(1 point) Find the volume of the solid generated by revolving the described region about the given axis:

The region bounded below by the curve 𝑦=𝑥2+3
 and above by the curve 𝑦=𝑥+5
, rotated about the 𝑥
-axis.

Answer: = 

To find the volume of the solid generated by revolving the region bounded by the curves \( y = x^2 + 3 \) and \( y = x + 5 \) about the \( x \)-axis, we can use the method of disks or washers. Here, we'll use the washer method, as the region between two curves is being revolved.

### Step 1: Find the points of intersection
First, we need to find the points where the curves intersect. Set the equations equal to each other:

\[
x^2 + 3 = x + 5
\]

Rearrange to form a quadratic equation:

\[
x^2 - x - 2 = 0
\]

Factor the quadratic:

\[
(x - 2)(x + 1) = 0
\]

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

### Step 2: Set up the integral
The volume \( V \) of the solid of revolution is given by:

\[
V = \pi \int_{a}^{b} \left[ R(x)^2 - r(x)^2 \right] \, dx
\]

Where:
- \( R(x) \) is the outer radius (the distance from the \( x \)-axis to the curve further from it, which is \( y = x + 5 \)).
- \( r(x) \) is the inner radius (the distance from the \( x \)-axis to the curve closer to it, which is \( y = x^2 + 3 \)).
- \( a \) and \( b \) are the bounds of integration (in this case, from \( x = -1 \) to \( x = 2 \)).

Thus:

\[
V = \pi \int_{-1}^{2} \left[ (x + 5)^2 - (x^2 + 3)^2 \right] \, dx
\]

### Step 3: Simplify and compute the integral
Expand both squares:

\[
(x + 5)^2 = x^2 + 10x + 25
\]
\[
(x^2 + 3)^2 = x^4 + 6x^2 + 9
\]

So the integral becomes:

\[
V = \pi \int_{-1}^{2} \left( x^2 + 10x + 25 - x^4 - 6x^2 - 9 \right) \, dx
\]
\[
V = \pi \int_{-1}^{2} \left( -x^4 - 5x^2 + 10x + 16 \right) \, dx
\]

Now, integrate term by term:

\[
\int_{-1}^{2} -x^4 \, dx = \left[ -\frac{x^5}{5} \right]_{-1}^{2} = -\frac{32}{5} + \frac{1}{5} = -\frac{31}{5}
\]
\[
\int_{-1}^{2} -5x^2 \, dx = \left[ -\frac{5x^3}{3} \right]_{-1}^{2} = -\frac{40}{3} + \frac{5}{3} = -\frac{35}{3}
\]
\[
\int_{-1}^{2} 10x \, dx = \left[ 5x^2 \right]_{-1}^{2} = 20 - 5 = 15
\]
\[
\int_{-1}^{2} 16 \, dx = \left[ 16x \right]_{-1}^{2} = 32 + 16 = 48
\]

Add up the contributions:

\[
V = \pi \left( -\frac{31}{5} - \frac{35}{3} + 15 + 48 \right)
\]

To simplify, convert to a common denominator and sum:

\[
V = \pi \left( -\frac{186}{15} - \frac{175}{15} + \frac{225}{15} + \frac{720}{15} \right) = \pi \times \frac{584}{15}
\]

Finally:

\[
V = \frac{584\pi}{15} \text{ cubic units}
\]

### Final Answer:
\[
V = \frac{584\pi}{15}
\]

Would you like more details, or do you have any questions?

Here are 5 related questions:
1. How would the integral change if the region was rotated around the y-axis?
2. What if the region was bounded by different curves, such as \( y = x^2 + 1 \) and \( y = 3x + 2 \)?
3. How would the problem be solved if we used cylindrical shells instead of washers?
4. How do we find the volume if the region is rotated around a line other than the x-axis or y-axis?
5. Can the washer method be used for regions bounded by more complex functions like trigonometric or exponential functions?

**Tip:** When solving problems involving volumes of revolution, carefully sketch the region and visualize the axis of rotation. This helps in setting up the correct integral.

Learn how to find the volume of a solid generated by revolving the region bounded by the curves y=x^2+3 and y=x+5 around the x-axis. Detailed step-by-step solution using the washer method and integration. Suitable for advanced high school students studying calculus.

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