9.8 Find the Area of a Polar Region or the Area Bounded by a Single Polar Curve

Embarking on the adventure of AP Calculus, we often encounter the mesmerizing world of shapes, curves, and areas that seem to dance between dimensions. Among these, the concept of polar coordinates offers a fresh perspective on understanding the geometry of curves.

Our mission is to master the technique of calculating areas of regions defined by polar curves, using definite integrals. By the end of this guide, you'll be able to wrap your head around polar curves and the areas they enclose with confidence and curiosity.

🤓 Understanding Polar Coordinates

Before we dive into calculating areas, let's go over what polar coordinates are. Unlike the familiar Cartesian coordinates (x and y), which locate points through horizontal and vertical distances, polar coordinates use a radius (r) and an angle (θ) to pinpoint the location of a point in a plane. This system is incredibly useful for describing curves that are circles or spirals, which are difficult to express in Cartesian terms.

Transitioning to Polar Coordinates

To understand the area under a polar curve, we must first grasp how to express the concept of area in polar terms. The area of a sector (a pizza slice of a circle) is a fundamental building block. In polar coordinates, the area of a sector with radius $r$ and angle $\theta$ (in radians) is given by $\frac{1}{2}r^2\theta$.

Calculating Area in Polar Coordinates

The beauty of calculus shines when we apply the concept of integration to polar coordinates. To find the area $A$ enclosed by a polar curve $r=f(\theta)$ from $\theta=a$ to $\theta=b$, we use the definite integral:

$A = \frac{1}{2}\int_{a}^{b}[f(\theta)]^2d\theta$

This formula is a direct extension of finding the area of a sector, but instead of a single slice, we sum up infinitely small slices (sectors) between $a$ and $b$, each with its own radius determined by the polar function $f(\theta)$.

🚀 Different Methods for Calculating Area in Polar Regions

Sector Method for Simple Curves

Problem Statement

Find the area inside the circle $r=2cos(\theta)$ over the range $0\leq\theta\leq\pi$.

Here are the steps we must follow:

1. Set Up the Integral
   1. Recognize that the area can be found by integrating the square of the radius over the given range of $\theta$ :
   2. $A = \frac{1}{2}\int_{0}^{\pi}(2cos(\theta))^2d\theta$
2. Simplify the Integral
   1. Apply trigonometric identities to simplify the expression :
   2. $2\int_{0}^{\pi}cos^2(\theta)d\theta$
3. Solve the Integral
   1. Use the half-angle identity, $cos^2(\theta)=\frac{1+cos(2\theta)}{2}$, to find the integral :
   2. $A=2[\frac{\theta}{2}+\frac{sin(2\theta)}{4}]\vert_0^\pi=\pi$

So, the area inside the curve $r=2cos(\theta)$ over the range $0\leq\theta\leq\pi$, is $\pi$ square units.

Using Symmetry to Simplify Calculations

Calculate the area of one petal of the rose curve $r=sin(2\theta)$.

Here are the steps we must follow:

1. Identify the Range for One Petal
   1. Since the curve $r=sin(2\theta)$ forms a petal between $0$ and $\frac{\pi}{2}$, we calculate the area over this range.
2. Set Up the Integral
   1. The area of one petal is given by integrating $\frac{1}{2}r^2$ over the angle range :
   2. $A = \frac{1}{2}\int_{0}^{\frac{\pi}{2}}(sin(\theta))^2d\theta$
3. Solve the Integral
   1. Using trigonometric identities and integration techniques, solve the integral :
   2. $A=\frac{1}{2}[\frac{\theta}{2}-\frac{sin(4\theta)}{8}]\vert_0^\frac{\pi}{2}=\frac{\pi}{8}$

So, the area of one petal of the rose curve $r=sin(2\theta)$ is $\frac{\pi}{8}$ units squared.

📝 Practice Problem

Let’s try a practice problem to test your new skills!

Calculate the area enclosed by the polar curve $r=3+3sin(\theta)$ over the interval $0\le\theta\le2\pi$.

1. Sketch the Curve

   Always start by sketching the curve to understand its shape and symmetry. For $r=3+3sin(\theta)$, it’s a limacon with an inner loop.

2. Inner Symmetry

   Note that the curve is symmetrical about the horizontal axis; this, you can calculate the area for half the curve and then double it.

3. Set Up the Integral for Half the Curve

   Since you’re calculating the area for half the curve (from $0$ to $\pi$) and then doubling it, set up the integral as follows :

   1. $A=2 \cdot\frac{1}{2}\int_{0}^{\pi}(3+3sin(\theta))^2d\theta$

4. Simplify and Solve the Integral

   Simplify the integral expression first:

   1. $A=2 \cdot \frac{1}{2}\int_{0}^{{\pi}}(9+18sin(\theta)+9sin^2(\theta))d\theta$

   Then, solve each term of the integral separately:

   1. $\int_{0}^{2\pi}9d\theta =9\theta\vert_0^{2\pi}=18\pi$
   2. $\int_{0}^{2\pi}18sin(\theta)d\theta$ $=-18cos(\theta)\vert_0^{2\pi}=0$
   3. $\int_{0}^{2\pi}9\frac{1-cos(2\theta)}{2}d\theta=\frac{9}{2}(\theta-\frac{sin(2\theta)}{2})\vert_0^{2\pi}=9\pi$

5. Combine

   Combine all terms to find the total area:

   1. $A=\frac{1}{2}(18\pi+0+9\pi)=\frac{27\pi}{2}$

So, the area enclosed by the polar curve is $r=3+3sin(\theta)$ is $\frac{27\pi}{2}$ units squared.

⭐️ Conclusion

Understanding how to find the area of a polar region or the area bounded by a single polar curve expands our problem-solving toolkit in calculus. It allows us to tackle a variety of problems involving curves in polar form with precision and accuracy.