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12.3 Areas and Lengths in Polar Coordinates

In Cartesian

\[ \int_{a}^{b} f(x) dx \]

shrink rectangles then sum infinitely-many of them

Graph of a function f(x) with vertical rectangles under the curve from a to b on the x-axis.
Graph showing the area between two curves f(x) and g(x) filled with vertical rectangles from a to b.
\[ \int_{a}^{b} [f(x) - g(x)] dx \]

Sum of infinitely-many rectangles between \( f(x) \) and \( g(x) \)

In Polar, same idea, but instead of rectangles, we use thin slices of circles

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Polar region shaded in red bounded by r=f(theta) and rays at angles alpha and beta.

\( r = f(\theta) \) curve in polar equation

area of red region?

divide into thin slices

Detailed view of a polar sector with radius r and central angle delta theta.

when \( \Delta \theta \) is small, \( r_1 \approx r_2 = r \)

from geometry, the area of this segment is

\[ \frac{1}{2} r^2 \Delta \theta \]

this is the polar equivalent of a rectangle

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Sum up these slices from \(\alpha\) to \(\beta\), and shrink \(\Delta \theta \to d\theta\)

so, the entire region has area

\[ \int_{\alpha}^{\beta} \frac{1}{2} r^2 d\theta \]

function of \(\theta\)

if \(r = f(\theta)\)

\[ \int_{\alpha}^{\beta} \frac{1}{2} [f(\theta)]^2 d\theta \]

between curves

Polar graph with two curves f(theta) and g(theta) bounding a shaded area between angles alpha and beta.
\[ \int_{\alpha}^{\beta} \left\{ \frac{1}{2} [f(\theta)]^2 - \frac{1}{2} [g(\theta)]^2 \right\} d\theta \]
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example Find area of one petal of the rose \(r = \sin 2\theta\)

A four-petaled rose curve r = sin(2 theta) centered at the origin.

find area of any, for simplicity, let's find QI

A single petal of the rose curve in the first quadrant, with the lower half shaded.

notice symmetry, so we can find area of half (red portion) then double it

\[ \int_{\alpha}^{\beta} \frac{1}{2} r^2 d\theta \]

double \(\to\) to middle of petal

\[ 2 \int_{0}^{\pi/4} \frac{1}{2} (\sin 2\theta)^2 d\theta = \int_{0}^{\pi/4} \sin^2 2\theta d\theta = \int_{0}^{\pi/4} \frac{1 - \cos(4\theta)}{2} d\theta \]\[ = \frac{1}{2} \int_{0}^{\pi/4} 1 - \cos(4\theta) d\theta = \frac{1}{2} \left( \theta - \frac{1}{4} \sin 4\theta \right) \bigg|_0^{\pi/4} = \frac{1}{2} \left( \frac{\pi}{4} - 0 \right) = \boxed{\frac{\pi}{8}} \]
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Polar Area Calculations

Alternative: w/o using symmetry

\[ \int_{0}^{\pi/2} \frac{1}{2} (\sin 2\theta)^2 d\theta = \dots = \frac{\pi}{8} \]
Polar graph of a single rose petal in the first quadrant from theta equals 0 to pi over 2.

Start \( \alpha = 0 \), \( \beta = \pi/2 \)

Example: Area Bounded by Curves

Find the area bounded by \( r = \frac{1}{\sqrt{2}} \) and \( r = \sin 2\theta \) closer to the origin.

\( r = \frac{1}{\sqrt{2}} \): circle radius \( \frac{1}{\sqrt{2}} \)

\( r = \sin 2\theta \): rose we looked at

QI (Quadrant I) Detail:

Four-petaled rose curve with a circle of radius 1 over square root of 2 centered at the origin.
Close-up of the first quadrant showing the intersection of the rose petal and the circle.
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Area Subtraction Method

One approach: find area of rose petal outside circle, then subtract from area of one petal.

Shaded region of a rose petal outside of a circular arc in the first quadrant.
\[ \int_{\pi/8}^{3\pi/8} \left[ \frac{1}{2}(\sin 2\theta)^2 - \frac{1}{2}\left(\frac{1}{\sqrt{2}}\right)^2 \right] d\theta \]

Outside (rose)

Inside (circle)

Intersection Calculation

Intersection: \( r = \frac{1}{\sqrt{2}} \) and \( r = \sin 2\theta \) are equal.

Solve: \[ \frac{1}{\sqrt{2}} = \sin 2\theta \] \[ \vdots \] \[ \theta = \frac{\pi}{8} \]

Final Area Logic

Then, subtract from rose petal area:

Diagram showing a petal with a shaded inner region.
=
Diagram of a full shaded rose petal labeled last example.
-
Diagram of a rose petal with the outer portion shaded.
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Arc length:

To find the arc length of a curve defined in polar coordinates by \( r = f(\theta) \), we consider the curve between angles \( \alpha \) and \( \beta \).

Polar coordinate graph showing a curve r=f(theta) between angles alpha and beta in the first quadrant.

Consider a small segment of the curve. We can approximate this segment using a circular arc and a radial change.

  • Circular arc length \( = r \Delta \theta \)
  • When \( \Delta \theta \) is small, \( r_1 \approx r_2 = r \)
Diagram of a small sector showing radial distances r1, r2, radial change delta r, and arc length r delta theta.

The length of the small black curve segment is approximately:

\[ \sqrt{(\Delta r)^2 + (r \Delta \theta)^2} \approx \text{length of black curve} \]

This can be rewritten by factoring out \( (\Delta \theta)^2 \):

\[ \sqrt{(\Delta \theta)^2 \left[ \frac{(\Delta r)^2}{(\Delta \theta)^2} + r^2 \right]} \]
Right triangle approximation of the curve segment with legs delta r and r delta theta.
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Continuing the simplification:

\[ = \sqrt{\left[ \left( \frac{\Delta r}{\Delta \theta} \right)^2 + r^2 \right] (\Delta \theta)^2} = \sqrt{r^2 + \left( \frac{\Delta r}{\Delta \theta} \right)^2} \Delta \theta \]

Shrink:

\[ \frac{\Delta r}{\Delta \theta} \to \frac{dr}{d\theta} \]\[ \Delta \theta \to d\theta \]

Accumulate from \( \theta = \alpha \) to \( \theta = \beta \) by integration:

\[ \int_{\alpha}^{\beta} \sqrt{r^2 + \left( \frac{dr}{d\theta} \right)^2} d\theta \]
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Example: Length of a Logarithmic Spiral

Find the length of the curve defined by the polar equation:

\[ r = e^{\theta} \quad \text{for } 0 \le \theta \le 2\pi \]
A sketch of a logarithmic spiral on a Cartesian coordinate system, labeled as a logarithmic spiral.
Logarithmic spiral

Calculating the Arc Length

The formula for the arc length in polar coordinates is:

\[ \text{length} = \int_{0}^{2\pi} \sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2} \, d\theta \]

Given \( r = e^{\theta} \), we find the derivative:

\[ \frac{dr}{d\theta} = e^{\theta} \]

Substituting these into the integral:

\[ \int_{0}^{2\pi} \sqrt{(e^{\theta})^2 + (e^{\theta})^2} \, d\theta = \int_{0}^{2\pi} \sqrt{e^{2\theta} + e^{2\theta}} \, d\theta = \int_{0}^{2\pi} \sqrt{2e^{2\theta}} \, d\theta \]

Simplifying the integrand:

\[ = \int_{0}^{2\pi} \sqrt{2} \cdot \sqrt{e^{2\theta}} \, d\theta = \int_{0}^{2\pi} \sqrt{2} \, e^{\theta} \, d\theta = \sqrt{2} \int_{0}^{2\pi} e^{\theta} \, d\theta \]

Evaluating the definite integral:

\[ = \sqrt{2} \left( e^{\theta} \right) \Big|_{0}^{2\pi} = \sqrt{2} (e^{2\pi} - 1) \]
\[ \text{Final Answer: } \sqrt{2} (e^{2\pi} - 1) \]