7.4 Area and Arc Length in Polar Coordinates

Area in Polar Coordinates

Why the polar area formula looks different

In rectangular coordinates, you find area by stacking thin vertical rectangles. In polar coordinates, you're sweeping out thin "pie slices" (sectors) from the origin. Each tiny sector has radius $r$ and angular width $d\theta$, giving it an area of $\frac{1}{2}r^2\,d\theta$. Summing up all those slices from angle $a$ to angle $b$ gives the polar area formula:

$A = \frac{1}{2}\int_a^b r^2\,d\theta$

How to find the area enclosed by a polar curve

1. Identify the curve and limits. Write down $r = f(\theta)$ and determine the range of $\theta$ that traces the region exactly once. Sketch the curve if you can.
2. Substitute into the formula. Replace $r$ with $f(\theta)$ inside the integral.
3. Expand and evaluate. Expand $r^2$, then integrate term by term. Trig identities (especially power-reduction identities) come up constantly here.

Example: Area of the cardioid $r = 1 + \cos\theta$

The cardioid traces out once as $\theta$ goes from $0$ to $2\pi$:

$A = \frac{1}{2}\int_0^{2\pi}(1+\cos\theta)^2\,d\theta$

Expand: $(1+\cos\theta)^2 = 1 + 2\cos\theta + \cos^2\theta$

Use the identity $\cos^2\theta = \frac{1+\cos 2\theta}{2}$ to rewrite the integrand as $\frac{3}{2} + 2\cos\theta + \frac{\cos 2\theta}{2}$. The $\cos\theta$ and $\cos 2\theta$ terms integrate to zero over a full period, leaving:

$A = \frac{1}{2}\cdot\frac{3}{2}\cdot 2\pi = \frac{3\pi}{2}$

Area between two polar curves

When you need the area between an outer curve $r_{\text{out}}$ and an inner curve $r_{\text{in}}$:

$A = \frac{1}{2}\int_a^b\left(r_{\text{out}}^2 - r_{\text{in}}^2\right)d\theta$

The key steps:

1. Find intersection points by setting $r_1 = r_2$ and solving for $\theta$. These angles become your limits of integration. Also check the origin separately, since both curves can pass through $r = 0$ at different angles.
2. Determine which curve is farther from the origin on each interval. Plug in a test angle if you're unsure.
3. Set up and evaluate the integral on each sub-interval, then add the pieces.

Arc Length in Polar Coordinates

The arc length formula

For a polar curve $r = f(\theta)$ traced from $\theta = a$ to $\theta = b$, the arc length is:

$L = \int_a^b\sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2}\,d\theta$

This formula comes from the parametric arc length formula after converting $x = r\cos\theta$ and $y = r\sin\theta$ and simplifying.

How to compute arc length step by step

1. Write down $r = f(\theta)$ and the interval $[a, b]$.
2. Compute $\frac{dr}{d\theta}$.
3. Build the integrand: form $r^2 + \left(\frac{dr}{d\theta}\right)^2$, simplify, and take the square root.
4. Evaluate the integral. Many polar arc length integrals don't simplify to elementary functions, so expect to use trig identities, substitution, or numerical methods.

Example: Spiral of Archimedes $r = \theta$, from $\theta = 0$ to $\theta = 2\pi$

• $\frac{dr}{d\theta} = 1$
• Integrand: $\sqrt{\theta^2 + 1}$
• $L = \int_0^{2\pi}\sqrt{\theta^2+1}\,d\theta$

This requires a trig substitution ($\theta = \tan u$) or the formula $\int\sqrt{\theta^2+1}\,d\theta = \frac{1}{2}\left(\theta\sqrt{\theta^2+1}+\sinh^{-1}\theta\right)$. Evaluating gives $L \approx 21.26$ units.

Curves with multiple pieces: Some curves, like the lemniscate $r^2 = \cos(2\theta)$, only exist where the right side is non-negative. The lemniscate has two loops. Find the arc length of one loop using the appropriate $\theta$-interval, then multiply by 2 for the total.

Analysis of Polar Curve Behavior

Finding intersection points

To find where two polar curves $r_1 = f_1(\theta)$ and $r_2 = f_2(\theta)$ meet:

1. Set $f_1(\theta) = f_2(\theta)$ and solve for $\theta$.
2. Substitute back into either equation to get the $r$-values.
3. Check the origin separately. If $f_1(\alpha) = 0$ for some angle $\alpha$ and $f_2(\beta) = 0$ for some (possibly different) angle $\beta$, both curves pass through the origin even though they "arrive" there at different angles.

Example: $r = 2\cos\theta$ and $r = 1 + \cos\theta$

Setting them equal: $2\cos\theta = 1 + \cos\theta$, so $\cos\theta = 1$, giving $\theta = 0$ (and $r = 2$). That's only one algebraic solution. But $r = 2\cos\theta$ passes through the origin at $\theta = \pi/2$, and $r = 1 + \cos\theta$ passes through the origin at $\theta = \pi$. So the origin is a second intersection point that you'd miss if you only solved the equation algebraically.

Tangent lines to polar curves

To find $\frac{dy}{dx}$ for a polar curve, convert to parametric form using $x = r\cos\theta$ and $y = r\sin\theta$, then apply the chain rule:

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

1. Compute $\frac{dr}{d\theta}$.
2. Plug $r$, $\frac{dr}{d\theta}$, and $\theta_0$ into the formula above to get the slope.
3. Convert the point to rectangular coordinates: $x_0 = r_0\cos\theta_0$, $y_0 = r_0\sin\theta_0$.
4. Write the tangent line using point-slope form: $y - y_0 = m(x - x_0)$.