June 4, 2026 - Calculus Notes: Volumes of Solids of Revolution (Concise)

Conceptual Foundation: The volume of a solid of revolution is calculated by integrating the area of circular cross-sections (discs) from $a$ to $b$ .
Radius ( $R$ ): The height from the x-axis to the function $f(x)$ .
Area ( $A$ ): The area of a cross-section is $\pi \times [f(x)]^2$ .
Volume Formula: $\text{Volume} = \int_a^b \pi [f(x)]^2 \,dx$
Terminology: This technique is variously referred to as the "slicing method" or the "disc method," derived from the Riemann sum approach involving spinning approximated rectangles.

Quadratic Function: For $f(x) = x^2 - 4x + 5$ from $x = 1$ to $x = 4$ :
- Squaring the polynomial results in: $x^4 - 8x^3 + 26x^2 - 40x + 25$ .
- Integrated result after evaluation at bounds: $\frac{234 \pi}{15}$ , which simplifies to approximately $38$ .
Square Root Function: For $f(x) = \sqrt{x}$ from $1$ to $4$ :
- The integral becomes $\pi \int_1^4 x \,dx$ .
- Result: $\frac{15 \pi}{2} \, \text{units}^3$ .
Rational Function: For $f(x) = 1/x$ over the interval $[-2, -1]$ .

Methodology: Slice horizontally rather than vertically. This requires solving the function for $x$ in terms of $y$ (horizontal distance).
Area Formula: $A = \pi [g(y)]^2$ .
Example: For $y = 4 - x^2$ :
- Solve for $x$ to get $x = \sqrt{4 - y}$ .
- Determine bounds relative to the y-axis (e.g., $y = 0$ to $y = 4$ ).
- Volume integral: $\int_0^4 \pi [\sqrt{4 - y}]^2 \,dy = \int_0^4 \pi (4 - y) \,dy$ .

Definition: Used when the region between two functions, $f(x)$ and $g(x)$ , is revolved around an axis, creating a hollow center.
Cross-section: A washer (a large circle minus a small circle).
Area Formula: $A = \pi ([f(x)]^2 - [g(x)]^2)$
Requirement: One function must be consistently greater than the other ( $f(x) \geq g(x)$ ) over the interval.
Example Integration: For the area between $f(x) = x$ and $g(x) = 1/x$ from $1$ to $4$ , the volume is $\pi \int_1^4 (x^2 - \frac{1}{x^2}) \,dx$ .

Translation Principle: To revolve around a line like $y = -2$ , shift the entire system up by $2$ units so the axis of revolution effectively becomes the x-axis ( $y = 0$ ).
Adjusting the Radius:
- If revolving around $y = -2$ , the new radius becomes $R = f(x) + 2$ .
- Example: For a line where the distance to the axis is increased by $2$ , the integral uses $(f(x) + 2)^2$ .
General Rule: If revolving around $y = k$ , translations are applied to move the axis to the standard x-axis position.

Question: Should the units be cubed?
Answer: Yes, since we are measuring volume, the answer should be in $\text{units}^3$ .
Question: What was the most difficult part of the polynomial problem?
Answer: The algebra after integration (dealing with fractions and common denominators), rather than the calculus itself.
Question: Can you revolve around a slanted line like $y = x$ ?
Answer: Theoretically yes, but it is complex because measuring distance involves square roots and coordinate transformations, unlike the simple horizontal or vertical distances used with the axes.
Discussion on Trigonometric Regions: For functions like $\sin(x)$ and $\cos(x)$ between $0$ and $\pi$ , one must find intersection points (like $\pi / 4$ ) and split the integral based on which function is on top.