Find the area of the region ℛ enclosed by the curve 𝑥2 − 𝑦2 = 16 and the straight line
𝑥 = 5. If the region ℛ is revolved about the line 𝑥 = 5, find the volume of the solid
generated.

To solve this problem, we need to approach it in two main steps: first, find the area of the region \( \mathcal{R} \) enclosed by the curve \( x^2 - y^2 = 16 \) and the line \( x = 5 \); then, find the volume of the solid generated by revolving this region about the line \( x = 5 \).

### Step 1: Find the Area of Region \( \mathcal{R} \)

The given curve is a hyperbola defined by:
\[
x^2 - y^2 = 16
\]
This can be rewritten as:
\[
y^2 = x^2 - 16
\]
Thus, the equation for \( y \) becomes:
\[
y = \pm \sqrt{x^2 - 16}
\]

The line given is \( x = 5 \).

To find the area enclosed by the curve and the line, we first need to find the points of intersection between the curve and the line.

At \( x = 5 \), the equation of the hyperbola becomes:
\[
y^2 = 5^2 - 16 = 25 - 16 = 9
\]
So,
\[
y = \pm 3
\]
The points of intersection are \( (5, 3) \) and \( (5, -3) \).

Next, we calculate the area between the hyperbola and the vertical line \( x = 5 \) from \( y = -3 \) to \( y = 3 \).

The area of the region \( \mathcal{R} \) is given by integrating the function \( \sqrt{x^2 - 16} \) from the left boundary of the hyperbola to the vertical line \( x = 5 \).

The limits of integration start from the left vertex of the hyperbola, which is at \( x = -4 \) (since \( x^2 = 16 \) when \( y = 0 \)), and end at \( x = 5 \).

Thus, the area is:
\[
\text{Area} = 2 \int_{4}^{5} \sqrt{x^2 - 16} \, dx
\]
The factor of 2 accounts for the symmetry of the hyperbola about the x-axis.

### Step 2: Find the Volume of the Solid Generated by Revolving Region \( \mathcal{R} \) About the Line \( x = 5 \)

The volume of the solid formed by revolving the region \( \mathcal{R} \) around the line \( x = 5 \) can be found using the method of cylindrical shells. The formula for the volume using cylindrical shells is:
\[
V = 2\pi \int_{a}^{b} (r_{\text{shell}})(h_{\text{shell}}) \, dx
\]
Where:
- \( r_{\text{shell}} = 5 - x \) is the radius of the shell.
- \( h_{\text{shell}} = 2\sqrt{x^2 - 16} \) is the height of the shell (since it's the vertical distance between the upper and lower parts of the hyperbola).

Thus, the volume is:
\[
V = 2\pi \int_{4}^{5} (5 - x)(2\sqrt{x^2 - 16}) \, dx
\]

This integral gives us the volume of the solid of revolution.

Now, let's calculate both the area and the volume numerically.

I'll calculate these values for you now.The area of the region \( \mathcal{R} \) enclosed by the curve \( x^2 - y^2 = 16 \) and the line \( x = 5 \) is approximately \( 3.91 \) square units.

The volume of the solid generated by revolving this region about the line \( x = 5 \) is approximately \( 9.73 \) cubic units.

If you have any questions or need further details, feel free to ask!

Here are five related questions:
1. What is the significance of using cylindrical shells for volume calculation in this scenario?
2. How would the volume change if the region were revolved about a different line, such as \( x = 6 \)?
3. Can you calculate the centroid of the region \( \mathcal{R} \)?
4. How does the shape of the hyperbola influence the area and volume calculations?
5. What would the area be if the region were enclosed by the curve and the line \( x = 4 \) instead of \( x = 5 \)?

**Tip:** When dealing with areas and volumes of revolution, always check if symmetry can simplify your calculations!

Explore how to find the area of the region enclosed by the curve x^2 - y^2 = 16 and the line x = 5, and calculate the volume of the solid generated by revolving this region about the line x = 5. Detailed step-by-step solution including integration and cylindrical shells method. Suitable for advanced high school and college-level students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation