find the area of the region enclosed by the curves y^2-3x=9 and x-y=3

To find the area of the region enclosed by the curves \( y^2 - 3x = 9 \) and \( x - y = 3 \), we first need to express each equation in terms of \( y \) or \( x \) and then determine the points of intersection.

### Step 1: Rewrite the equations in a more standard form

1. **First curve:**  
   \[
   y^2 - 3x = 9 \implies y^2 = 3x + 9
   \implies y = \pm \sqrt{3x + 9}
   \]
   
2. **Second curve:**
   \[
   x - y = 3 \implies y = x - 3
   \]

### Step 2: Find the points of intersection

To find the points of intersection, substitute \( y = x - 3 \) into \( y^2 = 3x + 9 \):

\[
(x - 3)^2 = 3x + 9
\]
Expand and simplify:
\[
x^2 - 6x + 9 = 3x + 9
\]
Subtract \( 3x + 9 \) from both sides:
\[
x^2 - 9x = 0
\]
Factor the quadratic equation:
\[
x(x - 9) = 0
\]
So, \( x = 0 \) or \( x = 9 \).

For \( x = 0 \):
\[
y = 0 - 3 = -3
\]

For \( x = 9 \):
\[
y = 9 - 3 = 6
\]

Thus, the points of intersection are \( (0, -3) \) and \( (9, 6) \).

### Step 3: Set up the integral

The area enclosed by the curves can be found by integrating the difference between the upper curve and the lower curve from \( x = 0 \) to \( x = 9 \).

The upper curve is \( y = \sqrt{3x + 9} \), and the lower curve is \( y = x - 3 \).

The area \( A \) is given by:
\[
A = \int_{0}^{9} \left( \sqrt{3x + 9} - (x - 3) \right) dx
\]

### Step 4: Evaluate the integral

Let's break the integral into two parts:
\[
A = \int_{0}^{9} \sqrt{3x + 9} \, dx - \int_{0}^{9} (x - 3) \, dx
\]

1. **Evaluate the first integral:**
   \[
   \int \sqrt{3x + 9} \, dx
   \]
   Use the substitution \( u = 3x + 9 \), hence \( du = 3 \, dx \), and the integral becomes:
   \[
   \frac{1}{3} \int \sqrt{u} \, du = \frac{1}{3} \cdot \frac{2}{3} u^{3/2} = \frac{2}{9} (3x + 9)^{3/2}
   \]
   So,
   \[
   \left[ \frac{2}{9} (3x + 9)^{3/2} \right]_{0}^{9} = \frac{2}{9} \left[ (36)^{3/2} - (9)^{3/2} \right] = \frac{2}{9} \left[ 216 - 27 \right] = \frac{2}{9} \cdot 189 = 42
   \]

2. **Evaluate the second integral:**
   \[
   \int_{0}^{9} (x - 3) \, dx = \left[ \frac{x^2}{2} - 3x \right]_{0}^{9} = \left[ \frac{81}{2} - 27 \right] - 0 = \frac{81}{2} - 27 = \frac{27}{2}
   \]

### Step 5: Find the total area

Subtract the two results:
\[
A = 42 - \frac{27}{2} = \frac{84}{2} - \frac{27}{2} = \frac{57}{2} = 28.5
\]

Thus, the area of the region enclosed by the curves is \( \boxed{28.5} \) square units.

Do you want details or have any questions?

Here are some related questions to explore:

1. How do you determine which curve is on top or bottom within a specific range?
2. Can the method used here be applied to any pair of curves?
3. How does the choice of limits of integration affect the result?
4. What happens if the curves intersect at more than two points?
5. How would you calculate the area if the curves are given parametrically?

**Tip:** When solving problems involving areas between curves, always sketch the region first to understand the arrangement and identify the correct limits of integration.

Find the area of the region enclosed by the curves y^2 - 3x = 9 and x - y = 3.

Explore how to find the area enclosed by the curves y^2 - 3x = 9 and x - y = 3 using integral calculus. This detailed solution includes steps for solving intersections, setting up integrals, and calculating the area, suitable for Grade 12 or college-level students.

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