The curve 𝑦 = 𝑥3 + 1 is revolved through 360 degrees about the x-axis. Find the volume between the upper part of x axis and this volume between x = 0 and
x =1.

To find the volume of the solid generated when the curve \( y = x^3 + 1 \) is revolved around the x-axis between \( x = 0 \) and \( x = 1 \), we use the method of disks (or washers). The formula for the volume of a solid of revolution around the x-axis is given by:

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

Here, \( f(x) = x^3 + 1 \), \( a = 0 \), and \( b = 1 \).

### Step 1: Set up the integral
The volume \( V \) is:

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

### Step 2: Expand the integrand
First, we need to expand the expression inside the integral:

\[
(x^3 + 1)^2 = (x^3 + 1)(x^3 + 1) = x^6 + 2x^3 + 1
\]

So the integral becomes:

\[
V = \pi \int_{0}^{1} (x^6 + 2x^3 + 1) \, dx
\]

### Step 3: Integrate
Now, integrate term by term:

\[
\int x^6 \, dx = \frac{x^7}{7}, \quad \int 2x^3 \, dx = \frac{2x^4}{4} = \frac{x^4}{2}, \quad \int 1 \, dx = x
\]

Thus:

\[
V = \pi \left[ \frac{x^7}{7} + \frac{x^4}{2} + x \right]_{0}^{1}
\]

### Step 4: Evaluate the definite integral
Evaluate the expression at the bounds \( x = 1 \) and \( x = 0 \):

At \( x = 1 \):

\[
V(1) = \pi \left( \frac{1^7}{7} + \frac{1^4}{2} + 1 \right) = \pi \left( \frac{1}{7} + \frac{1}{2} + 1 \right) = \pi \left( \frac{1}{7} + \frac{1}{2} + 1 \right)
\]

To simplify:

\[
V(1) = \pi \left( \frac{2}{14} + \frac{7}{14} + \frac{14}{14} \right) = \pi \left( \frac{23}{14} \right)
\]

At \( x = 0 \):

\[
V(0) = \pi \left( 0 + 0 + 0 \right) = 0
\]

Thus, the volume is:

\[
V = \pi \left( \frac{23}{14} - 0 \right) = \frac{23\pi}{14}
\]

### Final Answer:
The volume of the solid is \( \frac{23\pi}{14} \) cubic units.

Would you like any details or have any further questions?

Here are some related questions for further exploration:

1. How would the volume change if the curve was \( y = x^3 \) instead?
2. What is the geometric significance of the disk method in this context?
3. How does the shell method differ from the disk method in calculating volumes?
4. What if the curve was revolved around the y-axis instead?
5. How would you find the surface area of the solid of revolution instead of the volume?
6. What role does the interval \([0, 1]\) play in the integration process?
7. Can you extend this method to finding the volume of revolution of more complex functions?
8. What would the volume be if the curve \( y = x^3 + 1 \) was revolved around the line \( y = -1 \)?

**Tip:** When expanding a polynomial squared in an integral, double-check your expansion to avoid errors in subsequent integration steps.

Learn how to find the volume of revolution generated by revolving the curve y = x^3 + 1 around the x-axis. Detailed step-by-step solution using the method of disks, including integral calculus. Suitable for advanced high school students and calculus enthusiasts.

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