Find the volume V of the solid obtained by rotating the region bounded by the given curves about the specified line.
y = x3,   y = 1,   x = 2;  
 about y = −4

To find the volume \( V \) of the solid obtained by rotating the region bounded by the curves \( y = x^3 \), \( y = 1 \), and \( x = 2 \) about the line \( y = -4 \), we can use the **method of cylindrical shells** or the **method of washers/disks**.

Since the problem involves rotation about a horizontal line that is **below the bounded region**, the **washer method** will be appropriate here.

### Step 1: Set up the general formula for the washer method
The volume \( V \) of a solid obtained by rotating a region around a horizontal line is given by:
\[
V = \pi \int_{a}^{b} \left[ R_{\text{outer}}(x)^2 - R_{\text{inner}}(x)^2 \right] \, dx
\]
where:
- \( R_{\text{outer}}(x) \) is the distance from the axis of rotation to the outer curve.
- \( R_{\text{inner}}(x) \) is the distance from the axis of rotation to the inner curve.
- \( a \) and \( b \) are the bounds of integration in the \( x \)-direction.

### Step 2: Identify the curves and distances to the axis
- The region is bounded by \( y = x^3 \) (the lower curve) and \( y = 1 \) (the upper horizontal line).
- The axis of rotation is \( y = -4 \).
  
For each \( x \), the distance from the axis of rotation to the curves is:
- **Outer radius** (upper curve, \( y = 1 \)): 
  \[
  R_{\text{outer}}(x) = 1 - (-4) = 5
  \]
- **Inner radius** (lower curve, \( y = x^3 \)): 
  \[
  R_{\text{inner}}(x) = x^3 - (-4) = x^3 + 4
  \]

### Step 3: Set up the integral
The limits of integration are from \( x = 0 \) to \( x = 2 \), as the region is bounded by \( x = 0 \) (the starting point of the curve \( y = x^3 \)) and \( x = 2 \) (where \( y = 1 \)).

The volume integral is:
\[
V = \pi \int_{0}^{2} \left[ 5^2 - (x^3 + 4)^2 \right] \, dx
\]
Simplify the expression inside the integral:
\[
V = \pi \int_{0}^{2} \left[ 25 - (x^6 + 8x^3 + 16) \right] \, dx
\]
\[
V = \pi \int_{0}^{2} \left[ 25 - x^6 - 8x^3 - 16 \right] \, dx
\]
\[
V = \pi \int_{0}^{2} \left[ 9 - x^6 - 8x^3 \right] \, dx
\]

### Step 4: Compute the integral
Now integrate term by term:
\[
V = \pi \left[ \int_{0}^{2} 9 \, dx - \int_{0}^{2} x^6 \, dx - \int_{0}^{2} 8x^3 \, dx \right]
\]

- \( \int_{0}^{2} 9 \, dx = 9x \big|_0^2 = 18 \)
- \( \int_{0}^{2} x^6 \, dx = \frac{x^7}{7} \big|_0^2 = \frac{128}{7} \)
- \( \int_{0}^{2} 8x^3 \, dx = 8 \times \frac{x^4}{4} \big|_0^2 = 2 \times 16 = 32 \)

Thus:
\[
V = \pi \left[ 18 - \frac{128}{7} - 32 \right]
\]
\[
V = \pi \left[ 18 - 32 - \frac{128}{7} \right]
\]
\[
V = \pi \left[ -14 - \frac{128}{7} \right] = \pi \left[ \frac{-98}{7} - \frac{128}{7} \right] = \pi \left( \frac{-226}{7} \right)
\]

### Step 5: Final answer
Thus, the volume is:
\[
V = \frac{-226\pi}{7}
\]
Since volume cannot be negative, we take the absolute value:
\[
V = \frac{226\pi}{7}
\]

Would you like more details on this method or have any questions?

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### 5 Related Questions:
1. How do you determine the bounds of integration when using the washer method?
2. What is the difference between the washer and cylindrical shell methods?
3. How does the axis of rotation affect the integral setup in volume problems?
4. Can you explain how to handle rotations about vertical lines?
5. How does the method change when rotating around a different axis, like \( x = 1 \)?

### Tip:
Always carefully identify the distance from the axis of rotation to the curves when setting up volume integrals.

Find the volume V of the solid obtained by rotating the region bounded by the given curves about the specified line. y = x^3, y = 1, x = 2; about y = -4

Learn how to calculate the volume of a solid obtained by rotating the region bounded by y = x^3, y = 1, and x = 2 about the line y = -4. This detailed solution involves the washer method, providing step-by-step guidance on setting up and solving the integral. Ideal for students in Grades 11-12 studying calculus.

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