Find the area of the region formed by `x^2+y^2-6x-4y+12 le 0`. `y le x` and `x le 5/2`.

Answer

To find the area of the region defined by the inequalities \(x^2 + y^2 - 6x - 4y + 12 \leq 0\), \(y \leq x\), and \(x \leq \frac{5}{2}\), we will follow these steps:

### Step 1: Rewrite the Circle Equation
We start with the equation of the circle:
\[ x^2 + y^2 - 6x - 4y + 12 = 0 \]
We can complete the square for \(x\) and \(y\):

1. For \(x\):
\[ x^2 - 6x = (x - 3)^2 - 9 \]
2. For \(y\):
\[ y^2 - 4y = (y - 2)^2 - 4 \]

Substituting these into the equation gives:
\[ (x - 3)^2 - 9 + (y - 2)^2 - 4 + 12 = 0 \]
This simplifies to:
\[ (x - 3)^2 + (y - 2)^2 - 1 = 0 \]
Thus, we have:
\[ (x - 3)^2 + (y - 2)^2 = 1 \]
This represents a circle with center at \((3, 2)\) and radius \(1\).

### Step 2: Identify the Region
The inequality \(x^2 + y^2 - 6x - 4y + 12 \leq 0\) describes the area inside or on the boundary of the circle. The lines \(y = x\) and \(x = \frac{5}{2}\) will help us define the area of interest.

### Step 3: Find Intersection Points
We need to find the intersection points of the line \(y = x\) with the circle:
Substituting \(y = x\) into the circle's equation:
\[ (x - 3)^2 + (x - 2)^2 = 1 \]
Expanding this:
\[ (x - 3)^2 + (x - 2)^2 = 1 \implies (x^2 - 6x + 9) + (x^2 - 4x + 4) = 1 \]
Combining terms:
\[ 2x^2 - 10x + 13 = 1 \implies 2x^2 - 10x + 12 = 0 \implies x^2 - 5x + 6 = 0 \]
Factoring gives:
\[ (x - 2)(x - 3) = 0 \implies x = 2 \text{ or } x = 3 \]
Thus, the intersection points are \((2, 2)\) and \((3, 3)\).

### Step 4: Determine the Area
We need to find the area under the line \(y = x\) from \(x = 2\) to \(x = \frac{5}{2}\) and subtract the area under the circle.

1. **Area under the line \(y = x\)** from \(x = 2\) to \(x = \frac{5}{2}\):
\[ \text{Area} = \int_{2}^{\frac{5}{2}} x \, dx = \left[\frac{x^2}{2}\right]_{2}^{\frac{5}{2}} = \frac{(\frac{5}{2})^2}{2} - \frac{2^2}{2} = \frac{25/4}{2} - 2 = \frac{25}{8} - 2 = \frac{25}{8} - \frac{16}{8} = \frac{9}{8} \]

2. **Area under the circle** from \(x = 2\) to \(x = \frac{5}{2}\):
The upper half of the circle can be expressed as:
\[ y = 2 + \sqrt{1 - (x - 3)^2} \]
Thus, the area under the circle is:
\[ \text{Area} = \int_{2}^{\frac{5}{2}} \left(2 + \sqrt{1 - (x - 3)^2}\right) \, dx \]

### Step 5: Calculate the Area Under the Circle
Calculating this integral involves two parts:
1. The area of the rectangle formed by the line \(y = 2\).
2. The area under the semicircle.

The area of the rectangle from \(x = 2\) to \(x = \frac{5}{2}\) is:
\[ \text{Area}_{\text{rectangle}} = \text{base} \times \text{height} = \left(\frac{5}{2} - 2\right) \times 2 = \frac{1}{2} \times 2 = 1 \]

The area under the semicircle can be calculated using the formula for the area of a circle:
\[ \text{Area}_{\text{semicircle}} = \frac{1}{2} \pi r^2 = \frac{1}{2} \pi (1^2) = \frac{\pi}{2} \]

### Step 6: Total Area Calculation
Finally, the total area of the shaded region is:
\[ \text{Total Area} = \text{Area under line} - \text{Area under circle} \]
\[ \text{Total Area} = \frac{9}{8} - \left(1 + \frac{\pi}{2}\right) \]
This gives us the final area of the region defined by the inequalities.