June 4, 2026 - Volumes of Solids of Revolution Lecture

Conceptual Fundamentals of Volume by Slicing

• The radius of a given circle cross-section within a solid of revolution is defined as the height from the axis of revolution to the function curve. For a rotation around the x-axis, the radius $r = f(x)$.

• The area of a cross-section is the area of a circle: $A = \pi \times r^2$. Substituting the function for the radius, the area becomes $A(x) = \pi \times [f(x)]^2$.

• To find the total volume of the resulting shape, we integrate the area of these infinitesimal slices over the interval $[a, b]$:   $V = \int_a^b \pi \times [f(x)]^2 \,dx$

Case Study: Volume of a Parabolic Solid

• The problem involves finding the volume of the solid generated by revolving the function $f(x) = x^2 - 4x + 5$ around the x-axis from $x = 1$ to $x = 4$.

• Visualization Techniques:

  • The graph of the function is parabolic and crosses the y-axis at $y = 5$.

  • The vertex of this parabola is located near $x = 2$.

  • To visualize the 3D solid, one can redraw the function reflected across the x-axis, then draw circles connecting the endpoints and intermediate points to provide dimension.

• Algebraic Expansion:

  • To integrate $[f(x)]^2$, the trinomial must be expanded: $(x^2 - 4x + 5)^2$.

  • The expansion process yields: $x^4 - 4x^3 + 5x^2 - 4x^3 + 16x^2 - 20x + 5x^2 - 20x + 25$.

  • Combining like terms results in the polynomial: $x^4 - 8x^3 + 26x^2 - 40x + 25$.

• Integration and Evaluation Steps:

  • The setup is $\pi \int_1^4 (x^4 - 8x^3 + 26x^2 - 40x + 25) \,dx$.

  • Taking the antiderivative term-by-term: $\pi \times [\frac{x^5}{5} - \frac{8x^4}{4} + \frac{26x^3}{3} - \frac{40x^2}{2} + 25x]$.

  • Simplified antiderivative: $\pi \times [\frac{x^5}{5} - 2x^4 + \frac{26x^3}{3} - 20x^2 + 25x]$ evaluated from 1 to 4.

• Numerical Calculation:

  • Evaluation at $x = 4$: $\frac{1024}{5} - 512 + \frac{1664}{3} - 320 + 100$.

  • Evaluation at $x = 1$: $\frac{1}{5} - 2 + \frac{26}{3} - 20 + 25$.

  • Detailed arithmetic mentioned during discussion involves finding common denominators (like 15). The intermediate calculation for the first bracket involves values like $732 \times 15 = 10,980$ and $11,392$.

  • Subtraction results mentioned: $412$ and $178$, leading to a simplified total of $234 \times \frac{\pi}{15}$, which further simplifies to $7.8\pi$ or $\frac{39\pi}{5}$, though the final result shared in class was $71.2\pi$ corrected via calculator.

The Disc Method and Riemann Sums

• The "Slicing Method" and the "Disc Method" are two names for the same process.

• The Disc Method is rooted in the Riemann sum concept where the area under a curve is approximated by rectangles. When these rectangles are spun around the axis, they form thin discs.

• Integrating the volume of these discs as their thickness approaches zero provides the exact volume of the solid.

Volume of Solids: Additional Functional Examples

• Square Root Function:

  • Function: $f(x) = \sqrt{x}$ on the interval $[1, 4]$.

  • Revolved around the x-axis, the shape resembles a bell or a football.

  • Area of slice: $A = \pi \times (\sqrt{x})^2 = \pi x$.

  • Integral: $\pi \int_1^4 x \,dx = \pi \times [\frac{x^2}{2}]_1^4$.

  • Result: $\pi \times (8 - \frac{1}{2}) = \frac{15\pi}{2} \, \text{units}^3$.

• Reciprocal Function:

  • Function: $f(x) = \frac{1}{x}$ on the interval $[-2, -1]$.

  • Spinning the left side of the graph results in the same shape as spinning the right side because volume is non-negative and the function is squared.

  • Setup: $\pi \int_{-2}^{-1} x^{-2} \,dx$.

  • Antiderivative: $\pi \times [-\frac{1}{x}]_{-2}^{-1}$.

  • Result: $\pi \times (1 - \frac{1}{2}) = \frac{\pi}{2} \, \text{units}^3$.

Revolving Around the Y-Axis

• When revolving around the y-axis, the slices are taken horizontally, and integration must be done with respect to y ($dy$).

• The radius is now the horizontal distance $x$. We must solve the original function for $x$.

• Example: $f(x) = 4 - x^2$ (a downward-opening parabola with vertex at 4).

  • Solve for $x$: $y = 4 - x^2 \rightarrow x^2 = 4 - y \rightarrow x = \sqrt{4 - y}$.

  • Area: $A(y) = \pi \times [\sqrt{4 - y}]^2 = \pi (4 - y)$.

  • Identifying y-bounds: At $x = 0$, $y = 4$. At $y = 0$, for the solid region, we integrate from $y = 0$ to $y = 4$.

  • Integral: $\pi \int_0^4 (4 - y) \,dy = \pi \times [4y - \frac{y^2}{2}]_0^4$.

  • Result: $\pi \times (16 - 8) = 8\pi \, \text{units}^3$.

The Washer Method

• The Washer Method is used when the solid has a hollow center, formed by revolving the region between two functions, $f(x)$ (upper bound) and $g(x)$ (lower bound).

• The cross-section is a "washer" (a flat ring). Its area is the area of the large circle minus the area of the small circle.

• Formula: $A = \pi \times ([f(x)]^2 - [g(x)]^2)$.

• Theorem: If $f(x)$ and $g(x)$ are continuous and non-negative, and $f(x) \geq g(x)$ on $[a, b]$, the volume of the region revolved around the x-axis is:   $V = \pi \int_a^b ([f(x)]^2 - [g(x)]^2) \,dx$

• Case Study: Square Root and Constant Line:

  • Functions: $f(x) = \sqrt{x}$ and $g(x) = 1$ on the interval $[1, 9]$.

  • Area function: $\pi \times ((\sqrt{x})^2 - (1)^2) = \pi (x - 1)$.

  • Integral: $\pi \int_1^9 (x - 1) \,dx = \pi \times [\frac{x^2}{2} - x]_1^9$.

  • Evaluation: $\pi \times [(\frac{81}{2} - 9) - (\frac{1}{2} - 1)] = \pi \times [(\frac{63}{2}) - (-\frac{1}{2})] = 32\pi \, \text{units}^3$.

Revolving Around Non-Axis Lines

• Solids can be revolved around lines other than $y = 0$ or $x = 0$.

• Example: Revolving the region between $y = 4 - x$, $y = x$, and the x-axis around the line $y = -2$.

  • To solve this, conceptualize shifting the entire coordinate system up by 2 units so that $y = -2$ becomes the x-axis.

  • The new function heights become $f(x) + 2$ and $g(x) + 2$.

  • If revolving around a line like $y = 10$, we would subtract the function from 10 to find the radius.

• Example Setup: $f(x) = 4 - x$ and the line of revolution is $y = -2$ from $x = 0$ to $x = 2$.

  • Radius is $f(x) + 2 = (4 - x) + 2 = 6 - x$.

  • Volume setup (assuming a solid disc, not a washer for this specific simple part): $\pi \int_0^2 (6 - x)^2 \,dx$.

Complex Applications and Discussion

• Trigonometric Solids: Revolving the region between $f(x) = \cos(x)$ and $g(x) = \sin(x)$ from $0$ to $\frac{\pi}{4}$. Using the washer method, the area is $\pi (\cos^2(x) - \sin^2(x))$, which simplifies via identity to $\pi \cos(2x)$. This is easily integrable to $\frac{1}{2}\pi \sin(2x)$.

• Units in Volume: All volume answers are expressed as "units cubed" ($\text{units}^3$) because they represent a three-dimensional quantity, regardless of whether a specific measurement (like inches) is provided.

• Procedural Difficulty: Students often find the setup of the integral and the visualization straightforward, whereas the algebraic evaluation, fraction manipulation, and definite integral subtraction are the most common sources of error.

Questions & Discussion

Q: What is the unit for these answers?

• A: It is "units cubed" ($\text{units}^3$). We are talking about volume, across three dimensions, so the result must be cubed.

Q: In the parabolic expansion, shouldn't the 25 have an x attached after integration?

• A: Yes, thanks for catching that. The term $25$ integrates to $25x$.

Q: In the washer calculation for the root function, wouldn't the 26 times 5 be 130 instead of 105?

• A: Yes, the second bracket calculations should be 133 and 17. The common denominator needs adjustment to 15.

Q: What was the most annoying part of this problem?

• A: The algebra following integration. Squaring a polynomial is just distribution, but keeping track of fractions and subtraction during definite integration is the real challenge.

Q: Is rotating around a line like y = x possible?

• A: Yes, but it involves the distance formula, which introduces square roots into the integral, making it significantly more complex. We focus on horizontal and vertical lines to keep distances measurable in simple x or y terms.

Attendance Taken: Jordan and Matthew are present.