Complete the area of the region bounded by the parabolas `y^(2)+8x=16` and `y^(2)-24x=48`.

Answer

To find the area of the region bounded by the parabolas \( y^2 + 8x = 16 \) and \( y^2 - 24x = 48 \), we will follow these steps:

### Step 1: Rewrite the equations of the parabolas
The first parabola can be rewritten as:
\[ y^2 = 16 - 8x \implies y^2 = -8(x - 2) \]
This is a parabola that opens to the left with vertex at \( (2, 0) \).

The second parabola can be rewritten as:
\[ y^2 = 24x + 48 \implies y^2 = 24(x + 2) \]
This is a parabola that opens to the right with vertex at \( (-2, 0) \).

### Step 2: Find the points of intersection
To find the points where the two parabolas intersect, we set their equations equal to each other:
\[ 16 - 8x = 24x + 48 \]
Rearranging gives:
\[ 16 - 48 = 24x + 8x \implies -32 = 32x \implies x = -1 \]
Now, substituting \( x = -1 \) back into either equation to find \( y \):
\[ y^2 = 16 - 8(-1) = 16 + 8 = 24 \implies y = \pm \sqrt{24} = \pm 2\sqrt{6} \]
Thus, the points of intersection are \( (-1, 2\sqrt{6}) \) and \( (-1, -2\sqrt{6}) \).

### Step 3: Set up the integral for the area
The area \( A \) between the curves can be calculated using the formula:
\[ A = \int_{y_1}^{y_2} (x_{\text{right}} - x_{\text{left}}) \, dy \]
Here, \( y_1 = -2\sqrt{6} \) and \( y_2 = 2\sqrt{6} \).

From the first parabola:
\[ x_{\text{left}} = \frac{16 - y^2}{8} \]
From the second parabola:
\[ x_{\text{right}} = \frac{y^2 - 48}{24} \]

Thus, the area can be expressed as:
\[ A = \int_{-2\sqrt{6}}^{2\sqrt{6}} \left( \frac{y^2 - 48}{24} - \frac{16 - y^2}{8} \right) dy \]

### Step 4: Simplify the integrand
First, we simplify the expression inside the integral:
\[ A = \int_{-2\sqrt{6}}^{2\sqrt{6}} \left( \frac{y^2 - 48}{24} - \frac{2(16 - y^2)}{16} \right) dy \]
Combining the fractions gives:
\[ A = \int_{-2\sqrt{6}}^{2\sqrt{6}} \left( \frac{y^2 - 48}{24} - \frac{32 - 2y^2}{16} \right) dy \]

### Step 5: Evaluate the integral
Now, we can evaluate the integral:
\[ A = \int_{-2\sqrt{6}}^{2\sqrt{6}} \left( \frac{y^2 - 48}{24} - \frac{32 - 2y^2}{16} \right) dy \]
This integral can be computed by finding the antiderivative and evaluating it at the limits.

### Step 6: Final calculation
After performing the integration and substituting the limits, we can find the area.

### Final Answer
The area of the region bounded by the two parabolas is \( \frac{32\sqrt{6}}{3} \).

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