We are given two inequalities that define a region in the $xy$-plane:

1. $x - 3y^2 \geq 0$
2. $2 - x - 5|y| \geq 0$

Step 1: Analyze the First Inequality

The inequality $x - 3y^2 \geq 0$ can be rewritten as: $x \geq 3y^2$ This represents a region to the right of the parabola $x = 3y^2$.

Step 2: Analyze the Second Inequality

The inequality $2 - x - 5|y| \geq 0$ can be rewritten as: $x \leq 2 - 5|y|$ This describes a region to the left of the line $x = 2 - 5|y|$, where $|y|$ represents the absolute value of $y$, so we need to treat the cases for $y \geq 0$ and $y < 0$.

Step 3: Find the Boundary Points

To find the points where these curves intersect, we set the expressions for $x$ equal to each other:

$3y^2 = 2 - 5|y|$

Case 1: $y \geq 0$

For $y \geq 0$, the absolute value simplifies to $|y| = y$, so the equation becomes: $3y^2 = 2 - 5y$ Rearranging this: $3y^2 + 5y - 2 = 0$ Solving this quadratic equation using the quadratic formula: $y = \frac{-5 \pm \sqrt{5^2 - 4(3)(-2)}}{2(3)} = \frac{-5 \pm \sqrt{25 + 24}}{6} = \frac{-5 \pm \sqrt{49}}{6} = \frac{-5 \pm 7}{6}$ This gives two solutions: $y = \frac{-5 + 7}{6} = \frac{2}{6} = \frac{1}{3}, \quad y = \frac{-5 - 7}{6} = \frac{-12}{6} = -2$

Case 2: $y < 0$

For $y < 0$, $|y| = -y$, so the equation becomes: $3y^2 = 2 + 5y$ Rearranging this: $3y^2 - 5y - 2 = 0$ Solving this quadratic equation: $y = \frac{5 \pm \sqrt{(-5)^2 - 4(3)(-2)}}{2(3)} = \frac{5 \pm \sqrt{25 + 24}}{6} = \frac{5 \pm \sqrt{49}}{6} = \frac{5 \pm 7}{6}$ This gives two solutions: $y = \frac{5 + 7}{6} = \frac{12}{6} = 2, \quad y = \frac{5 - 7}{6} = \frac{-2}{6} = \frac{-1}{3}$ However, since this case is for $y < 0$, only $y = -\frac{1}{3}$ is valid.

Step 4: Set the Limits and Calculate the Area

The region is symmetric about the $x$-axis due to the absolute value in the second inequality, so we can calculate the area for $y \geq 0$ and double it.

The limits of integration for $y \geq 0$ are from $y = 0$ to $y = \frac{1}{3}$. The area is determined by the difference between the curves: $x = 2 - 5y \quad \text{and} \quad x = 3y^2$ Thus, the area for $y \geq 0$ is: $\text{Area} = \int_0^{1/3} \left( (2 - 5y) - 3y^2 \right) \, dy$

Now, let's compute this integral.The total area of the region defined by the inequalities $x - 3y^2 \geq 0$ and $2 - x - 5|y| \geq 0$ is approximately $0.704$ square units.

Would you like more details, or do you have any questions?

Here are 5 related questions to consider:

1. How does symmetry simplify the calculation of areas in regions like this?
2. What is the geometric interpretation of the curves involved in the inequalities?
3. How can the quadratic formula help in solving intersection points between curves?
4. How would the result change if the coefficients in the inequalities were different?
5. What other methods can be used to find areas under curves besides integration?

Tip: Symmetry is a powerful tool in integration, often allowing you to compute only half the area and then multiply by two to get the total area.