Find the volume of the solid obtained by rotating the region between y=x^(2)+1 and y-x-3=0 about the x-axis.

To find the volume of the solid obtained by rotating the region between the curves \( y = x^2 + 1 \) and \( y = x + 3 \) about the \( x \)-axis, we can use the method of cylindrical shells or the disk method. Here, we will use the disk method.

### Steps:

1. **Find the points of intersection:**

Set the two functions equal to each other to find the points of intersection:
\[ x^2 + 1 = x + 3 \]
\[ x^2 - x - 2 = 0 \]
\[ (x - 2)(x + 1) = 0 \]

Thus, the points of intersection are \( x = 2 \) and \( x = -1 \).

2. **Set up the integral using the disk method:**

The volume \( V \) of the solid of revolution about the \( x \)-axis is given by:
\[ V = \pi \int_{a}^{b} \left[ R(x)^2 - r(x)^2 \right] dx \]
where \( R(x) \) is the outer radius and \( r(x) \) is the inner radius. For this problem, \( R(x) \) is the upper function \( y = x + 3 \) and \( r(x) \) is the lower function \( y = x^2 + 1 \).

3. **Determine the limits of integration:**

The limits of integration are from \( x = -1 \) to \( x = 2 \).

4. **Set up and evaluate the integral:**

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

Expanding the terms inside the integral:

\[ (x + 3)^2 = x^2 + 6x + 9 \]
\[ (x^2 + 1)^2 = x^4 + 2x^2 + 1 \]

Thus, the integral becomes:

\[ V = \pi \int_{-1}^{2} [x^2 + 6x + 9 - (x^4 + 2x^2 + 1)] \, dx \]
\[ = \pi \int_{-1}^{2} [-x^4 - x^2 + 6x + 8] \, dx \]

Now, we integrate each term separately:

\[ \int_{-1}^{2} -x^4 \, dx = \left[ -\frac{x^5}{5} \right]_{-1}^{2} \]
\[ \int_{-1}^{2} -x^2 \, dx = \left[ -\frac{x^3}{3} \right]_{-1}^{2} \]
\[ \int_{-1}^{2} 6x \, dx = \left[ 3x^2 \right]_{-1}^{2} \]
\[ \int_{-1}^{2} 8 \, dx = \left[ 8x \right]_{-1}^{2} \]

Evaluating each integral at the bounds:

\[ \left[ -\frac{x^5}{5} \right]_{-1}^{2} = \left( -\frac{2^5}{5} \right) - \left( -\frac{(-1)^5}{5} \right) = -\frac{32}{5} + \frac{1}{5} = -\frac{31}{5} \]
\[ \left[ -\frac{x^3}{3} \right]_{-1}^{2} = \left( -\frac{2^3}{3} \right) - \left( -\frac{(-1)^3}{3} \right) = -\frac{8}{3} + \frac{1}{3} = -\frac{7}{3} \]
\[ \left[ 3x^2 \right]_{-1}^{2} = 3(2^2) - 3((-1)^2) = 12 - 3 = 9 \]
\[ \left[ 8x \right]_{-1}^{2} = 8(2) - 8(-1) = 16 + 8 = 24 \]

Adding these results together:

\[ V = \pi \left( -\frac{31}{5} - \frac{7}{3} + 9 + 24 \right) \]

Finding a common denominator (15):

\[ -\frac{31}{5} = -\frac{93}{15} \]
\[ -\frac{7}{3} = -\frac{35}{15} \]
\[ 9 = \frac{135}{15} \]
\[ 24 = \frac{360}{15} \]

So the integral evaluates to:

\[ V = \pi \left( -\frac{93}{15} - \frac{35}{15} + \frac{135}{15} + \frac{360}{15} \right) \]
\[ = \pi \left( \frac{-93 - 35 + 135 + 360}{15} \right) \]
\[ = \pi \left( \frac{367}{15} \right) \]
\[ = \frac{367\pi}{15} \]

Thus, the volume of the solid is:

\[ \boxed{\frac{367\pi}{15}} \]

If you have any questions or need further details, let me know!

Here are 5 questions you might want to ask next:
1. How do you find the surface area of a solid of revolution?
2. What is the method of integration by parts in calculus?
3. Can you explain the Mean Value Theorem?
4. How do you solve a differential equation using separation of variables?
5. What are the properties of indefinite integrals?

**Tip:** When rotating regions around the axis, always visualize the solid and set up the correct limits of integration to ensure accurate results.

Learn how to find the volume of the solid obtained by rotating the region between the curves y = x^2 + 1 and y = x + 3 about the x-axis. Step-by-step solution using the disk method in integral calculus.

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