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This problem involves finding the volume of a solid generated by revolving a region bounded by the given curves around the vertical line \(x = 8\).

### Step 1: Understand the Problem
The curves are:
- \(x = y^2 + 3\) (a parabola)
- \(x = 7\) (a vertical line)

We are rotating the region between these curves about \(x = 8\). To solve this, we will use the **method of cylindrical shells**.

### Step 2: Set up the Volume Integral Using Cylindrical Shells
Using cylindrical shells, the formula for the volume is:

\[
V = 2\pi \int_{a}^{b} (r)(h) \, dy
\]

Where:
- \(r\) is the radius of each shell (distance from the axis of rotation, \(x = 8\))
- \(h\) is the height of the shell (difference between the bounding functions)

#### Radius \(r\):
Since we are revolving around \(x = 8\), the radius is the horizontal distance from \(x = 8\) to the curve \(x = y^2 + 3\):

\[
r = 8 - (y^2 + 3) = 5 - y^2
\]

#### Height \(h\):
The height is the horizontal distance between the line \(x = 7\) and the curve \(x = y^2 + 3\):

\[
h = 7 - (y^2 + 3) = 4 - y^2
\]

### Step 3: Determine the Bounds
The curves intersect where \(x = y^2 + 3 = 7\), solving for \(y\):

\[
y^2 + 3 = 7 \implies y^2 = 4 \implies y = \pm 2
\]

So, the bounds of integration are from \(y = -2\) to \(y = 2\).

### Step 4: Set up the Integral
Now we set up the volume integral:

\[
V = 2\pi \int_{-2}^{2} (5 - y^2)(4 - y^2) \, dy
\]

### Step 5: Simplify and Compute
We will now expand the integrand:

\[
(5 - y^2)(4 - y^2) = 20 - 5y^2 - 4y^2 + y^4 = 20 - 9y^2 + y^4
\]

Thus, the integral becomes:

\[
V = 2\pi \int_{-2}^{2} \left( 20 - 9y^2 + y^4 \right) \, dy
\]

Since the integrand is an even function (only even powers of \(y\)), we can double the integral from 0 to 2:

\[
V = 4\pi \int_{0}^{2} \left( 20 - 9y^2 + y^4 \right) \, dy
\]

Now, integrate term by term:

\[
\int 20 \, dy = 20y
\]
\[
\int -9y^2 \, dy = -3y^3
\]
\[
\int y^4 \, dy = \frac{y^5}{5}
\]

Evaluating from 0 to 2:

\[
V = 4\pi \left[ 20(2) - 3(2^3) + \frac{2^5}{5} \right]
= 4\pi \left[ 40 - 24 + \frac{32}{5} \right]
= 4\pi \left[ 16 + \frac{32}{5} \right]
= 4\pi \left[ \frac{80}{5} + \frac{32}{5} \right]
= 4\pi \left[ \frac{112}{5} \right]
\]

Finally:

\[
V = \frac{448\pi}{5}
\]

### Conclusion:
The volume of the solid generated by revolving the region is \(\frac{448\pi}{5}\) cubic units.

Would you like more details or have any questions? Here are some related questions:

1. How does the method of cylindrical shells compare to the washer method?
2. Why is the integrand considered an even function in this problem?
3. What changes if the axis of rotation is shifted horizontally or vertically?
4. How would the problem change if you rotated around a different line, such as \(x = 7\)?
5. Can the volume be solved using the washer or disk method instead?

**Tip**: Always double-check whether the function and axis of rotation form an appropriate cylindrical shell or disk before applying any method!

Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the line x = 8. x = y^2 + 3, x = 7

Learn how to find the volume of a solid generated by revolving a region bounded by the graphs of the equations x = y^2 + 3 and x = 7 around the line x = 8 using the method of cylindrical shells. This problem involves setting up and solving integrals for volume in calculus.

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