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The problem in the image asks to use the shell method to calculate the volume of the solid formed by revolving the region bounded by the curves \( x = 0 \), \( y = 0 \), and \( y = -y^2 + 6y + 7 \) around the **x-axis**, between \( y = 0 \) and \( y = 7 \).

### Step-by-Step Approach

We will use the shell method for this problem. The shell method formula for volume is given by:

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

Here:
- The **radius** is the distance from the x-axis, which is simply \( y \) for each shell.
- The **height** is the horizontal distance between the curves, which is \( x = 35 - y^2 \) (since the other boundary is \( x = 0 \)).

We revolve the region around the x-axis, so the limits of integration are from \( y = 0 \) to \( y = 7 \).

The volume integral becomes:

\[
V = 2\pi \int_{0}^{7} y (35 - y^2) \, dy
\]

### Solving the Integral

Let's expand and integrate:

\[
V = 2\pi \int_{0}^{7} (35y - y^3) \, dy
\]

Now, integrate term by term:

\[
\int 35y \, dy = \frac{35y^2}{2}
\]
\[
\int y^3 \, dy = \frac{y^4}{4}
\]

Now, substitute these into the integral:

\[
V = 2\pi \left[ \frac{35y^2}{2} - \frac{y^4}{4} \right]_{0}^{7}
\]

### Evaluating at the limits:

At \( y = 7 \):

\[
\frac{35(7)^2}{2} - \frac{(7)^4}{4} = \frac{35 \times 49}{2} - \frac{2401}{4}
= \frac{1715}{2} - \frac{2401}{4} = \frac{3430}{4} - \frac{2401}{4} = \frac{1029}{4}
\]

At \( y = 0 \):

\[
\frac{35(0)^2}{2} - \frac{(0)^4}{4} = 0
\]

Thus, the volume is:

\[
V = 2\pi \times \frac{1029}{4} = \frac{2058\pi}{4} = 514.5\pi
\]

### Final Answer:

The volume of the solid is:

\[
V = 514.5\pi \, \text{cubic units}
\]

---

Would you like further details or have any questions? Here are some related problems to expand your understanding:

1. How would the integral change if the region were revolved around the **y-axis**?
2. What if the region was bounded by a different curve, such as \( x = y^3 \)?
3. How would the volume change if we changed the limits of integration from 0 to 5 instead of 0 to 7?
4. Can we calculate the same volume using the disk method instead of the shell method?
5. What are the geometrical interpretations of the shell and disk methods?

**Tip:** When using the shell method, always check if the height of the shell is along the x-axis or y-axis, as this determines how to set up the integral.

Use the shell method to write and evaluate the definite integral that represents the volume of the solid formed by revolving the region between x=0, y=0, and y = -y^2 + 6y + 7 between 0 and 7 about the x-axis.

Learn how to use the shell method to find the volume of a solid formed by revolving the region bounded by y = -y^2 + 6y + 7, x = 0, and y = 0 between y = 0 and y = 7 about the x-axis. This detailed step-by-step solution covers integral calculus and the shell method formula. Suitable for college-level students in Calculus II.

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