Mathematics / Analysis Integral Calculus & Series 100% Free Open Access
Chapter 3 โ€ข Theory & Derivations

Geometric Applications of Integration in Cartesian Coordinates

ยง3.1 Plane Areas Between Intersecting Curves: Vertical vs Horizontal Slicing

1. General Area Formulation as an Integral

Let $f, g: [a, b] \to \mathbb{R}$ be two continuous functions on the closed interval $[a, b]$ such that $f(x) \ge g(x)$ for all $x \in [a, b]$. The area $A$ of the planar region bounded above by $y = f(x)$, below by $y = g(x)$, and laterally by vertical lines $x = a$ and $x = b$ is the Riemann integral of the differential height element:

$$dA = \Big( f(x) - g(x) \Big) \, dx \implies \mathbf{A = \int_a^b \Big( f(x) - g(x) \Big) \, dx}$$

2. Signed Area vs Total Geometric Area

If the curves intersect multiple times within $[a, b]$ at roots $c_1, c_2, \dots \in (a, b)$, the relative ordering $f(x) \ge g(x)$ flips. The total geometric area is given by the integral of the absolute difference:

$$A_{\text{total}} = \int_a^b |f(x) - g(x)| \, dx = \sum_{k} \int_{c_k}^{c_{k+1}} \Big| f(x) - g(x) \Big| \, dx$$

3. Horizontal Slicing ($dy$ Integration)

When the boundary curves are more naturally parameterized as functions of $y$, say $x = w_R(y)$ (right boundary) and $x = w_L(y)$ (left boundary) for $y \in [c, d]$ with $w_R(y) \ge w_L(y)$, horizontal slicing avoids piecewise decomposition:

$$dA = \Big( w_R(y) - w_L(y) \Big) \, dy \implies \mathbf{A = \int_c^d \Big( w_R(y) - w_L(y) \Big) \, dy}$$

ยง3.2 Volumes of Revolution: The Disk and Washer Methods

1. Rotational Symmetry & The Differential Disk Element

Let a region bounded by $y = f(x) \ge 0$, the $x$-axis, and lines $x = a, x = b$ be rotated through $2\pi$ radians about the $x$-axis. Slicing perpendicular to the axis of rotation produces circular cross-sectional disks of radius $R(x) = f(x)$ and thickness $dx$.

$$dV = A(x) \, dx = \pi [R(x)]^2 \, dx \implies \mathbf{V = \pi \int_a^b [f(x)]^2 \, dx}$$

2. The Washer Method for Annular Cross-Sections

When the planar region is bounded between an outer curve $y = f(x)$ and an inner curve $y = g(x)$ ($0 \le g(x) \le f(x)$), rotation generates an annular washer with outer radius $R(x) = f(x)$ and inner hole radius $r(x) = g(x)$:

$$dV = \Big( \pi R(x)^2 - \pi r(x)^2 \Big) dx = \pi \Big( [f(x)]^2 - [g(x)]^2 \Big) dx$$
$$\mathbf{V = \pi \int_a^b \Big( [f(x)]^2 - [g(x)]^2 \Big) \, dx}$$

3. Rotation About Non-Origin Axes

If the rotation axis is shifted to the horizontal line $y = k$:

$$R(x) = |f(x) - k|, \qquad r(x) = |g(x) - k| \implies V = \pi \int_a^b \Big( [R(x)]^2 - [r(x)]^2 \Big) \, dx$$

Similarly, for rotation about a vertical line $x = h$ using horizontal slices perpendicular to the axis of rotation:

$$V = \pi \int_c^d \Big( [R(y)]^2 - [r(y)]^2 \Big) \, dy$$

ยง3.3 Volumes of Revolution: The Method of Cylindrical Shells

1. When Washer Integration is Pathological

When solving for $x$ in terms of $y$ is algebraically intractable (e.g., $y = 3x^2 - 2x^5$) or requires decomposing the domain into difficult sub-regions, revolving about a vertical axis is vastly superior using cylindrical shells parallel to the axis of revolution.

2. Unrolling the Differential Cylindrical Shell

Consider revolving a thin vertical strip of width $dx$ at distance $x$ from the vertical axis of rotation $x = 0$, having height $h(x) = f(x) - g(x)$. When revolved through $2\pi$, this strip traces a thin hollow cylindrical shell of radius $r = x$, height $h = f(x) - g(x)$, and wall thickness $dx$.

Unrolling this shell into a flat rectangular prism yields the volume element:

$$dV = (\text{Circumference}) \times (\text{Height}) \times (\text{Thickness}) = 2\pi x \cdot h(x) \cdot dx$$
$$\mathbf{V = 2\pi \int_a^b x \Big( f(x) - g(x) \Big) \, dx}$$

3. General Axis of Rotation ($x = h$)

If the axis of rotation is the vertical line $x = h$:

$$\text{Radius } r(x) = |x - h| \implies \mathbf{V = 2\pi \int_a^b |x - h| \Big( f(x) - g(x) \Big) \, dx}$$

For horizontal axis rotation revolving about $y = k$ with horizontal strips of thickness $dy$:

$$\mathbf{V = 2\pi \int_c^d |y - k| \Big( w_R(y) - w_L(y) \Big) \, dy}$$

ยง3.4 Volumes by Slicing with General Cross-Sections & Cavalieri's Principle

1. Cavalieri's Principle

If two three-dimensional solids have the same height and equal cross-sectional areas at every cutting plane parallel to their bases, then their total volumes are identical.

2. General Volume by Parallel Slicing

Let a solid extend along the $x$-axis from $x = a$ to $x = b$. If the area of the cross-section perpendicular to the $x$-axis at point $x$ is given by a continuous function $A(x)$, the differential slab volume is $dV = A(x) \, dx$:

$$\mathbf{V = \int_a^b A(x) \, dx}$$

3. Common Cross-Sectional Geometry over a Planar Base

Suppose the base of a solid is bounded between $y = f(x)$ and $y = g(x)$, so the cross-sectional base length is $s(x) = f(x) - g(x)$. Standard geometric cross-sections perpendicular to the $x$-axis have areas:

  • Squares: $A(x) = [s(x)]^2$
  • Equilateral Triangles: $A(x) = \frac{\sqrt{3}}{4} [s(x)]^2$
  • Semicircles with diameter $s(x)$: $A(x) = \frac{\pi}{8} [s(x)]^2$
  • Isosceles Right Triangles (hypotenuse on base): $A(x) = \frac{1}{4} [s(x)]^2$

ยง3.5 Arc Length of Smooth Curves, Surface Areas of Revolution & Pappus's Theorems

1. Differential Arc Length Element $ds$

By the Pythagorean theorem applied to an infinitesimal segment on a smooth curve $y = f(x) \in C^1([a, b])$:

$$(ds)^2 = (dx)^2 + (dy)^2 = (dx)^2 \left( 1 + \left(\frac{dy}{dx}\right)^2 \right) \implies ds = \sqrt{1 + [f'(x)]^2} \, dx$$

The total Arc Length of the curve from $x = a$ to $x = b$ is:

$$\mathbf{L = \int_a^b \sqrt{1 + [f'(x)]^2} \, dx}$$

For a parametric curve $(x(t), y(t))$ for $t \in [t_0, t_1]$, $ds = \sqrt{\dot{x}(t)^2 + \dot{y}(t)^2} \, dt$, so $L = \int_{t_0}^{t_1} \sqrt{\dot{x}^2 + \dot{y}^2} \, dt$.

2. Area of a Surface of Revolution

Revolving the differential arc element $ds$ about an axis sweeps a frustum of a cone of surface area $dS = 2\pi r \, ds$, where $r$ is the perpendicular distance from the curve to the axis of rotation:

  • Rotation About the $x$-axis ($r = y = f(x)$): $$\mathbf{S = 2\pi \int_a^b f(x) \sqrt{1 + [f'(x)]^2} \, dx}$$
  • Rotation About the $y$-axis ($r = x$): $$\mathbf{S = 2\pi \int_a^b x \sqrt{1 + [f'(x)]^2} \, dx}$$

3. The Centroid Theorems of Pappus

Let a planar curve of length $L$ and centroid distance $\bar{y}$ from an axis in its plane (not crossing the curve) be revolved through $2\pi$:

$$\mathbf{S = 2\pi \bar{y} L}$$

Let a planar region of area $A$ and centroid distance $\bar{y}$ be revolved through $2\pi$:

$$\mathbf{V = 2\pi \bar{y} A}$$
TIERED UNIVERSITY HONORS PROBLEMS

Step-by-Step Solved Examination Problems

Comprehensive analytical derivations, multi-tier solutions (Foundational, Intermediate Exam, and Honors/Proof Challenge) with complete line-by-line verification.

Tier 1: Foundational Example 3.1: Cylindrical Shells vs Washer Method Verification

Find the volume of the solid generated by revolving the region bounded by $y = \sqrt{x}$, the $x$-axis ($y = 0$), and the line $x = 4$ about the line $x = 6$ using the method of cylindrical shells.

Step 1: Identify Shell Geometry

The region is bounded for $x \in [0, 4]$. A vertical strip at position $x$ has:

  • Height: $h(x) = \sqrt{x} - 0 = \sqrt{x}$
  • Radius of rotation: The distance from $x$ to the rotation line $x = 6$ is $r(x) = 6 - x$
  • Thickness: $dx$

Step 2: Formulate the Shell Volume Integral

$$V = 2\pi \int_0^4 r(x) h(x) \, dx = 2\pi \int_0^4 (6 - x)\sqrt{x} \, dx = 2\pi \int_0^4 \Big( 6x^{1/2} - x^{3/2} \Big) dx$$

Step 3: Evaluate the Antiderivative

$$\begin{aligned} V &= 2\pi \left[ 6 \cdot \frac{2}{3} x^{3/2} - \frac{2}{5} x^{5/2} \right]_0^4 = 2\pi \left[ 4 x^{3/2} - \frac{2}{5} x^{5/2} \right]_0^4 \ &= 2\pi \left( 4(4^{3/2}) - \frac{2}{5}(4^{5/2}) \right) = 2\pi \left( 4(8) - \frac{2}{5}(32) \right) \ &= 2\pi \left( 32 - \frac{64}{5} \right) = 2\pi \left( \frac{160 - 64}{5} \right) = 2\pi \left( \frac{96}{5} \right) = \frac{192\pi}{5} \end{aligned}$$
Final Answer & Physical Insight

$V = \frac{192\pi}{5}$

Tier 2: Intermediate Exam Example 3.2: Arc Length and Surface Area of the Catenary

Consider the catenary curve $y = a \cosh(x/a)$ on the symmetric interval $[-a, a]$. (a) Compute the exact arc length $L$ of this catenary arc. (b) Compute the surface area $S$ generated by revolving this catenary about the $x$-axis.

Part (a): Compute the Arc Length $L$

Differentiating $y = a\cosh(x/a)$ with respect to $x$:

$$y' = a \cdot \frac{1}{a} \sinh(x/a) = \sinh(x/a)$$

Using the fundamental hyperbolic identity $1 + \sinh^2(u) = \cosh^2(u)$:

$$ds = \sqrt{1 + (y')^2} \, dx = \sqrt{1 + \sinh^2(x/a)} \, dx = \cosh(x/a) \, dx$$

Integrating from $-a$ to $a$ (using symmetry about $x = 0$):

$$L = \int_{-a}^a \cosh(x/a) \, dx = 2 \int_0^a \cosh(x/a) \, dx = 2 \Big[ a \sinh(x/a) \Big]_0^a = 2a \sinh(1)$$

Part (b): Surface Area of Revolution about $x$-axis

The differential surface area element is $dS = 2\pi y \, ds$:

$$dS = 2\pi \Big( a \cosh(x/a) \Big) \Big( \cosh(x/a) \, dx \Big) = 2\pi a \cosh^2(x/a) \, dx$$

Applying the hyperbolic power-reduction identity $\cosh^2(u) = \frac{\cosh(2u) + 1}{2}$:

$$\begin{aligned} S &= 2\pi a \int_{-a}^a \left( \frac{\cosh(2x/a) + 1}{2} \right) dx = 2\pi a \int_0^a \Big( \cosh(2x/a) + 1 \Big) dx \ &= 2\pi a \left[ \frac{a}{2}\sinh(2x/a) + x \right]_0^a = 2\pi a \left( \frac{a}{2}\sinh(2) + a \right) = \pi a^2 \Big( \sinh(2) + 2 \Big) \end{aligned}$$
Final Answer & Physical Insight

$L = 2a\sinh(1)$, \quad $S = \pi a^2 (\sinh(2) + 2)$

Tier 3: Honors / Proof Challenge Example 3.3: Gabriel's Horn (Torricelli's Trumpet) Paradox

Consider the curve $y = \frac{1}{x}$ for $x \in [1, \infty)$ revolved about the $x$-axis. (a) Prove that the resulting solid has finite volume $V = \pi$. (b) Prove that its surface area $S$ is strictly infinite ($S \to \infty$). (c) Resolve the physical painter's paradox: How can a horn hold a finite volume of paint ($\pi$ units) yet require an infinite amount of paint to coat its inner surface?

Part (a): Volume Evaluation

Using the disk method with $R(x) = \frac{1}{x}$ on $[1, \infty)$:

$$V = \pi \int_1^\infty [R(x)]^2 \, dx = \pi \lim_{M \to \infty} \int_1^M \frac{1}{x^2} \, dx = \pi \lim_{M \to \infty} \left[ -\frac{1}{x} \right]_1^M = \pi \lim_{M \to \infty} \left( 1 - \frac{1}{M} \right) = \mathbf{\pi}$$

The volume is strictly finite and equal to $\pi$.

Part (b): Surface Area Divergence Proof

The derivative is $y' = -\frac{1}{x^2}$. The surface area is:

$$S = 2\pi \int_1^\infty y \sqrt{1 + (y')^2} \, dx = 2\pi \int_1^\infty \frac{1}{x} \sqrt{1 + \frac{1}{x^4}} \, dx$$

Since $\sqrt{1 + \frac{1}{x^4}} > 1$ for all $x \ge 1$:

$$S > 2\pi \int_1^\infty \frac{1}{x} \, dx = 2\pi \lim_{M \to \infty} \Big[ \ln(x) \Big]_1^M = 2\pi \lim_{M \to \infty} \ln(M) = \mathbf{\infty}$$

By the Direct Comparison Test, the surface area $S$ diverges to infinity.

Part (c): Resolution of the Painter's Paradox

The paradox arises from treating "paint" as a mathematical zero-thickness 2D coating versus a physical 3D fluid with non-zero molecular thickness $\delta > 0$:

  1. In physical reality, paint molecules have a finite non-zero diameter $\delta$. Once the neck of the trumpet narrows such that the radius $r(x) = 1/x < \delta$ (i.e. for $x > 1/\delta$), no paint molecules can enter the throat, making it physically impossible to coat the entire infinite surface.
  2. Mathematically, the volume of a 3D coating of constant normal thickness $\delta$ is an integral over an expanded 3D shell, which itself diverges to infinity. Thus, a finite volume of paint can only produce a coating whose thickness vanishes as $x \to \infty$ at a rate faster than $1/x$, which is impossible for physical matter.
Final Answer & Physical Insight

$V = \pi$ (finite), while $S = \infty$ (divergent); resolved because physical paint has finite molecular thickness $\delta > 0$.