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

To find the volume 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 will use the **cylindrical shell method**, which is generally useful when rotating around horizontal lines like \( y = -4 \).

### Step 1: Visualize the region and the rotation
We are rotating the region enclosed by:
- The curve \( y = x^3 \) from \( x = 0 \) to \( x = 2 \) (since \( x = 2 \) is given as a boundary).
- The horizontal line \( y = 1 \), which serves as a top boundary of the region.

We are rotating this region about the line \( y = -4 \), which is below the region.

### Step 2: Set up the cylindrical shell formula
The volume of the solid using the cylindrical shell method is given by:

\[
V = 2\pi \int_{a}^{b} ( \text{radius of shell} ) ( \text{height of shell} ) \, dx
\]

Here:
- **Radius** of the shell is the distance from the line \( y = -4 \) to the curve \( y = x^3 \), which is \( x^3 - (-4) = x^3 + 4 \).
- **Height** of the shell is the length from \( y = x^3 \) to \( y = 1 \), which is \( 1 - x^3 \).

### Step 3: Define limits of integration
The limits of integration are from \( x = 0 \) to \( x = 2 \), since the region is bounded by these \( x \)-values.

### Step 4: Set up the integral
Now, using the cylindrical shell method, we have:

\[
V = 2\pi \int_{0}^{2} (x^3 + 4)(1 - x^3) \, dx
\]

### Step 5: Simplify the integrand
Expand the product:

\[
(x^3 + 4)(1 - x^3) = x^3 - x^6 + 4 - 4x^3
\]

So the integral becomes:

\[
V = 2\pi \int_{0}^{2} ( -x^6 - 3x^3 + 4 ) \, dx
\]

### Step 6: Compute the integral
Now, integrate each term separately:

\[
\int_{0}^{2} (-x^6 - 3x^3 + 4) \, dx = \left[ -\frac{x^7}{7} - \frac{3x^4}{4} + 4x \right]_{0}^{2}
\]

Evaluating at \( x = 2 \):

\[
-\frac{2^7}{7} - \frac{3(2^4)}{4} + 4(2) = -\frac{128}{7} - \frac{48}{4} + 8 = -\frac{128}{7} - 12 + 8 = -\frac{128}{7} - 4
\]

Evaluating at \( x = 0 \) yields 0.

So, the integral evaluates to:

\[
-\frac{128}{7} - 4 = -\frac{128 + 28}{7} = -\frac{156}{7}
\]

### Step 7: Multiply by \( 2\pi \)
Now, multiply the result by \( 2\pi \) to get the volume:

\[
V = 2\pi \left( -\frac{156}{7} \right) = -\frac{312\pi}{7}
\]

Since the volume must be positive, we take the absolute value:

\[
V = \frac{312\pi}{7}
\]

Thus, the volume of the solid is \( \frac{312\pi}{7} \) cubic units.

---

Would you like a more detailed explanation of any step or further clarification?

### Related Questions:
1. How does the cylindrical shell method differ from the disk method for finding volumes of revolution?
2. What are the geometric implications of rotating around a line other than the x- or y-axis?
3. How would the result change if we rotated around the line \( y = 1 \) instead of \( y = -4 \)?
4. Can we apply the disk or washer method to this problem, and how would it differ?
5. How would we find the surface area of the solid generated by this rotation?

### Tip:
Always visualize the region and the axis of rotation to determine whether the cylindrical shell or disk method is more appropriate for solving volume of revolution problems.

Find the volume of a 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 using the cylindrical shell method. This solution includes detailed steps, key formulas, and calculus concepts, suitable for college-level calculus students.

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