Question

Let the area of the region $\\{(x, y) : x - 2y + 4 \geq 0, x + 2y^2 \geq 0, x + 4y^2 \leq 8, y \geq 0\\}$ be $\frac{m}{n}$, where $m$ and $n$ are coprime numbers. Then $m + n$ is equal to ______.

Answer 1

Given the region defined by:

$x - 2y + 4 \geq 0, \quad x + 2y^2 \geq 0, \quad x + 4y^2 \leq 8, \quad y \geq 0$

We need to find the area $A$ of this region and express it in the form $\frac{m}{n}$ where $m$ and $n$ are coprime numbers.

Step 1. Setting Up the Integral: The area is given by:
$A = \int_0^{\sqrt{8}} \left[ (8 - 4y^2) - (-2y^2) \right] \, dy + \int_{\sqrt{8}}^2 \left[ (8 - 4y^2) - (2y - 4) \right] \, dy$

Step 2. Evaluating the First Integral:

$\int_0^{\sqrt{2}} \left[ (8 - 4y^2) - (-2y^2) \right] \, dy = \int_0^{\sqrt{2}} (8 - 2y^2) \, dy$

Integrating term by term:

$\int_0^{\sqrt{2}} (8 - 2y^2) \, dy = \left[ 8y - \frac{2y^3}{3} \right]_0^{\sqrt{2}}$

Substituting the limits:

$= \left( 8 \times \sqrt{2} - \frac{2 \cdot (\sqrt{2})^3}{3} \right) - (0 - 0) = \frac{16\sqrt{2}}{3}$

Step 3. Evaluating the Second Integral:

$\int_{\sqrt{2}}^2 \left[ (8 - 4y^2) - (2y - 4) \right] \, dy$

Simplifying the integrand:

$= \int_{\sqrt{2}}^2 (8 - 4y^2 - 2y + 4) \, dy = \int_{\sqrt{2}}^2 (12 - 4y^2 - 2y) \, dy$

Integrating term by term:

$= \left[ 12y - \frac{4y^3}{3} - y^2 \right]_{\sqrt{2}}^2$
Substituting the limits:

$= \left( 24 - \frac{32}{3} - 4 \right) - \left( 12\sqrt{2} - \frac{16\sqrt{2}}{3} - 2 \right) = \frac{107}{12}$

Step 4. Final Area Calculation: The total area is:

$A = \frac{16\sqrt{2}}{3} + \frac{107}{12}$

Expressing $A$ in the form $\frac{m}{n}$ where $m$ and $n$ are coprime, we have $m + n = 119$.