FInd the area bounded by the curves `y^2=9xandx^2=9y.`

Options

A. 26 sq. units.
B. 27 sq. units.
C. 28 sq. units.
D. 29 sq. units.

Answer

To find the area bounded by the curves \( y^2 = 9x \) and \( x^2 = 9y \), we can follow these steps:

### Step 1: Identify the curves
The equations given are:
1. \( y^2 = 9x \) (which is a parabola opening to the right)
2. \( x^2 = 9y \) (which is a parabola opening upwards)

### Step 2: Find the points of intersection
To find the points of intersection, we can solve the two equations simultaneously.

From \( y^2 = 9x \), we can express \( x \) in terms of \( y \):
\[ x = \frac{y^2}{9} \]

Substituting this into the second equation \( x^2 = 9y \):
\[ \left(\frac{y^2}{9}\right)^2 = 9y \]
\[ \frac{y^4}{81} = 9y \]
Multiplying through by 81 to eliminate the fraction:
\[ y^4 = 729y \]
Rearranging gives:
\[ y^4 - 729y = 0 \]
Factoring out \( y \):
\[ y(y^3 - 729) = 0 \]
This gives us \( y = 0 \) or \( y^3 = 729 \), which leads to \( y = 9 \).

### Step 3: Find corresponding \( x \) values
For \( y = 0 \):
\[ x = \frac{0^2}{9} = 0 \]
For \( y = 9 \):
\[ x = \frac{9^2}{9} = 9 \]

Thus, the points of intersection are \( (0, 0) \) and \( (9, 9) \).

### Step 4: Set up the integral for the area
The area between the curves from \( x = 0 \) to \( x = 9 \) can be found by integrating the difference of the upper curve and the lower curve.

From \( y^2 = 9x \), we can express \( y \) as:
\[ y = 3\sqrt{x} \]
From \( x^2 = 9y \), we can express \( y \) as:
\[ y = \frac{x^2}{9} \]

The area \( A \) can be calculated as:
\[ A = \int_0^9 \left( 3\sqrt{x} - \frac{x^2}{9} \right) \, dx \]

### Step 5: Calculate the integral
Calculating the integral:
\[ A = \int_0^9 3\sqrt{x} \, dx - \int_0^9 \frac{x^2}{9} \, dx \]

Calculating the first integral:
\[ \int 3\sqrt{x} \, dx = 3 \cdot \frac{2}{3} x^{3/2} = 2x^{3/2} \]
Evaluating from 0 to 9:
\[ 2[9^{3/2} - 0^{3/2}] = 2[27 - 0] = 54 \]

Calculating the second integral:
\[ \int \frac{x^2}{9} \, dx = \frac{1}{9} \cdot \frac{x^3}{3} = \frac{x^3}{27} \]
Evaluating from 0 to 9:
\[ \frac{9^3}{27} - \frac{0^3}{27} = \frac{729}{27} = 27 \]

### Step 6: Find the area
Now, substituting back into the area calculation:
\[ A = 54 - 27 = 27 \]

### Final Answer
The area bounded by the curves \( y^2 = 9x \) and \( x^2 = 9y \) is \( 27 \) square units.

---