1. Find the Boundaries: Solve $f(x) = g(x)$ to find $a$ and $b$.
2. Identify Radii: $R$ is the Top (furthest from axis), $r$ is the Bottom (closest to axis).
3. The Difference: Set up the integral as $\pi \int [R^2 - r^2] dx$.
4. Sign Rule: Volume must be positive! If it's negative, you flipped $R$ and $r$.
Find the volume of the hollow solid generated by rotating the region between $f(x)$ and $g(x)$ about the X-axis.
$f(x) = x$ and $g(x) = x^2$ from $x=0$ to $x=1$.
$f(x) = \sqrt{x}$ and $g(x) = 1$ from $x=1$ to $x=4$.
$f(x) = 4$ and $g(x) = 2$ from $x=0$ to $x=10$.
Note: This creates a Pipe!
In the "Pipe" problem, calculate the volume using simple geometry: $\pi R^2 h - \pi r^2 h$. Does the integral match? Why is the Washer method better when the walls of the pipe are curved (like a vase)?
_________________________________________________________________________
_________________________________________________________________________
Rotate the region between $y = e^x$ and $y = 1$ from $x=0$ to $x=1$ around the X-axis.
The Ring: $y = x$ and $y = x^2$ rotated around the Y-axis.
1. Rewrite: $x = y$ and $x = \sqrt{y}$.
2. Between $y=0$ and $y=1$, which is further from the Y-axis?
3. $R = \sqrt{y}$ and $r = y$.
4. Integrate $\pi \int (y - y^2) dy$.
Take the region $y = \sqrt{x}$ from $x=0$ to $x=4$.
Rotate it around the line **$y = -1$** (a line 1 unit below the X-axis).
Task: Find the new Outer Radius ($R = \sqrt{x} + 1$) and Inner Radius ($r = 1$). Calculate the Volume.
Objective: Explain the Washer Method to a younger sibling using two cups.
The Activity:
1. Take a large cup. Tell them it's the "Whole Life" ($R$).
2. Put a smaller cup inside it. Tell them it's the "Self" ($r$).
3. Pour water into the space between the cups.
4. "The water is the Spirit. To hold more water, do we need a bigger outer cup or a smaller inner cup?"
The Lesson: "Our life holds more of God when we make the 'Hole of Self' smaller."
Response: ___________________________________________________________