find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis. y= 1/ (x^4), y=0, x=2, x=9; about the y-axis. Try using the shell method, washer method, and disk method

Let's solve this problem of finding the volume of the solid obtained by rotating the region bounded by the curves \( y = \frac{1}{x^4} \), \( y = 0 \), \( x = 2 \), and \( x = 9 \) around the **y-axis**, using the **shell method**, **washer method**, and **disk method** (where applicable). 

### Step 1: **Shell Method**
The shell method involves slicing the region into cylindrical shells, and integrating along the axis of rotation. The formula for the volume using the shell method is:

\[
V = 2\pi \int_{a}^{b} x \cdot f(x) \, dx
\]

For the given curves, we use \( y = \frac{1}{x^4} \) as the function, and rotate about the **y-axis**. The radius of each shell is \( x \), and the height of the shell is \( f(x) = \frac{1}{x^4} \).

The limits for \( x \) are from 2 to 9.

\[
V = 2\pi \int_{2}^{9} x \cdot \frac{1}{x^4} \, dx = 2\pi \int_{2}^{9} \frac{1}{x^3} \, dx
\]

Now, solve the integral:

\[
V = 2\pi \int_{2}^{9} x^{-3} \, dx = 2\pi \left[ \frac{x^{-2}}{-2} \right]_{2}^{9}
\]

\[
V = 2\pi \left[ \frac{-1}{2x^2} \right]_{2}^{9}
\]

\[
V = 2\pi \left( \frac{-1}{2 \cdot 9^2} - \frac{-1}{2 \cdot 2^2} \right)
\]

\[
V = 2\pi \left( \frac{-1}{162} + \frac{1}{8} \right)
\]

\[
V = 2\pi \left( \frac{-1}{162} + \frac{21}{168} \right)
\]

\[
V = 2\pi \cdot \frac{41}{168}
\]

\[
V = \frac{41\pi}{84}
\]

Thus, the volume using the **shell method** is:

\[
V = \frac{41\pi}{84}
\]

### Step 2: **Washer Method**

The washer method requires slicing the solid perpendicular to the axis of rotation (in this case, the y-axis). However, the washer method is generally better suited for regions rotated about a horizontal axis, such as the x-axis, and doesn't apply directly for a rotation about the y-axis without significant changes to the method.

For this specific problem, **the washer method is not the optimal choice** for rotation about the y-axis. Instead, we use either the shell method (as done) or disk method for appropriate rotations.

### Step 3: **Disk Method**

The disk method is also typically used for rotations around the x-axis or other horizontal axes. For a rotation around the **y-axis**, the **disk method is not easily applicable** because we would need to express \( x \) as a function of \( y \), and this would become cumbersome for \( y = \frac{1}{x^4} \).

Therefore, the disk method is not useful in this case.

---

### Final Answer
The volume of the solid obtained by rotating the region bounded by \( y = \frac{1}{x^4} \), \( y = 0 \), \( x = 2 \), and \( x = 9 \) around the **y-axis** is:

\[
V = \frac{41\pi}{84}
\]

Would you like any more details on this process or further clarifications?

### Related Questions:
1. How does the method of cylindrical shells compare to the disk/washer method for this type of problem?
2. Can you provide more examples where the washer method would be useful?
3. What happens if we rotate around a different axis, like the x-axis?
4. How would the problem change if the region was unbounded on one side (e.g., \( x \to \infty \))?
5. Could we use numerical methods to approximate the volume in more complex cases?

#### Tip:
The **shell method** is often the best choice when rotating around the y-axis, as it simplifies the calculation of cylindrical shells formed parallel to the axis of rotation.

Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis. y = 1/(x^4), y = 0, x = 2, x = 9; about the y-axis. Try using the shell method, washer method, and disk method.

Learn how to find the volume of the solid formed by rotating the region bounded by y = 1/x^4, y = 0, x = 2, and x = 9 around the y-axis using the shell method. This calculus problem covers key concepts in integral calculus and solids of revolution. Suitable for advanced high school and college calculus students.

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