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

Correct Answer: 119

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 \( m + n = 119 \).