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.