The area bounded between the parabolas `x^(2)=(y)/(4) and x^(2)=9y` and the straight line y=2 is

Options

A. `20sqrt(2)`
B. `(10sqrt(2))/(3)`
C. `(20sqrt(2))/(3)`
D. `10sqrt(2)`

Explanation

To find the area bounded between the parabolas \( x^2 = \frac{y}{4} \) and \( x^2 = 9y \) and the straight line \( y = 2 \), we can follow these steps:

### Step 1: Understand the curves
The equations of the curves are:
1. \( x^2 = \frac{y}{4} \) (which can be rewritten as \( y = 4x^2 \))
2. \( x^2 = 9y \) (which can be rewritten as \( y = \frac{x^2}{9} \))

We also have the line \( y = 2 \).

### Step 2: Find the points of intersection
To find the area between these curves, we first need to find the points where they intersect. We can set \( y = 2 \) in both parabolas.

1. For \( y = 4x^2 \):
\[ 2 = 4x^2 \implies x^2 = \frac{1}{2} \implies x = \pm \frac{1}{\sqrt{2}} \]

2. For \( y = \frac{x^2}{9} \):
\[ 2 = \frac{x^2}{9} \implies x^2 = 18 \implies x = \pm 3\sqrt{2} \]

### Step 3: Set up the integral
The area between the curves from \( x = -\frac{1}{\sqrt{2}} \) to \( x = 3\sqrt{2} \) can be calculated using the integral of the difference of the functions:

\[ \text{Area} = \int_{-\frac{1}{\sqrt{2}}}^{3\sqrt{2}} \left( \text{Upper curve} - \text{Lower curve} \right) \, dx \]

In this case, the upper curve is \( y = 2 \) and the lower curves are \( y = 4x^2 \) and \( y = \frac{x^2}{9} \).

### Step 4: Calculate the area
We will calculate the area in two parts, from \( -\frac{1}{\sqrt{2}} \) to \( 0 \) and from \( 0 \) to \( 3\sqrt{2} \).

1. From \( -\frac{1}{\sqrt{2}} \) to \( 0 \):
\[ \text{Area}_1 = \int_{-\frac{1}{\sqrt{2}}}^{0} (2 - 4x^2) \, dx \]

2. From \( 0 \) to \( 3\sqrt{2} \):
\[ \text{Area}_2 = \int_{0}^{3\sqrt{2}} (2 - \frac{x^2}{9}) \, dx \]

### Step 5: Evaluate the integrals
1. For \( \text{Area}_1 \):
\[ \text{Area}_1 = \int_{-\frac{1}{\sqrt{2}}}^{0} (2 - 4x^2) \, dx = \left[ 2x - \frac{4x^3}{3} \right]_{-\frac{1}{\sqrt{2}}}^{0} \]
Evaluating this gives:
\[ = \left( 0 - 0 \right) - \left( 2\left(-\frac{1}{\sqrt{2}}\right) - \frac{4\left(-\frac{1}{\sqrt{2}}\right)^3}{3} \right) \]

2. For \( \text{Area}_2 \):
\[ \text{Area}_2 = \int_{0}^{3\sqrt{2}} (2 - \frac{x^2}{9}) \, dx = \left[ 2x - \frac{x^3}{27} \right]_{0}^{3\sqrt{2}} \]
Evaluating this gives:
\[ = \left( 2(3\sqrt{2}) - \frac{(3\sqrt{2})^3}{27} \right) - (0 - 0) \]

### Step 6: Combine the areas
Finally, add \( \text{Area}_1 \) and \( \text{Area}_2 \) to find the total area.

### Final Answer
The total area bounded between the parabolas and the line is:
\[ \text{Total Area} = \text{Area}_1 + \text{Area}_2 \]