The region bounded by the given curves is rotated about the specified axis. Find the volume of the resulting solid by any method.
x = (y − 7)2,   x = 4;   about y = 5

We are asked to find the volume of the solid obtained by rotating the region bounded by the curves \( x = (y - 7)^2 \) and \( x = 4 \) about the axis \( y = 5 \).

To solve this, we'll use the **shell method** since the axis of rotation is horizontal and the functions are written in terms of \( y \). Here's the process broken down:

### 1. Sketch the curves:
- \( x = (y - 7)^2 \) is a sideways parabola, opening to the right, with its vertex at \( (0, 7) \).
- \( x = 4 \) is a vertical line.
- The bounds for \( y \) are found by solving where the curves intersect. Set \( (y - 7)^2 = 4 \):
  \[
  (y - 7)^2 = 4 \implies y - 7 = \pm 2 \implies y = 9 \text{ or } y = 5.
  \]
So the region of interest is between \( y = 5 \) and \( y = 9 \).

### 2. Use the shell method formula:
The shell method for a region rotated around a horizontal axis \( y = c \) is given by:
\[
 V = 2\pi \int_{a}^{b} ( \text{radius} ) \cdot (\text{height}) \, dy
\]
Where:
- **Radius**: Distance from the axis of rotation \( y = 5 \). So the radius is \( |y - 5| \).
- **Height**: The horizontal distance between the two curves, which is \( 4 - (y - 7)^2 \).

### 3. Set up the integral:
The limits of integration are from \( y = 5 \) to \( y = 9 \). Thus, the volume integral becomes:
\[
 V = 2\pi \int_{5}^{9} (y - 5) \cdot \left[ 4 - (y - 7)^2 \right] \, dy
\]

### 4. Solve the integral:
First, expand the integrand:
\[
(y - 5) \left[ 4 - (y - 7)^2 \right] = (y - 5) \left[ 4 - (y^2 - 14y + 49) \right] = (y - 5) \left[ -y^2 + 14y - 45 \right]
\]
Now distribute:
\[
= -(y - 5)y^2 + 14(y - 5)y - 45(y - 5)
\]
\[
= -y^3 + 5y^2 + 14y^2 - 70y - 45y + 225
\]
\[
= -y^3 + 19y^2 - 115y + 225
\]

### 5. Integrate term by term:
\[
 V = 2\pi \int_{5}^{9} \left( -y^3 + 19y^2 - 115y + 225 \right) \, dy
\]
\[
= 2\pi \left[ \frac{-y^4}{4} + \frac{19y^3}{3} - \frac{115y^2}{2} + 225y \right]_{5}^{9}
\]

### 6. Evaluate the definite integral:
We need to compute the expression at \( y = 9 \) and \( y = 5 \), then subtract.

At \( y = 9 \):
\[
\frac{-(9)^4}{4} + \frac{19(9)^3}{3} - \frac{115(9)^2}{2} + 225(9) = \frac{-6561}{4} + \frac{19 \cdot 729}{3} - \frac{115 \cdot 81}{2} + 2025
\]
At \( y = 5 \):
\[
\frac{-(5)^4}{4} + \frac{19(5)^3}{3} - \frac{115(5)^2}{2} + 225(5) = \frac{-625}{4} + \frac{19 \cdot 125}{3} - \frac{115 \cdot 25}{2} + 1125
\]

Let me compute these values for you.The volume of the solid obtained by rotating the region bounded by the curves \( x = (y - 7)^2 \) and \( x = 4 \) about the axis \( y = 5 \) is approximately **134.04 cubic units**.

Would you like further clarification or details on any part of this solution?

Here are 5 related questions to expand on this topic:
1. How would the method change if the axis of rotation were vertical instead of horizontal?
2. How would you calculate the volume using the disk/washer method instead of the shell method?
3. What if the curves were different, such as \( x = (y - 6)^2 \) and \( x = 5 \)?
4. How would the solution change if the axis of rotation were \( y = 7 \) instead of \( y = 5 \)?
5. How do you determine whether to use the disk/washer method versus the shell method in general?

**Tip**: When choosing between the shell and disk/washer methods, consider the axis of rotation: shell method is often easier for vertical slices if the axis is horizontal, and disk/washer method is simpler for horizontal slices when the axis is vertical.

The region bounded by the given curves is rotated about the specified axis. Find the volume of the resulting solid by any method. x = (y − 7)^2, x = 4; about y = 5

Learn how to calculate the volume of a solid generated by rotating the region bounded by x = (y − 7)^2 and x = 4 around the axis y = 5. This problem covers the shell method and definite integrals. Suitable for undergraduate calculus students.

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