💡 Key Idea: Slice the solid into thin disks (like coins), find the volume of each disk, then integrate!

The Disk Method Formula

Rotating around the x-axis

When rotating $y = f(x)$ around the x-axis from $x = a$ to $x = b$:

$V = \int_a^b \pi [f(x)]^2\,dx = \pi \int_a^b [f(x)]^2\,dx$

Think: Volume = $\pi$ × (radius)² × thickness, summed up!

Where Does This Come From?

Volume of a Disk

A disk (thin cylinder) has:

• Radius: $r = f(x)$
• Thickness: $\Delta x$
• Volume: $\pi r^2 \cdot \Delta x = \pi [f(x)]^2 \Delta x$

Summing the Disks

$V \approx \sum_{i=1}^n \pi [f(x_i)]^2 \Delta x$

Taking the limit as $n \to \infty$:

$V = \lim_{n \to \infty} \sum_{i=1}^n \pi [f(x_i)]^2 \Delta x = \int_a^b \pi [f(x)]^2\,dx$

Example 1: Basic Disk Method

Find the volume when $y = \sqrt{x}$ from $x = 0$ to $x = 4$ is rotated around the x-axis.

Step 1: Set up the integral

$V = \pi \int_0^4 [\sqrt{x}]^2\,dx$

$= \pi \int_0^4 x\,dx$

Step 2: Integrate

$= \pi \left[\frac{x^2}{2}\right]_0^4$

$= \pi \left(\frac{16}{2} - 0\right)$

$= \pi \cdot 8 = 8\pi$

Answer: $8\pi$ cubic units

Example 2: Polynomial Function

Find the volume when $y = x^2$ from $x = 0$ to $x = 2$ is rotated around the x-axis.

Step 1: Set up the integral

$V = \pi \int_0^2 [x^2]^2\,dx$

$= \pi \int_0^2 x^4\,dx$

Step 2: Integrate

$= \pi \left[\frac{x^5}{5}\right]_0^2$

$= \pi \left(\frac{32}{5} - 0\right) = \frac{32\pi}{5}$

Answer: $\frac{32\pi}{5}$ cubic units

Rotating Around the y-axis

Formula for y-axis rotation

When rotating $x = g(y)$ around the y-axis from $y = c$ to $y = d$:

$V = \pi \int_c^d [g(y)]^2\,dy$

Key: Swap $x$ and $y$, integrate with respect to $y$!

Example 3: Rotation Around y-axis

Find the volume when $x = y^2$ from $y = 0$ to $y = 2$ is rotated around the y-axis.

Step 1: Set up the integral

$V = \pi \int_0^2 [y^2]^2\,dy$

$= \pi \int_0^2 y^4\,dy$

Step 2: Integrate

$= \pi \left[\frac{y^5}{5}\right]_0^2$

$= \pi \left(\frac{32}{5} - 0\right) = \frac{32\pi}{5}$

Answer: $\frac{32\pi}{5}$ cubic units

Rewriting Functions

Sometimes you need to solve for $x$ in terms of $y$ (or vice versa).

Example: Rotate $y = \sqrt{x}$ around the y-axis from $y = 0$ to $y = 2$.

Step 1: Rewrite as $x = g(y)$

$y = \sqrt{x}$ $y^2 = x$ $x = y^2$

Step 2: Set up and integrate

$V = \pi \int_0^2 [y^2]^2\,dy = \pi \int_0^2 y^4\,dy$

$= \pi \left[\frac{y^5}{5}\right]_0^2 = \frac{32\pi}{5}$

Special Solids

Sphere

Rotate $y = \sqrt{r^2 - x^2}$ (semicircle) around the x-axis from $x = -r$ to $x = r$:

$V = \pi \int_{-r}^r (r^2 - x^2)\,dx = \frac{4\pi r^3}{3}$

This is the famous sphere volume formula!

Cone

Rotate $y = \frac{h}{r}x$ (line) from $x = 0$ to $x = r$ around the x-axis:

$V = \pi \int_0^r \left(\frac{h}{r}x\right)^2\,dx = \frac{\pi h^2}{r^2} \int_0^r x^2\,dx$

$= \frac{\pi h^2}{r^2} \cdot \frac{r^3}{3} = \frac{\pi r h^2}{3}$

With base radius $r$ and height $h$, we get $V = \frac{1}{3}\pi r^2 h$ ✓

Example 4: Trigonometric Function

Find the volume when $y = \sin x$ from $x = 0$ to $x = \pi$ is rotated around the x-axis.

Step 1: Set up the integral

$V = \pi \int_0^{\pi} [\sin x]^2\,dx$

$= \pi \int_0^{\pi} \sin^2 x\,dx$

Step 2: Use trig identity

$\sin^2 x = \frac{1 - \cos 2x}{2}$

$V = \pi \int_0^{\pi} \frac{1 - \cos 2x}{2}\,dx$

$= \frac{\pi}{2} \int_0^{\pi} (1 - \cos 2x)\,dx$

Step 3: Integrate

$= \frac{\pi}{2} \left[x - \frac{\sin 2x}{2}\right]_0^{\pi}$

$= \frac{\pi}{2} \left[\left(\pi - \frac{\sin 2\pi}{2}\right) - \left(0 - \frac{\sin 0}{2}\right)\right]$

$= \frac{\pi}{2}[\pi - 0] = \frac{\pi^2}{2}$

Answer: $\frac{\pi^2}{2}$ cubic units

The Washer Method (Preview)

What if there's a hole in the middle?

Example: Rotate the region between $y = x$ and $y = x^2$ around the x-axis.

This creates a washer (disk with hole)!

Formula: $V = \pi \int_a^b [(R(x))^2 - (r(x))^2]\,dx$

where $R(x)$ is outer radius and $r(x)$ is inner radius.

We'll cover this in detail next!

⚠️ Common Mistakes

Mistake 1: Forgetting to Square

WRONG: $V = \pi \int_a^b f(x)\,dx$

RIGHT: $V = \pi \int_a^b [f(x)]^2\,dx$

The radius must be squared!

Mistake 2: Forgetting π

Volume of revolution always involves $\pi$!

$V = \pi \int_a^b [f(x)]^2\,dx$

Mistake 3: Wrong Variable

Rotating around x-axis: integrate with respect to $x$

Rotating around y-axis: integrate with respect to $y$

Match the axis to the variable!

Mistake 4: Using Diameter Instead of Radius

The disk method uses radius, not diameter!

Radius = $f(x)$ (distance from axis to curve)

Summary of Disk Method

Around x-axis

$V = \pi \int_a^b [f(x)]^2\,dx$

• Radius: $r = f(x)$
• Limits: $x = a$ to $x = b$

Around y-axis

$V = \pi \int_c^d [g(y)]^2\,dy$

• Radius: $r = g(y)$
• Limits: $y = c$ to $y = d$

Visualizing the Solid

Steps to visualize:

1. Sketch the 2D region
2. Identify the axis of rotation
3. Imagine spinning the region
4. See the 3D solid formed
5. Picture a cross-section (disk)

Practice this mental rotation - it helps tremendously!

📝 Practice Strategy

1. Sketch the region to be rotated
2. Identify the axis of rotation (x or y)
3. Find the radius function $r = f(x)$ or $r = g(y)$
4. Set up: $V = \pi \int [\text{radius}]^2\,d(\text{variable})$
5. Don't forget to square the radius!
6. Integrate and evaluate
7. Include $\pi$ in your final answer
8. Check units: volume should be cubic units

Here, $R(x) = \sqrt{x}$ (radius of disk at position $x$)

Bounds: $a = 0$ to $b = 4$

$V = \pi \int_0^4 (\sqrt{x})^2 \, dx$

$= \pi \int_0^4 x \, dx$

$= \pi \left[\frac{x^2}{2}\right]_0^4$

$= \pi \left(\frac{16}{2} - 0\right)$

$= 8\pi$ cubic units

Rotating this semicircle around the x-axis creates a sphere of radius 2!

Step 2: Set up the integral

$V = \pi \int_{-2}^2 [\sqrt{4-x^2}]^2\,dx$

$= \pi \int_{-2}^2 (4-x^2)\,dx$

Step 3: Integrate

$= \pi \left[4x - \frac{x^3}{3}\right]_{-2}^2$

At $x = 2$: $4(2) - \frac{8}{3} = 8 - \frac{8}{3} = \frac{16}{3}$

At $x = -2$: $4(-2) - \frac{-8}{3} = -8 + \frac{8}{3} = -\frac{16}{3}$

Step 4: Subtract

$V = \pi \left(\frac{16}{3} - \left(-\frac{16}{3}\right)\right)$

$= \pi \cdot \frac{32}{3} = \frac{32\pi}{3}$

Answer: $\frac{32\pi}{3}$ cubic units

Check: Sphere formula with $r = 2$: $V = \frac{4\pi(2)^3}{3} = \frac{32\pi}{3}$ ✓

$= \pi \int_0^3 4x^2 \, dx$

$= 4\pi \left[\frac{x^3}{3}\right]_0^3$

$= 4\pi \cdot \frac{27}{3}$

$= 4\pi \cdot 9 = 36\pi$ cubic units

$= \pi \int_0^1 e^{2x}\,dx$

Step 2: Integrate

For $\int e^{2x}\,dx$, use substitution or remember: $\int e^{ax}\,dx = \frac{e^{ax}}{a}$

$= \pi \left[\frac{e^{2x}}{2}\right]_0^1$

Step 3: Evaluate

$= \pi \left(\frac{e^2}{2} - \frac{e^0}{2}\right)$

$= \pi \left(\frac{e^2}{2} - \frac{1}{2}\right)$

$= \frac{\pi}{2}(e^2 - 1)$

Answer: $\frac{\pi(e^2 - 1)}{2}$ cubic units