Probability via Area: Point Above a Parabola in a Unit Circle

Worked Solution

How to Think About It: For a uniform distribution on a region, probability is just area divided by total area. The unit disk has area $\pi$. You need the area of the region inside the disk and above $y = x^2$. This means finding where $y = x^2$ intersects $x^2 + y^2 = 1$, then integrating to find the area between the parabola and the part of the disk above it.

Quick sanity check: the parabola $y = x^2$ passes through the origin and curves upward. The region above it (inside the disk) is the upper part of the disk minus the area under the parabola. Since the parabola is below $y = 0$ only for... wait, $y = x^2 \geq 0$ always, so the parabola only touches the lower half indirectly. The region $y > x^2$ is a subset of the upper half of the disk (since $x^2 \geq 0$). Rough guess: maybe 30-40% of the disk area.

Quick Estimate: The region $y > x^2$ inside the unit disk is bounded above by the circle and below by the parabola. The circle area is $\pi \approx 3.14$. The area of the upper semicircle is $\pi/2 \approx 1.57$. The parabola cuts a chunk from below this region. The integral of $x^2$ from $-1$ to $1$ is $2/3$, so very roughly the region has area $\pi/2 - 2/3 \approx 1.57 - 0.67 = 0.90$ (very rough), giving probability $\approx 0.90/3.14 \approx 28\%$. The actual answer will be close.

Approach: Find the intersection points of $y = x^2$ and $x^2 + y^2 = 1$, then integrate. The probability is the area of the region $\{(x,y): x^2 + y^2 \leq 1,\, y > x^2\}$ divided by $\pi$.

Formal Solution:

Step 1 -- Intersection points:

Substitute $y = x^2$ into $x^2 + y^2 = 1$: $$x^2 + x^4 = 1 \implies x^4 + x^2 - 1 = 0$$

Let $u = x^2$: $u^2 + u - 1 = 0 \implies u = \dfrac{-1 + \sqrt{5}}{2}$ (taking the positive root).

So $x^2 = \dfrac{\sqrt{5}-1}{2} \approx 0.618$ (the golden ratio minus 1), giving $x = \pm x_0$ where $x_0 = \sqrt{(\sqrt{5}-1)/2} \approx 0.786$.

At the intersection, $y_0 = x_0^2 = (\sqrt{5}-1)/2 \approx 0.618$.

Step 2 -- Set up the area integral:

For $x \in [-x_0, x_0]$, the region $y > x^2$ inside the circle has $y$ ranging from $x^2$ up to $\sqrt{1-x^2}$. Outside $[-x_0, x_0]$, the parabola $y = x^2$ lies outside (or on) the circle, so there is no region to contribute.

$$A = \int_{-x_0}^{x_0} \left(\sqrt{1-x^2} - x^2\right) dx$$

By symmetry in $x$: $$A = 2\int_{0}^{x_0} \left(\sqrt{1-x^2} - x^2\right) dx$$

Step 3 -- Evaluate the integral:

Split into two parts:

Part I: $\displaystyle 2\int_0^{x_0} \sqrt{1-x^2}\,dx$

Using the formula $\int_0^a \sqrt{1-x^2}\,dx = \dfrac{1}{2}\left(a\sqrt{1-a^2} + \arcsin(a)\right)$:

With $a = x_0$ and $x_0^2 = (\sqrt{5}-1)/2$, so $1 - x_0^2 = (3-\sqrt{5})/2$ and $\sqrt{1-x_0^2} = \sqrt{(3-\sqrt{5})/2} = y_0'$ (the $y$-coordinate at intersection is $y_0 = x_0^2$; note $\sqrt{1-x_0^2} \neq y_0$).

Actually $\sqrt{1-x_0^2} = \sqrt{1-(\sqrt{5}-1)/2} = \sqrt{(3-\sqrt{5})/2}$.

Also $\arcsin(x_0) = \arcsin\!\left(\sqrt{(\sqrt{5}-1)/2}\right)$.

So Part I $= 2 \cdot \dfrac{1}{2}\left(x_0\sqrt{1-x_0^2} + \arcsin(x_0)\right) = x_0\sqrt{1-x_0^2} + \arcsin(x_0)$.

Part II: $\displaystyle 2\int_0^{x_0} x^2\,dx = 2 \cdot \frac{x_0^3}{3} = \frac{2x_0^3}{3}$

So: $$A = x_0\sqrt{1-x_0^2} + \arcsin(x_0) - \frac{2x_0^3}{3}$$

Numerically: $x_0 \approx 0.7862$, $\sqrt{1-x_0^2} \approx \sqrt{0.3820} \approx 0.6181$, $\arcsin(0.7862) \approx 0.9046$ rad.

$$A \approx 0.7862 \times 0.6181 + 0.9046 - \frac{2 \times 0.4864}{3} \approx 0.4861 + 0.9046 - 0.3243 \approx 1.0664$$

$$P(y > x^2) = \frac{A}{\pi} \approx \frac{1.0664}{3.1416} \approx \boxed{0.3394}$$

Answer: $P(y > x^2) = \dfrac{1}{\pi}\left(x_0\sqrt{1-x_0^2} + \arcsin(x_0) - \dfrac{2x_0^3}{3}\right) \approx 33.9\%$, where $x_0 = \sqrt{(\sqrt{5}-1)/2}$.