Mathematics / Analysis Integral Calculus & Series 100% Free Open Access
Chapter 4 • Theory & Derivations

Graphing in Polar Coordinates & Geometric Applications

§4.1 The Polar Coordinate Frame & Classical Curve Tracing

1. Coordinate Transformation Equations

In the plane $\mathbb{R}^2$, a point $P$ is determined by its directed distance $r$ from the origin (pole) and counterclockwise angle $\theta$ from the positive horizontal axis (polar axis):

$$x = r \cos\theta, \qquad y = r \sin\theta \quad \Longleftrightarrow \quad r^2 = x^2 + y^2, \qquad \tan\theta = \frac{y}{x}$$

2. Symmetry Criteria for Polar Curves $r = f(\theta)$

  1. Symmetry about Polar Axis (Horizontal Axis): The equation is unchanged when replacing $(r, \theta)$ with $(r, -\theta)$ or $(-r, \pi - \theta)$.
  2. Symmetry about Normal Axis $\theta = \pi/2$ (Vertical Axis): The equation is unchanged when replacing $(r, \theta)$ with $(r, \pi - \theta)$ or $(-r, -\theta)$.
  3. Symmetry about the Pole (Origin): The equation is unchanged when replacing $(r, \theta)$ with $(-r, \theta)$ or $(r, \theta + \pi)$.

3. Zoology of Classical Polar Curves

  • Cardioids: $r = a(1 \pm \cos\theta)$ or $r = a(1 \pm \sin\theta)$ ($a > 0$). Heart-shaped curves passing through the pole with a cusp at $r = 0$.
  • Limaçons: $r = a + b\cos\theta$. If $a < b$, it features an inner loop; if $a = b$, it is a cardioid; if $a > b$, it is dimpled or convex.
  • Rose Curves: $r = a\cos(n\theta)$ or $r = a\sin(n\theta)$. If $n \in \mathbb{N}$ is odd, the curve has $n$ petals; if $n$ is even, the curve has $2n$ petals.
  • Lemniscates: $r^2 = a^2\cos(2\theta)$. Figure-eight curves with nodal tangents at $\theta = \pm \pi/4$.

§4.2 Tangents to Polar Curves & Angle Between Radius Vector and Tangent

1. Slope of the Cartesian Tangent $\frac{dy}{dx}$

Treating $\theta$ as a parametric variable with $x(\theta) = r(\theta)\cos\theta$ and $y(\theta) = r(\theta)\sin\theta$:

$$\frac{dx}{d\theta} = \frac{dr}{d\theta}\cos\theta - r\sin\theta, \qquad \frac{dy}{d\theta} = \frac{dr}{d\theta}\sin\theta + r\cos\theta$$
$$\mathbf{\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{\frac{dr}{d\theta}\sin\theta + r\cos\theta}{\frac{dr}{d\theta}\cos\theta - r\sin\theta}}$$

2. The Polar Angle $\psi$ (Angle Between Radius Vector and Tangent)

Let $\phi$ be the inclination angle of the tangent line with the polar axis ($\tan\phi = \frac{dy}{dx}$) and $\theta$ be the vectorial angle of $P$. The angle $\psi = \phi - \theta$ between the radius vector $OP$ and the tangent line satisfies:

$$\tan\psi = \tan(\phi - \theta) = \frac{\tan\phi - \tan\theta}{1 + \tan\phi \tan\theta}$$

Substituting $\tan\phi = \frac{r' \sin\theta + r \cos\theta}{r' \cos\theta - r \sin\theta}$ and $\tan\theta = \frac{\sin\theta}{\cos\theta}$:

$$\tan\psi = \frac{\frac{r' \sin\theta + r \cos\theta}{r' \cos\theta - r \sin\theta} - \frac{\sin\theta}{\cos\theta}}{1 + \left(\frac{r' \sin\theta + r \cos\theta}{r' \cos\theta - r \sin\theta}\right)\frac{\sin\theta}{\cos\theta}} = \frac{r(\cos^2\theta + \sin^2\theta)}{r'(\cos^2\theta + \sin^2\theta)} = \frac{r}{r'}$$
$$\mathbf{\tan\psi = \frac{r}{\frac{dr}{d\theta}} = r \frac{d\theta}{dr}}$$

This remarkably elegant identity proves that the geometry of polar tangents depends solely on $r$ and its angular rate of change.

§4.3 Areas Enclosed by Polar Curves & Multi-Looped Sectors

1. Derivation of the Polar Area Element

Consider an infinitesimal sector between rays $\theta$ and $\theta + d\theta$ bounded by $r = f(\theta)$. Approximating this sector by a circular sector of radius $r$ and central angle $d\theta$:

$$dA = \frac{1}{2} r^2 \, d\theta$$

Integrating from $\theta = \alpha$ to $\theta = \beta$ yields the total swept area:

$$\mathbf{A = \frac{1}{2} \int_\alpha^\beta [r(\theta)]^2 \, d\theta}$$

2. Area Between Two Polar Curves

If $r_{\text{outer}}(\theta) \ge r_{\text{inner}}(\theta) \ge 0$ for $\theta \in [\alpha, \beta]$:

$$\mathbf{A = \frac{1}{2} \int_\alpha^\beta \Big( [r_{\text{outer}}(\theta)]^2 - [r_{\text{inner}}(\theta)]^2 \Big) \, d\theta}$$

3. Critical Precaution: Multi-Loop & Overlapping Limits

For rose curves $r = a\cos(n\theta)$, a single petal is bounded between adjacent roots where $r = 0$. For $r = a\cos(2\theta)$, setting $2\theta = \pm \frac{\pi}{2} \implies \theta \in [-\frac{\pi}{4}, \frac{\pi}{4}]$ bounds exactly one petal. Integrating over $[0, 2\pi]$ directly without symmetry can count overlapping loops multiple times.

§4.4 Arc Length in Polar Coordinates

1. Analytical Derivation of $ds$

Recall $dx = (r'\cos\theta - r\sin\theta)d\theta$ and $dy = (r'\sin\theta + r\cos\theta)d\theta$. Squaring and summing:

$$\begin{aligned} (dx)^2 + (dy)^2 &= \left[ (r')^2 \cos^2\theta - 2r r' \sin\theta\cos\theta + r^2 \sin^2\theta \right] (d\theta)^2 \ &\quad + \left[ (r')^2 \sin^2\theta + 2r r' \sin\theta\cos\theta + r^2 \cos^2\theta \right] (d\theta)^2 \ &= \Big( r^2 (\sin^2\theta + \cos^2\theta) + (r')^2 (\sin^2\theta + \cos^2\theta) \Big) (d\theta)^2 \ &= \left( r^2 + \left(\frac{dr}{d\theta}\right)^2 \right) (d\theta)^2 \end{aligned}$$
$$\mathbf{ds = \sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2} \, d\theta}$$

2. Total Arc Length Integral

$$\mathbf{L = \int_\alpha^\beta \sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2} \, d\theta}$$

§4.5 Surfaces and Volumes of Revolution in Polar Coordinates

1. Revolution About the Polar Axis ($x$-axis)

When revolving about the polar axis ($y = 0$), the perpendicular distance to the rotation axis is $y = r\sin\theta$.

  • Surface Area of Revolution: $$\mathbf{S = 2\pi \int_\alpha^\beta r\sin\theta \sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2} \, d\theta}$$
  • Volume of Revolution: Revolving differential triangular sectors $dA$ gives conical rings: $$\mathbf{V = \frac{2}{3}\pi \int_\alpha^\beta r^3 \sin\theta \, d\theta}$$

2. Revolution About the Normal Axis $\theta = \pi/2$ ($y$-axis)

The perpendicular distance is $x = r\cos\theta$:

$$\mathbf{S = 2\pi \int_\alpha^\beta r\cos\theta \sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2} \, d\theta}$$
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 4.1: Tangent and Radius Angle for the Cardioid

For the cardioid $r = 2(1 + \cos\theta)$: (a) Find the angle $\psi$ between the radius vector and the tangent line at $\theta = \frac{\pi}{3}$. (b) Find the Cartesian slope $\frac{dy}{dx}$ of the tangent line at this point.

Part (a): Compute the Angle $\psi$

Differentiating $r = 2(1 + \cos\theta)$ with respect to $\theta$:

$$\frac{dr}{d\theta} = -2\sin\theta$$

Using the polar tangent identity $\tan\psi = \frac{r}{dr/d\theta}$:

$$\tan\psi = \frac{2(1 + \cos\theta)}{-2\sin\theta} = -\frac{2\cos^2(\theta/2)}{2\sin(\theta/2)\cos(\theta/2)} = -\cot\left(\frac{\theta}{2}\right) = \tan\left( \frac{\pi}{2} + \frac{\theta}{2} \right)$$

Thus, $\psi = \frac{\pi}{2} + \frac{\theta}{2}$. Evaluating at $\theta = \frac{\pi}{3}$:

$$\psi = \frac{\pi}{2} + \frac{\pi/3}{2} = \frac{\pi}{2} + \frac{\pi}{6} = \frac{2\pi}{3} = 120^\circ$$

Part (b): Compute the Cartesian Slope $\frac{dy}{dx}$

The inclination angle of the tangent is $\phi = \theta + \psi = \frac{\pi}{3} + \frac{2\pi}{3} = \pi$.

Therefore, the Cartesian slope is:

$$\frac{dy}{dx} = \tan\phi = \tan(\pi) = \mathbf{0}$$

The tangent line to the cardioid at $\theta = \frac{\pi}{3}$ is perfectly horizontal.

Final Answer & Physical Insight

$\psi = \frac{2\pi}{3}$ ($120^\circ$); \quad $\frac{dy}{dx} = 0$ (horizontal tangent)

Tier 2: Intermediate Exam Example 4.2: Area and Perimeter of a Rose Petal

Consider the four-petaled rose $r = 4\cos(2\theta)$. (a) Find the exact area of one complete petal. (b) Find the total area enclosed by all four petals.

Part (a): Area of One Petal

The tip of the petal on the positive polar axis occurs at $\theta = 0$ where $r = 4$. The boundaries of this petal occur where $r = 0$:

$$\cos(2\theta) = 0 \implies 2\theta = \pm \frac{\pi}{2} \implies \theta = -\frac{\pi}{4} \text{ and } \theta = \frac{\pi}{4}$$

The area of this petal is:

$$A_{\text{petal}} = \frac{1}{2} \int_{-\pi/4}^{\pi/4} [4\cos(2\theta)]^2 \, d\theta = \frac{16}{2} \int_{-\pi/4}^{\pi/4} \cos^2(2\theta) \, d\theta = 8 \int_{-\pi/4}^{\pi/4} \left( \frac{1 + \cos(4\theta)}{2} \right) d\theta$$
$$A_{\text{petal}} = 4 \left[ \theta + \frac{\sin(4\theta)}{4} \right]_{-\pi/4}^{\pi/4} = 4 \left( \left(\frac{\pi}{4} + 0\right) - \left(-\frac{\pi}{4} + 0\right) \right) = 4\left(\frac{\pi}{2}\right) = \mathbf{2\pi}$$

Part (b): Total Area of All Four Petals

By fourfold rotational symmetry, the total enclosed area is:

$$A_{\text{total}} = 4 \times A_{\text{petal}} = 4 \times 2\pi = \mathbf{8\pi}$$
Final Answer & Physical Insight

$A_{\text{petal}} = 2\pi$; \quad $A_{\text{total}} = 8\pi$

Tier 3: Honors / Proof Challenge Example 4.3: Total Perimeter and Surface Area of the Revolved Cardioid

For the complete cardioid $r = a(1 + \cos\theta)$ ($a > 0$): (a) Prove that its total perimeter is $L = 8a$. (b) Find the total surface area generated by revolving this cardioid about the initial polar axis.

Part (a): Total Perimeter of the Cardioid

We compute $r' = -a\sin\theta$. The arc length element is:

$$\begin{aligned} ds &= \sqrt{r^2 + (r')^2} \, d\theta = \sqrt{a^2(1 + \cos\theta)^2 + a^2\sin^2\theta} \, d\theta \ &= a \sqrt{1 + 2\cos\theta + \cos^2\theta + \sin^2\theta} \, d\theta = a \sqrt{2 + 2\cos\theta} \, d\theta \ &= a \sqrt{4\cos^2(\theta/2)} \, d\theta = 2a |\cos(\theta/2)| \, d\theta \end{aligned}$$

Using symmetry about the polar axis, integrate from $\theta = 0$ to $\pi$ where $\cos(\theta/2) \ge 0$:

$$L = 2 \int_0^\pi 2a \cos(\theta/2) \, d\theta = 4a \Big[ 2\sin(\theta/2) \Big]_0^\pi = 8a \Big( \sin(\pi/2) - 0 \Big) = \mathbf{8a} \quad \blacksquare$$

Part (b): Surface Area Revolved About the Polar Axis

The surface area element is $dS = 2\pi y \, ds = 2\pi (r\sin\theta) ds$:

$$dS = 2\pi \Big( a(1 + \cos\theta)\sin\theta \Big) \Big( 2a\cos(\theta/2) \, d\theta \Big)$$

Using $1 + \cos\theta = 2\cos^2(\theta/2)$ and $\sin\theta = 2\sin(\theta/2)\cos(\theta/2)$:

$$dS = 2\pi a^2 \left( 2\cos^2\frac{\theta}{2} \right) \left( 2\sin\frac{\theta}{2}\cos\frac{\theta}{2} \right) \left( 2\cos\frac{\theta}{2} \right) d\theta = 16\pi a^2 \cos^4\left(\frac{\theta}{2}\right)\sin\left(\frac{\theta}{2}\right) d\theta$$

Integrating over $\theta \in [0, \pi]$ with substitution $u = \cos(\theta/2) \implies du = -\frac{1}{2}\sin(\theta/2)d\theta$:

$$S = 16\pi a^2 \int_0^\pi \cos^4\left(\frac{\theta}{2}\right) \sin\left(\frac{\theta}{2}\right) d\theta = 16\pi a^2 \int_0^1 u^4 (2 du) = 32\pi a^2 \left[ \frac{u^5}{5} \right]_0^1 = \mathbf{\frac{32}{5}\pi a^2}$$
Final Answer & Physical Insight

$L = 8a$; \quad $S = \frac{32}{5}\pi a^2$