Find the area of the region formed by the segment cut off from the parabola `x^(2)=8y` by the line `x-2y+8=0`.

Answer

To find the area of the region formed by the segment cut off from the parabola \( x^2 = 8y \) by the line \( x - 2y + 8 = 0 \), we will follow these steps:

### Step 1: Find the intersection points of the parabola and the line.

1. **Equation of the parabola**: \( x^2 = 8y \) (Equation 1)
2. **Equation of the line**: Rearranging \( x - 2y + 8 = 0 \) gives \( y = \frac{x + 8}{2} \) (Equation 2)

Now, substitute Equation 2 into Equation 1:

\[ x^2 = 8\left(\frac{x + 8}{2}\right) \]

This simplifies to:

\[ x^2 = 4(x + 8) \]

Rearranging gives:

\[ x^2 - 4x - 32 = 0 \]

### Step 2: Solve the quadratic equation.

Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

Here, \( a = 1, b = -4, c = -32 \):

\[ x = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot (-32)}}{2 \cdot 1} \]

Calculating the discriminant:

\[ = \frac{4 \pm \sqrt{16 + 128}}{2} = \frac{4 \pm \sqrt{144}}{2} = \frac{4 \pm 12}{2} \]

This gives us:

\[ x = \frac{16}{2} = 8 \quad \text{and} \quad x = \frac{-8}{2} = -4 \]

### Step 3: Find the corresponding y-values.

For \( x = 8 \):

\[ y = \frac{8 + 8}{2} = 8 \]

For \( x = -4 \):

\[ y = \frac{-4 + 8}{2} = 2 \]

Thus, the intersection points are \( (8, 8) \) and \( (-4, 2) \).

### Step 4: Set up the area integral.

The area \( A \) between the curve and the line from \( x = -4 \) to \( x = 8 \) is given by:

\[ A = \int_{-4}^{8} \left( \text{(line)} - \text{(parabola)} \right) \, dx \]

Substituting the equations:

\[ A = \int_{-4}^{8} \left( \frac{x + 8}{2} - \frac{x^2}{8} \right) \, dx \]

### Step 5: Simplify and compute the integral.

The integral becomes:

\[ A = \int_{-4}^{8} \left( \frac{x + 8}{2} - \frac{x^2}{8} \right) \, dx = \int_{-4}^{8} \left( \frac{1}{2}x + 4 - \frac{1}{8}x^2 \right) \, dx \]

### Step 6: Calculate the integral.

Calculating the integral:

\[ A = \left[ \frac{1}{4}x^2 + 4x - \frac{1}{24}x^3 \right]_{-4}^{8} \]

Calculating at the bounds:

1. For \( x = 8 \):

\[ A(8) = \frac{1}{4}(8^2) + 4(8) - \frac{1}{24}(8^3) = \frac{64}{4} + 32 - \frac{512}{24} = 16 + 32 - \frac{64}{3} = 48 - \frac{64}{3} = \frac{144 - 64}{3} = \frac{80}{3} \]

2. For \( x = -4 \):

\[ A(-4) = \frac{1}{4}(-4^2) + 4(-4) - \frac{1}{24}(-4^3) = \frac{16}{4} - 16 + \frac{64}{24} = 4 - 16 + \frac{8}{3} = -12 + \frac{8}{3} = -\frac{36}{3} + \frac{8}{3} = -\frac{28}{3} \]

### Step 7: Combine the results.

Thus, the area \( A \) is:

\[ A = \left( \frac{80}{3} - \left( -\frac{28}{3} \right) \right) = \frac{80 + 28}{3} = \frac{108}{3} = 36 \]

### Final Answer:

The area of the region formed by the segment cut off from the parabola by the line is \( 36 \) square units.

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