dx = \left[9x - \frac{x^3}{3}\right]_{0}^{3}$$ $$A = \left(9(3) - \frac{(3)^3}{3}\right) - \left(9(0) - \frac{(0)^3}{3}\right)$$ $$A = 27 - \frac{27}{3} = 27 - 9 = 18$$
The area of the region bounded by $y=9$, $x=0$, and $y=x^2$ is 18 square units.
Final Answer: 18