If the shell method is used, then since the line y=10 is parallel to the x-axis, it should be in terms of y. Please adjust

You are correct. When revolving around a line that is horizontal (parallel to the x-axis) using the shell method, the integral should be set up in terms of \( y \), not \( x \). Let's correct the setup accordingly.

### Corrected Approach: Shell Method in Terms of \( y \)

Given that the region is revolved around the line \( y = 10 \), and that the equations describe the curves in terms of \( x \) as a function of \( y \), we express the boundaries and set up the integral in terms of \( y \).

#### Step 1: Rewrite the function
The curve \( y = \frac{10}{x^2} \) can be rewritten as:
\[
x = \sqrt{\frac{10}{y}}
\]

#### Step 2: Define the limits for \( y \)
The limits for \( y \) are from \( y = 10/5^2 = 0.4 \) (when \( x = 5 \)) to \( y = 10/1^2 = 10 \) (when \( x = 1 \)).

#### Step 3: Set up the Shell Method Integral
- **Radius**: The distance from a point on the curve to the line \( y = 10 \) is \( 10 - y \).
- **Height**: The height of the shell is the distance between the vertical lines \( x = 1 \) and \( x = 5 \), which is represented by \( x \) in terms of \( y \): \( 2\sqrt{\frac{10}{y}} \).

Thus, the volume is given by:
\[
V = 2\pi \int_{0.4}^{10} (10 - y) \cdot 2\sqrt{\frac{10}{y}} \, dy
\]

Now, let's compute this integral.The volume of the solid generated by revolving the region bounded by the curves \( y = \frac{10}{x^2} \), \( y = 0 \), \( x = 1 \), and \( x = 5 \) around the line \( y = 10 \), using the shell method in terms of \( y \), is approximately:

\[
V \approx 2\pi \left( \frac{800}{3} - 24.96 \cdot \sqrt{10} \right) \text{ cubic units.}
\]

This simplifies numerically to:

\[
V \approx 837.76 \text{ cubic units.}
\]

Would you like further clarification or more details on the setup?

Here are some related questions:
1. How does the shell method compare to the disk method for this problem?
2. What are the common errors to avoid when using the shell method?
3. How do you derive the limits of integration for similar problems?
4. How does changing the axis of rotation affect the result?
5. Can this setup be generalized to other curves and axes of rotation?

**Tip:** When rotating around a horizontal line that is not the x-axis, it's essential to adjust the radius expression accordingly to account for the shift in the axis.

Understand how to apply the shell method in calculus when revolving around the line y=10. Learn the correct setup using integral calculus in terms of y. Includes a step-by-step solution and numerical computation.

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