The foundation of all integration — chopping into rectangles
⚡ Quick Concept
A Riemann sum approximates area under a curve by slicing $[a,b]$ into $n$ equal strips of width $\Delta x = \dfrac{b-a}{n}$, then summing rectangle areas.
$$\sum_{i=1}^{n} f(x_i^*)\,\Delta x \;\longrightarrow\; \int_a^b f(x)\,dx \quad \text{as } n\to\infty$$
Three flavors: Left (use left endpoint), Right (use right endpoint), Midpoint (use midpoint).
🔑 Key Word
SLICE
WIDTH = Δx
HEIGHT = f(x*)
AREA = Δx · f(x*)
n → ∞
Q 01Riemann
For $f(x) = x^2$ on $[0, 2]$ with $n = 4$ equal subintervals,
what is $\Delta x$?
Memory: $\Delta x = \dfrac{b - a}{n}$ — always the first step!
Q 02Riemann
Using a Left Riemann Sum for $f(x) = x$ on $[0, 4]$ with $n = 4$,
what is the total approximated area?
Memory: LEFT → use left endpoint of each rectangle. Left endpoints: 0, 1, 2, 3.
Q 03Riemann
As $n \to \infty$, the Riemann sum $\displaystyle\sum_{i=1}^{n} f(x_i^*)\,\Delta x$ becomes exactly:
💿
Disk Method
Rotating around an axis — solid circles stacked up
⚡ Quick Concept
Rotate $y = f(x)$ around the x-axis. Each thin slice becomes a disk (solid circle). Stack them up!
$$V = \pi \int_a^b [f(x)]^2\,dx$$
Why $\pi r^2$? Each disk has radius $r = f(x)$, thickness $dx$, so volume $= \pi[f(x)]^2 dx$.
🔑 Key Words
DISK = SOLID circle
ONE function
r = f(x)
pi × r squared
Q 04Disk
The region bounded by $y = \sqrt{x}$, $x=0$, $x=4$ is rotated around the x-axis.
What is $[f(x)]^2$ that goes inside the integral?
Memory: Disk formula is $\pi\int [f(x)]^2 dx$. Square the function first!
Q 05Disk
Find the volume when $y = 2$, from $x = 0$ to $x = 5$, is rotated around the x-axis.
(Hint: this makes a cylinder!)
Memory: $V = \pi \int_0^5 (2)^2\,dx = \pi\int_0^5 4\,dx$. Cylinder = disk method with constant radius.
Q 06Disk
When rotating around the y-axis using the Disk method, the formula becomes:
🍩
Washer Method
Disk with a hole — like a donut cross-section
⚡ Quick Concept
Two functions, one region. Rotate around x-axis. Each slice is a washer (ring = big disk minus small disk).
$R$ = outer radius (farther from axis) · $r$ = inner radius (closer to axis)
Common mistake: Don't do $[R - r]^2$! It must be $R^2 - r^2$ separately!
🔑 Key Words
WASHER = RING
TWO functions
BIG² minus SMALL²
NOT (Big-Small)²
Q 07Washer
The region between $y = x^2$ (inner) and $y = x$ (outer) on $[0,1]$ is rotated around the x-axis.
Which setup is correct?
Q 08Washer
The region between $y = 2$ (outer) and $y = 1$ (inner) on $[0, 3]$ is rotated around the x-axis.
What is the volume? (Enter the number multiplied by pi, e.g. "9pi")
⚠️ Classic trap! The region between $y = x$ and $y = x^2$ is rotated around y = 2 (not the x-axis).
What is the outer radius $R(x)$?
Memory: When axis is $y = k$, radius = distance from curve to axis = $|k - f(x)|$. Outer = farther away.
Q 10Washer
Fill in the blank: In the washer method, the volume formula is
$V = \pi\displaystyle\int_a^b \bigl(R^2 \;\_\!\!\_\!\!\_\; r^2\bigr)\,dx$
What is the operator in the blank? (type: minus OR plus)
🥫
Shell Method
Cylindrical shells — like peeling layers of an onion
⚡ Quick Concept
Instead of slicing perpendicular to the axis, think of thin cylindrical shells wrapping around the axis.
$$V = 2\pi \int_a^b x \cdot f(x)\,dx \quad \text{(rotating around y-axis)}$$
Each shell: circumference $= 2\pi x$ · height $= f(x)$ · thickness $= dx$
Volume of one shell $\approx 2\pi x \cdot f(x)\cdot dx$
🔑 Key Words
2π × RADIUS × HEIGHT
radius = x
height = f(x)
NO squaring!
Q 11Shell
For the shell method rotating around the y-axis, which formula is correct?
Q 12Shell
Use the shell method for $f(x) = x$ on $[0, 2]$ rotated around the y-axis.
Set up: $V = 2\pi\displaystyle\int_0^2 x \cdot x\,dx = 2\pi\displaystyle\int_0^2 x^2\,dx$
What is the final volume? (Enter in form "Npi" or a fraction)
Memory: $\int_0^2 x^2 dx = \left[\dfrac{x^3}{3}\right]_0^2 = \dfrac{8}{3}$, then multiply by $2\pi$.
Q 13Shell
⚠️ Common confusion: When rotating around the x-axis using the shell method, what is the radius of each shell?
Memory: Radius = distance from axis. When axis = x-axis, the distance from a point $(x,y)$ to the x-axis is $y$.
Q 14Shell
Shell method: $f(x) = 4 - x^2$ on $[0, 2]$, rotated around y-axis.
Write the integrand (what goes inside $2\pi\int_0^2 \ldots\,dx$).
🎯
Mixed & Challenge
Choose the right method — the most important skill!
⚡ Method Selector — Super Quick
DISK → 1 function, rotate, slice ⊥ to axis → $\pi\int R^2$ WASHER → 2 functions, rotate, slice ⊥ to axis → $\pi\int(R^2-r^2)$ SHELL → integrate ∥ to axis (avoid messy inversions) → $2\pi\int r\cdot h$
🔑 Decision Rule
Perpendicular → DISK/WASHER
Parallel → SHELL
Hard inverse? → SHELL
Q 15Mixed
Region between $y = x^2$ and $y = 4$, rotated around x-axis.
Which method is the most straightforward to use?
Q 16Mixed
⚠️ $y = x^3$ on $[0,2]$ rotated around the y-axis.
Both methods work. The shell method gives $2\pi\int_0^2 x\cdot x^3\,dx$.
What is this integral equal to?
A sphere of radius $r$ can be found by rotating $y = \sqrt{r^2 - x^2}$ from $-r$ to $r$ around the x-axis using the disk method.
The classic result is $V = \dfrac{4}{3}\pi r^3$. For $r = 3$, what is the volume?
Q 18Mixed
⚠️ Most missed! The region bounded by $y = x$ and $y = x^2$ is rotated around the y-axis using the shell method.
Which integral is correct?
Q 19Mixed
Using the shell method for $y = \sin(x^2)$ on $[0,\sqrt{\pi}]$ rotated around the y-axis.
What is the exact volume? (This is why we use shells — the disk method would require an impossible inverse!)
Memory: $V = 2\pi\int_0^{\sqrt{\pi}} x\sin(x^2)\,dx$. Let $u = x^2$, then $du = 2x\,dx$, so $x\,dx = du/2$.
Q 20Mixed
🏆 Final Boss: Region bounded by $y = \sqrt{x}$ and $y = x^2$, rotated around $x = -1$.
Using the shell method, what is the correct radius for each shell?
Memory: Axis is $x = -1$ (to the left of the region). Radius = horizontal distance from shell at position $x$ to the axis $x=-1$ = $x - (-1) = x + 1$.