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

3 min readjanuary 25, 2023

Sumi Vora

Sumi Vora

Jed Quiaoit

Jed Quiaoit

Sumi Vora

Sumi Vora

Jed Quiaoit

Jed Quiaoit

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 rr and angle θ\theta (in radians) is given by 12r2θ\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 AA enclosed by a polar curve r=f(θ)r=f(\theta) from θ=a\theta=a to θ=b\theta=b, we use the definite integral:

A=12ab[f(θ)]2dθA = \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 aa and bb, each with its own radius determined by the polar function f(θ)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(θ)r=2cos(\theta) over the range 0θπ0\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 θ\theta :
    2. A=120π(2cos(θ))2dθA = \frac{1}{2}\int_{0}^{\pi}(2cos(\theta))^2d\theta
  2. Simplify the Integral
    1. Apply trigonometric identities to simplify the expression :
    2. 20πcos2(θ)dθ2\int_{0}^{\pi}cos^2(\theta)d\theta
  3. Solve the Integral
    1. Use the half-angle identity, cos2(θ)=1+cos(2θ)2cos^2(\theta)=\frac{1+cos(2\theta)}{2}, to find the integral :
    2. A=2[θ2+sin(2θ)4]0π=πA=2[\frac{\theta}{2}+\frac{sin(2\theta)}{4}]\vert_0^\pi=\pi

So, the area inside the curve r=2cos(θ)r=2cos(\theta) over the range 0θπ0\leq\theta\leq\pi, is π\pi square units.

Using Symmetry to Simplify Calculations

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

Here are the steps we must follow:

  1. Identify the Range for One Petal
    1. Since the curve r=sin(2θ)r=sin(2\theta) forms a petal between 00 and π2\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 12r2\frac{1}{2}r^2 over the angle range :
    2. A=120π2(sin(θ))2dθA = \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=12[θ2sin(4θ)8]0π2=π8A=\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(2θ)r=sin(2\theta) is π8\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(θ)r=3+3sin(\theta) over the interval 0θ2π0\le\theta\le2\pi.

  1. Sketch the Curve

    Always start by sketching the curve to understand its shape and symmetry. For r=3+3sin(θ)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 00 to π\pi) and then doubling it, set up the integral as follows :

    1. A=2120π(3+3sin(θ))2dθA=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=2120π(9+18sin(θ)+9sin2(θ))dθA=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. 02π9dθ=9θ02π=18π\int_{0}^{2\pi}9d\theta =9\theta\vert_0^{2\pi}=18\pi
    2. 02π18sin(θ)dθ\int_{0}^{2\pi}18sin(\theta)d\theta =18cos(θ)02π=0=-18cos(\theta)\vert_0^{2\pi}=0
    3. 02π91cos(2θ)2dθ=92(θsin(2θ)2)02π=9π\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=12(18π+0+9π)=27π2A=\frac{1}{2}(18\pi+0+9\pi)=\frac{27\pi}{2}

So, the area enclosed by the polar curve is r=3+3sin(θ)r=3+3sin(\theta) is 27π2\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.

Key Terms to Review (8)

Area under the curve

: The area under the curve refers to the total amount of space enclosed between a function and the x-axis on a given interval.

cosθ

: The cosine of an angle θ is the ratio of the length of the adjacent side to θ in a right triangle to the length of the hypotenuse.

Enclosed area between polar curves

: The enclosed area between two polar curves is the region in the plane that is bounded by these curves. It represents the area of overlap between the two curves.

Integration

: Integration is an operation in calculus that finds antiderivatives, or reverse derivatives, allowing us to calculate areas, volumes, and other quantities related to curves and functions.

Polar Coordinates

: Polar coordinates are a two-dimensional coordinate system used to locate points in space using radial distance (r) and angular displacement (θ) from a reference point called the pole.

Polar Functions

: Polar functions are mathematical equations that describe curves in terms of a distance from the origin and an angle. They are often used to represent shapes that have radial symmetry.

r^n

: In mathematics, r^n refers to raising a number r (the base) to an exponent n. It means multiplying r by itself n times.

sinθ

: The sine of an angle θ is the ratio of the length of the side opposite to θ in a right triangle to the length of the hypotenuse.

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

3 min readjanuary 25, 2023

Sumi Vora

Sumi Vora

Jed Quiaoit

Jed Quiaoit

Sumi Vora

Sumi Vora

Jed Quiaoit

Jed Quiaoit

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 rr and angle θ\theta (in radians) is given by 12r2θ\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 AA enclosed by a polar curve r=f(θ)r=f(\theta) from θ=a\theta=a to θ=b\theta=b, we use the definite integral:

A=12ab[f(θ)]2dθA = \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 aa and bb, each with its own radius determined by the polar function f(θ)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(θ)r=2cos(\theta) over the range 0θπ0\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 θ\theta :
    2. A=120π(2cos(θ))2dθA = \frac{1}{2}\int_{0}^{\pi}(2cos(\theta))^2d\theta
  2. Simplify the Integral
    1. Apply trigonometric identities to simplify the expression :
    2. 20πcos2(θ)dθ2\int_{0}^{\pi}cos^2(\theta)d\theta
  3. Solve the Integral
    1. Use the half-angle identity, cos2(θ)=1+cos(2θ)2cos^2(\theta)=\frac{1+cos(2\theta)}{2}, to find the integral :
    2. A=2[θ2+sin(2θ)4]0π=πA=2[\frac{\theta}{2}+\frac{sin(2\theta)}{4}]\vert_0^\pi=\pi

So, the area inside the curve r=2cos(θ)r=2cos(\theta) over the range 0θπ0\leq\theta\leq\pi, is π\pi square units.

Using Symmetry to Simplify Calculations

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

Here are the steps we must follow:

  1. Identify the Range for One Petal
    1. Since the curve r=sin(2θ)r=sin(2\theta) forms a petal between 00 and π2\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 12r2\frac{1}{2}r^2 over the angle range :
    2. A=120π2(sin(θ))2dθA = \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=12[θ2sin(4θ)8]0π2=π8A=\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(2θ)r=sin(2\theta) is π8\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(θ)r=3+3sin(\theta) over the interval 0θ2π0\le\theta\le2\pi.

  1. Sketch the Curve

    Always start by sketching the curve to understand its shape and symmetry. For r=3+3sin(θ)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 00 to π\pi) and then doubling it, set up the integral as follows :

    1. A=2120π(3+3sin(θ))2dθA=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=2120π(9+18sin(θ)+9sin2(θ))dθA=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. 02π9dθ=9θ02π=18π\int_{0}^{2\pi}9d\theta =9\theta\vert_0^{2\pi}=18\pi
    2. 02π18sin(θ)dθ\int_{0}^{2\pi}18sin(\theta)d\theta =18cos(θ)02π=0=-18cos(\theta)\vert_0^{2\pi}=0
    3. 02π91cos(2θ)2dθ=92(θsin(2θ)2)02π=9π\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=12(18π+0+9π)=27π2A=\frac{1}{2}(18\pi+0+9\pi)=\frac{27\pi}{2}

So, the area enclosed by the polar curve is r=3+3sin(θ)r=3+3sin(\theta) is 27π2\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.

Key Terms to Review (8)

Area under the curve

: The area under the curve refers to the total amount of space enclosed between a function and the x-axis on a given interval.

cosθ

: The cosine of an angle θ is the ratio of the length of the adjacent side to θ in a right triangle to the length of the hypotenuse.

Enclosed area between polar curves

: The enclosed area between two polar curves is the region in the plane that is bounded by these curves. It represents the area of overlap between the two curves.

Integration

: Integration is an operation in calculus that finds antiderivatives, or reverse derivatives, allowing us to calculate areas, volumes, and other quantities related to curves and functions.

Polar Coordinates

: Polar coordinates are a two-dimensional coordinate system used to locate points in space using radial distance (r) and angular displacement (θ) from a reference point called the pole.

Polar Functions

: Polar functions are mathematical equations that describe curves in terms of a distance from the origin and an angle. They are often used to represent shapes that have radial symmetry.

r^n

: In mathematics, r^n refers to raising a number r (the base) to an exponent n. It means multiplying r by itself n times.

sinθ

: The sine of an angle θ is the ratio of the length of the side opposite to θ in a right triangle to the length of the hypotenuse.


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AP® and SAT® are trademarks registered by the College Board, which is not affiliated with, and does not endorse this website.


© 2024 Fiveable Inc. All rights reserved.

AP® and SAT® are trademarks registered by the College Board, which is not affiliated with, and does not endorse this website.