The region bounded by the given curves is rotated about the specified axis. Find the volume of the resulting solid by any method.
y2 − x2 = 9,   y = 4;   about the x-axis

We are tasked with finding the volume of the solid formed by rotating the region bounded by the curves \( y^2 - x^2 = 9 \) and \( y = 4 \) around the x-axis.

### Step 1: Identify the Region

The first equation \( y^2 - x^2 = 9 \) represents a hyperbola, which can be rewritten as:

\[
y^2 = x^2 + 9 \quad \text{or} \quad y = \pm \sqrt{x^2 + 9}
\]

This means that the hyperbola opens upwards and downwards.

The second equation is the line \( y = 4 \), which is horizontal.

We will focus on the region where \( y^2 - x^2 = 9 \) is between \( y = 0 \) and \( y = 4 \).

### Step 2: Set up the Volume Integral

Since we are rotating the region around the x-axis, we can use the **method of cylindrical shells** or the **disk/washer method**. For this problem, the washer method seems appropriate because we have explicit functions in terms of \( y \).

For the washer method, the volume of the solid is given by the formula:

\[
V = \pi \int_{y_{\text{min}}}^{y_{\text{max}}} \left[ R_{\text{outer}}(y)^2 - R_{\text{inner}}(y)^2 \right] \, dy
\]

Here:
- \( R_{\text{outer}}(y) \) is the distance from the x-axis to the outer curve (i.e., the curve \( y = 4 \)),
- \( R_{\text{inner}}(y) \) is the distance from the x-axis to the inner curve (i.e., \( x \) in terms of \( y \) from the hyperbola).

#### Outer radius:
Since \( y = 4 \) is the upper bound, the outer radius is constant:

\[
R_{\text{outer}}(y) = 4
\]

#### Inner radius:
From \( y^2 - x^2 = 9 \), solving for \( x \) in terms of \( y \), we get:

\[
x = \pm \sqrt{y^2 - 9}
\]

Thus, the inner radius is \( R_{\text{inner}}(y) = \sqrt{y^2 - 9} \).

### Step 3: Define the Limits of Integration

The region is bounded vertically from \( y = 3 \) (where the hyperbola intersects the x-axis, i.e., when \( x = 0 \)) to \( y = 4 \).

### Step 4: Write the Volume Integral

Now, we can set up the volume integral:

\[
V = \pi \int_3^4 \left[ 4^2 - \left( \sqrt{y^2 - 9} \right)^2 \right] \, dy
\]

Simplifying:

\[
V = \pi \int_3^4 \left[ 16 - (y^2 - 9) \right] \, dy
\]
\[
V = \pi \int_3^4 \left[ 16 - y^2 + 9 \right] \, dy
\]
\[
V = \pi \int_3^4 \left[ 25 - y^2 \right] \, dy
\]

### Step 5: Solve the Integral

Now, we compute the integral:

\[
V = \pi \left[ 25y - \frac{y^3}{3} \right]_3^4
\]

First, evaluate at \( y = 4 \):

\[
25(4) - \frac{4^3}{3} = 100 - \frac{64}{3} = 100 - 21.33 = 78.67
\]

Now, evaluate at \( y = 3 \):

\[
25(3) - \frac{3^3}{3} = 75 - \frac{27}{3} = 75 - 9 = 66
\]

Finally, subtract the results:

\[
V = \pi \left( 78.67 - 66 \right) = \pi (12.67)
\]

Thus, the volume is:

\[
V \approx 12.67 \pi \quad \text{cubic units}.
\]

### Conclusion:
The volume of the solid is approximately \( 12.67\pi \) cubic units.

Would you like more details or clarification on any step?

Here are five related questions for practice:
1. What is the volume of the solid formed when \( y = 4 \) is rotated about the y-axis?
2. How does the method of cylindrical shells differ from the disk/washer method?
3. What is the general formula for the volume of solids of revolution using the washer method?
4. How would the volume change if the region were rotated about the line \( y = 5 \) instead?
5. How would you solve the volume problem if the curves were rotated around the line \( x = 1 \)?

**Tip:** Always carefully check the limits of integration when setting up volume integrals for solids of revolution.

The region bounded by the given curves is rotated about the specified axis. Find the volume of the resulting solid by any method. y^2 − x^2 = 9, y = 4; about the x-axis.

Explore the volume of a solid formed by rotating the region bounded by y^2 − x^2 = 9 and y = 4 around the x-axis. This problem demonstrates the washer method and provides a step-by-step solution suitable for Grade 12 and college calculus students.

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