The area of the region : `R = {(x, y) : 5x^2 le y le 2x^2 + 9}` is :

Options

A. `11sqrt3` square units
B. `12sqrt3` square units
C. `9sqrt3` square units
D. `6sqrt3` square units

Answer

To find the area of the region \( R = \{(x, y) : 5x^2 \leq y \leq 2x^2 + 9\} \), we will follow these steps:

### Step 1: Identify the curves
We have two curves:
1. \( y = 5x^2 \) (Curve 1)
2. \( y = 2x^2 + 9 \) (Curve 2)

### Step 2: Find the points of intersection
To find the area between these two curves, we first need to find their points of intersection by setting the equations equal to each other:

\[ 5x^2 = 2x^2 + 9 \]

Rearranging gives:

\[ 5x^2 - 2x^2 - 9 = 0 \implies 3x^2 - 9 = 0 \implies 3x^2 = 9 \implies x^2 = 3 \implies x = \pm \sqrt{3} \]

Thus, the points of intersection are \( x = -\sqrt{3} \) and \( x = \sqrt{3} \).

### Step 3: Set up the integral for the area
The area \( A \) between the curves from \( x = -\sqrt{3} \) to \( x = \sqrt{3} \) can be calculated using the integral:

\[ A = \int_{-\sqrt{3}}^{\sqrt{3}} \left( (2x^2 + 9) - (5x^2) \right) \, dx \]

This simplifies to:

\[ A = \int_{-\sqrt{3}}^{\sqrt{3}} (2x^2 + 9 - 5x^2) \, dx = \int_{-\sqrt{3}}^{\sqrt{3}} (9 - 3x^2) \, dx \]

### Step 4: Calculate the integral
Now we can compute the integral:

\[ A = \int_{-\sqrt{3}}^{\sqrt{3}} (9 - 3x^2) \, dx \]

This can be split into two separate integrals:

\[ A = \int_{-\sqrt{3}}^{\sqrt{3}} 9 \, dx - \int_{-\sqrt{3}}^{\sqrt{3}} 3x^2 \, dx \]

Calculating the first integral:

\[ \int_{-\sqrt{3}}^{\sqrt{3}} 9 \, dx = 9 \left[ x \right]_{-\sqrt{3}}^{\sqrt{3}} = 9 \left( \sqrt{3} - (-\sqrt{3}) \right) = 9 \cdot 2\sqrt{3} = 18\sqrt{3} \]

Calculating the second integral:

\[ \int_{-\sqrt{3}}^{\sqrt{3}} 3x^2 \, dx = 3 \left[ \frac{x^3}{3} \right]_{-\sqrt{3}}^{\sqrt{3}} = \left[ x^3 \right]_{-\sqrt{3}}^{\sqrt{3}} = (\sqrt{3})^3 - (-\sqrt{3})^3 = 3\sqrt{3} - (-3\sqrt{3}) = 6\sqrt{3} \]

### Step 5: Combine the results
Now, substituting back into the area formula:

\[ A = 18\sqrt{3} - 6\sqrt{3} = 12\sqrt{3} \]

Thus, the area of the region \( R \) is:

\[ \boxed{12\sqrt{3}} \]