Volume of Solids of Revolution Detailed Study Guide

Introduction to Solids of Revolution

• Conceptual Overview: Determining the volume of a solid involves a challenge similar to finding the area of a region under a curve. While there is an intuitive understanding of volume, calculus provides the necessary precision to define it exactly.
• Solid of Revolution Definition: A solid of revolution is a three-dimensional object obtained by rotating a two-dimensional region in a plane around a specific axis.
• Common Examples:
  • Sphere: Generated by rotating a semicircular region about its diameter.
  • Right Circular Cone: Generated by rotating a right triangle about one of its legs.
• Visualizing the Revolution: If a region $R$ under the graph of a continuous function $f$ on the interval $[a, b]$ is revolved around the $x$-axis, the resulting solid $S$ features vertical cross-sections that are all circles.

The Fundamental Definition of Volume

• Slicing and Cross-Sectional Area: Let $S$ be a solid lying between $x = a$ and $x = b$. If the cross-sectional area of $S$ in a plane $P$, passing through $x$ and perpendicular to the $x$-axis, is given by the continuous function $A(x)$, then the volume of the solid is defined as the integral of this area.
• Formula:

$V = \lim_{n \rightarrow \infty} \sum_{i=1}^{n} A(x_i) \Delta x = \int_a^b A(x) \, dx$

• Consistency with Known Formulas: For a cylinder, the cross-sectional area is constant ($A(x) = A$). Therefore, the definition yields $V = \int_a^b A \, dx = A(b - a)$, which matches the standard geometric formula $V = A \times h$.

The Disk Method

• Definition: The disk method is used when the region is rotated around an axis, and the cross-section perpendicular to that axis is a solid circle (a disk).
• Mathematical Context: Let $f$ be a continuous function such that $f(x) \ge 0$ for all $x$ in $[a, b]$. If the region bounded by $y = f(x)$, the $x$-axis, and the vertical lines $x = a$ and $x = b$ is revolved about the $x$-axis, the volume $V$ is:

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

• Key Detail: In this method, the radius of the circular cross-section at any point $x$ is $R = f(x)$. Thus, the area of the vertical cross-section is $A(x) = \pi R^2 = \pi [f(x)]^2$.

• Example 1: Rotation Bounded by $y = x^2$:

  • Scenario: Rotate the region bounded by $y = x^2$, the $x$-axis, and the line $x = 2$ about the $x$-axis.
  • Formula for Area: $A(x) = \pi [f(x)]^2 = \pi (x^2)^2 = \pi x^4$.
  • Limits: From $x = 0$ to $x = 2$.
  • Integration:

$V = \pi \int_0^2 x^4 \, dx = \pi \left[ \frac{1}{5} x^5 \right]_0^2 = \frac{32}{5} \pi \, \text{cu. units}$

• Example 2: Rotation Bounded by $x = 2/y$:
  • Scenario: Find the volume generated by revolving the region between the $y$-axis and the curve $x = \frac{2}{y}$ for $1 \le y \le 4$ about the $y$-axis.
  • Formula for Area: $A(y) = \pi [f(y)]^2 = \pi \left( \frac{2}{y} \right)^2 = \frac{4\pi}{y^2}$.
  • Limits: From $y = 1$ to $y = 4$.
  • Integration:

$V = \pi \int_1^4 \frac{4}{y^2} \, dy = 4\pi \int_1^4 y^{-2} \, dy = 4\pi \left[ -\frac{1}{y} \right]_1^4$

$V = 4\pi \left( -\frac{1}{4} - (-1) \right) = 4\pi \left( \frac{3}{4} \right) = 3\pi \, \text{cu. units}$

The Washer Method

• Definition: When the region being revolved is between two curves, the cross-sections perpendicular to the axis of revolution are annular rings, known as washers.
• Mathematical Context: Let $f(x)$ and $g(x)$ be continuous on $[a, b]$ with $f(x) \ge g(x) \ge 0$. When the region between these curves is rotated about the $x$-axis, the volume is defined by the outer radius $R_{\text{outer}}$ and the inner radius $R_{\text{inner}}$.
  • $R_{\text{outer}} = f(x)$
  • $R_{\text{inner}} = g(x)$
• Formula:

$V = \pi \int_a^b ([R_{\text{outer}}]^2 - [R_{\text{inner}}]^2) \, dx = \pi \int_a^b ([f(x)]^2 - [g(x)]^2) \, dx$

• Example 3a: Basic Washer Strategy:
  • Problem: Find the volume for the region bounded by $y = x^2 + 4$, $y = 2$, $x = 1$, and $x = 3$ about the $x$-axis.
  • Setup: $R_{\text{outer}} = x^2 + 4$; $R_{\text{inner}} = 2$.
  • Integration:

$V = \pi \int_1^3 ([x^2 + 4]^2 - [2]^2) \, dx = \pi \int_1^3 (x^4 + 8x^2 + 16 - 4) \, dx$

$V = \pi \int_1^3 (x^4 + 8x^2 + 12) \, dx = \pi \left[ \frac{x^5}{5} + \frac{8x^3}{3} + 12x \right]_1^3 = \frac{2126}{15} \pi \, \text{cu. units}$

• Example 3b: Rotation about Horizontal Line (Shifted Axis):
  • Problem: Rotate the region between $f(x) = x^2 + 2$ and $g(x) = 4 - x^2$ about the line $y = -3$.
  • Intersection: Set $x^2 + 2 = 4 - x^2 \Rightarrow 2x^2 = 2 \Rightarrow x = \pm 1$.
  • Adjusting Radii: Since the rotation is about $y = -3$, the radii are measured from the line. From the graph, $g(x) \ge f(x)$ on $[-1, 1]$.
    • $R_{\text{outer}} = [g(x) - (-3)] = (4 - x^2) + 3 = 7 - x^2$
    • $R_{\text{inner}} = [f(x) - (-3)] = (x^2 + 2) + 3 = x^2 + 5$
  • Integration:

$V = \pi \int_{-1}^1 ([7 - x^2]^2 - [x^2 + 5]^2) \, dx = \pi \int_{-1}^1 ([49 - 14x^2 + x^4] - [x^4 + 10x^2 + 25]) \, dx$

$V = \pi \int_{-1}^1 (24 - 24x^2) \, dx = \pi \left[ 24x - \frac{24x^3}{3} \right]_{-1}^1$

$V = \pi [ (24 - 8) - (-24 + 8) ] = \pi [ 16 - (-16) ] = 32\pi \, \text{cu. units}$

The Cylindrical Shell Method

• Concept: This method approximates the volume of a solid of revolution by summing the volumes of thin, concentric cylindrical shells (tubes).
• Derivation of Shell Volume:
  • Volume of shell = Volume of outer cylinder - Volume of inner cylinder.
  • Let $r_2$ be the outer radius, $r_1$ the inner radius, and $h$ the height.

$V = \pi r_2^2 h - \pi r_1^2 h = \pi h (r_2^2 - r_1^2) = \pi h (r_2 - r_1)(r_2 + r_1)$

*   Expressed using average radius $r = \frac{r_2 + r_1}{2}$ and thickness $\Delta r = r_2 - r_1$:

$\Delta V = 2\pi (\text{average radius}) (\text{height}) (\text{thickness}) = 2\pi rh \, \Delta r$

• Definition (Rotation about Y-axis): Let $f$ be continuous on $[a, b]$ where $a \ge 0$. If the region $R$ bounded by $y = f(x)$, the $x$-axis, and lines $x=a, b$ is revolved about the $y$-axis:

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

• Definition (Rotation about X-axis): When a region in the first quadrant bounded by $y=c, d$, $x=0$, and $x=g(y)$ is revolved about the $x$-axis:

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

• Example 5a: Rotation about Y-axis:
  • Problem: Region under $y = -x^3 + 3x^2$ on $[0, 3]$ revolved about the $y$-axis.
  • Components: Radius = $x$; Height = $-x^3 + 3x^2$.

$\Delta V = 2\pi x (-x^3 + 3x^2) \Delta x = 2\pi (-x^4 + 3x^3) \Delta x$

$V = \int_0^3 2\pi (-x^4 + 3x^3) \, dx = 2\pi \left[ -\frac{x^5}{5} + \frac{3x^4}{4} \right]_0^3$

$V = 2\pi \left( -\frac{243}{5} + \frac{243}{4} \right) = 2\pi \left( \frac{243}{20} \right) = \frac{243}{10} \pi \, \text{cu. units}$

• Example 5b: Rotation about X-axis:
  • Problem: Region bounded by $x = -y^2 + 6y$ and $x = 0$ revolved about the $x$-axis.
  • Components: Radius = $y$; Height = $x = -y^2 + 6y$.
  • Limits: Find roots of $x$: $-y^2 + 6y = 0 \Rightarrow -y(y - 6) = 0 \Rightarrow y = 0, y = 6$.

$\Delta V = 2\pi y (-y^2 + 6y) \Delta y = 2\pi (-y^3 + 6y^2) \Delta y$

$V = \int_0^6 2\pi (-y^3 + 6y^2) \, dy = 2\pi \left[ -\frac{y^4}{4} + \frac{6y^3}{3} \right]_0^6$

$V = 2\pi \left[ -\frac{1296}{4} + 2(216) \right] = 2\pi [-324 + 432] = 2\pi(108) = 216\pi \, \text{cu. units}$

Application Exercise Problems

1. Find the volume of the solid generated by revolving the region bounded by the $x$-axis and one arch of the curve $y = \sin(x)$ about the $x$-axis.
2. The line $y = x + 2$ and the parabola $y = x^2$ contain a bounded region $R$. Find the volume $V$ generated by revolving $R$ about the $x$-axis.
3. Find the volume of the solid obtained by rotating the region bounded by $y = x^2$ and $x = 2y$ about the $y$-axis.
4. Find the volume of the solid generated by revolving the region bounded by $y = \sqrt{\sin(x)}$, $y = 0$, and $x \le \frac{\pi}{2}$ about the $x$-axis.