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:
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:
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:
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 $dx$.
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 $r(x) = g(x)$:
3. Rotation About Non-Origin Axes
If the rotation axis is shifted to the horizontal line $y = k$:
Similarly, for rotation about a vertical line $x = h$ using horizontal slices perpendicular to the axis of rotation:
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:
3. General Axis of Rotation ($x = h$)
If the axis of rotation is the vertical line $x = h$:
For horizontal axis rotation revolving about $y = k$ with horizontal strips of thickness $dy$:
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 $dV = A(x) \, dx$:
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 $y = f(x) \in C^1([a, b])$:
The total Arc Length of the curve from $x = a$ to $x = b$ is:
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 $2\pi$:
Let a planar region of area $A$ and centroid distance $\bar{y}$ be revolved through $2\pi$:
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.
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
Step 3: Evaluate the Antiderivative
$V = \frac{192\pi}{5}$
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 $x$:
Using the fundamental hyperbolic identity $1 + \sinh^2(u) = \cosh^2(u)$:
Integrating from $-a$ to $a$ (using symmetry about $x = 0$):
Part (b): Surface Area of Revolution about $x$-axis
The differential surface area element is $dS = 2\pi y \, ds$:
Applying the hyperbolic power-reduction identity $\cosh^2(u) = \frac{\cosh(2u) + 1}{2}$:
$L = 2a\sinh(1)$, \quad $S = \pi a^2 (\sinh(2) + 2)$
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 $[1, \infty)$:
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:
Since $\sqrt{1 + \frac{1}{x^4}} > 1$ for all $x \ge 1$:
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 $\delta > 0$:
- 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.
- 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.
$V = \pi$ (finite), while $S = \infty$ (divergent); resolved because physical paint has finite molecular thickness $\delta > 0$.