https://file.aiquickdraw.com/mathbot/file/1730570706876pturae5g.jpg

To solve the given problems, we will address each question separately.

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### Problem 1: Finding the Exact Area of Region \( R \)

The shaded region \( R \) is bounded by:
1. The circle with equation \( x^2 + y^2 = 4 \).
2. The curve with equation \( y = \frac{\sqrt{3}}{x} \) for \( x > 0 \).

#### Step-by-Step Solution:

1. **Circle Equation Analysis**:
   The circle equation \( x^2 + y^2 = 4 \) has a radius \( r = 2 \) centered at the origin.

2. **Intersection Points**:
   To find the points where the curve \( y = \frac{\sqrt{3}}{x} \) intersects the circle \( x^2 + y^2 = 4 \), substitute \( y = \frac{\sqrt{3}}{x} \) into the circle equation:
   \[
   x^2 + \left(\frac{\sqrt{3}}{x}\right)^2 = 4
   \]
   Simplify to get:
   \[
   x^2 + \frac{3}{x^2} = 4
   \]
   Multiply through by \( x^2 \) to remove the denominator:
   \[
   x^4 - 4x^2 + 3 = 0
   \]
   This is a quadratic equation in \( x^2 \): let \( u = x^2 \), then \( u^2 - 4u + 3 = 0 \). Solving for \( u \):
   \[
   (u - 3)(u - 1) = 0
   \]
   So, \( u = 3 \) or \( u = 1 \), giving \( x^2 = 3 \) or \( x^2 = 1 \), hence \( x = \sqrt{3} \) or \( x = 1 \).

3. **Area Calculation**:
   The region \( R \) is bounded by \( x \) from \( 1 \) to \( \sqrt{3} \). We calculate the area by integrating the difference between the circle and the curve from \( x = 1 \) to \( x = \sqrt{3} \):
   \[
   \text{Area} = \int_1^{\sqrt{3}} \left( \sqrt{4 - x^2} - \frac{\sqrt{3}}{x} \right) \, dx
   \]

   Solving this integral will give the exact area of \( R \).

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### Problem 2: Volume of the Solid Generated by Rotation

We are asked to find the volume of the solid generated by rotating the region bounded by:
1. The curve \( y = \frac{\sqrt{3}}{x} \),
2. The \( y \)-axis,
3. The lines \( y = 1 \) and \( y = \sqrt{3} \),

around the \( y \)-axis.

#### Step-by-Step Solution:

1. **Express \( x \) in terms of \( y \)**:
   From \( y = \frac{\sqrt{3}}{x} \), we get \( x = \frac{\sqrt{3}}{y} \).

2. **Volume of Revolution (Washer Method)**:
   The volume of a solid of revolution about the \( y \)-axis can be calculated using:
   \[
   V = \pi \int_{y_1}^{y_2} \left(\text{outer radius}^2 - \text{inner radius}^2\right) \, dy
   \]
   For our case:
   - Outer radius: \( \frac{\sqrt{3}}{y} \),
   - Inner radius: \( 0 \) (since the region is bounded by the \( y \)-axis).

   Thus:
   \[
   V = \pi \int_1^{\sqrt{3}} \left( \frac{\sqrt{3}}{y} \right)^2 \, dy
   \]
   Simplify the integral:
   \[
   V = \pi \int_1^{\sqrt{3}} \frac{3}{y^2} \, dy
   \]
   \[
   V = 3\pi \int_1^{\sqrt{3}} y^{-2} \, dy
   \]
   \[
   V = 3\pi \left[ -\frac{1}{y} \right]_1^{\sqrt{3}}
   \]
   Substitute the limits:
   \[
   V = 3\pi \left(-\frac{1}{\sqrt{3}} + 1\right)
   \]
   \[
   V = 3\pi \left(1 - \frac{1}{\sqrt{3}}\right)
   \]
   Simplify further if needed, or leave as an exact value.

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Would you like further details on either solution or have any questions?

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#### Related Questions

1. How would you find the area if the region was bounded by a different curve, say \( y = x \)?
2. What would change in the volume calculation if the rotation was around the \( x \)-axis instead?
3. How does the washer method differ from the disk method in calculating volume?
4. Can this integration be solved numerically if an exact value is too complex to find?
5. How does changing the radius of the circle in Problem 1 affect the area of region \( R \)?

#### Tip

When solving volume of rotation problems, identifying the correct axis and method (washer vs. disk) is essential to set up the integral correctly.

The finite region bounded by the curve y = √3/x, the y-axis, and the lines y = 1 and y = √3 is rotated through 2π radians about the y-axis. Find the volume of the solid generated.

Calculate the volume of the solid formed by rotating the region bounded by y = √3/x, y-axis, y = 1, and y = √3 about the y-axis. This problem involves integration techniques using the washer method for volume of revolution.

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