Area between Polar Curves

Polar curves are represented using polar coordinates (r, θ), where r is the distance of a point from the pole (origin), and θ is the angle measured from the positive x-axis.

• Polar coordinates are particularly useful for describing curves such as circles, cardioids, roses, and spirals.
• Integration can be used to find the area enclosed by a polar curve or the area between two polar curves.

These coordinates (r, θ) are called polar coordinates. Many curves are easily described in this manner.

For example, the equation of the circle in the Cartesian system is given by, x² + y² = a²

The equation of the same curve is polar coordinates will be given by, r = a

This means that the curve includes all the points on the plane whose distance from the origin is “a”. Notice that there is no θ in the equation, which means that whatever the angle made by the point from the x-axis. The distance is independent of that angle and is given by “a”. 

Area defined by Polar Graphs

Polar graphs are the curves that are defined in the polar coordinate system.

Consider a polar curve r = f(θ). A small change in angle (dθ) forms a narrow sector whose area is approximately: dA=\frac{1}{2}r^2d\theta

Adding the areas of all such sectors from (θ = a) to (θ = b) gives the total area enclosed by the curve:

A=\frac{1}{2}\int_a^b r^2d\theta, where (r = f(θ).

Example: Find the area enclosed by the cardioid (r = 1-cosθ) from (θ = 0) to (θ = ℼ /2).

Using the polar area formula, A=\frac{1}{2}\int_0^{\pi/2}(1-\cos\theta)^2d\theta

Expanding the square: A=\frac{1}{2}\int_0^{\pi/2}(1+\cos^2\theta-2\cos\theta)d\theta

Using ,\cos^2\theta=\frac{1+\cos2\theta}{2}

we get, A=\frac{1}{2}\int_0^{\pi/2}\left(\frac{3}{2}+\frac{\cos2\theta}{2}-2\cos\theta\right)d\theta

Integrating, A=\frac{1}{2}\left[\frac{3}{2}\theta+\frac{1}{4}\sin2\theta-2\sin\theta\right]_0^{\pi/2}

Substituting the limits, A=\frac{1}{2}\left(\frac{3\pi}{4}-2\right)

Therefore, the required area is \frac{3\pi}{8}-1

Area Between Two Polar Curves

Steps to Find the Area Between Two Polar Curves are as follow:

• Find the points of intersection by setting (r₁=r₂).
• Determine the interval of integration.
• Identify the outer and inner curves.
• Substitute into the area formula.
• Evaluate the integral.

Suppose two polar curves are given by , r=r₁(θ) and r=r₂(θ)

where r₁(θ) is the outer curve and r₂(θ) is the inner curve on the interval [a,b].

The area between the two curves is: A=\frac{1}{2}\int_a^b\left(r_1^2-r_2^2\right)d\theta ,where (a) and (b) are the angles at which the curves intersect.

Example :Consider two polar graphs that are given by, r = 3sin(θ) and r = 3cos(θ). The goal is to calculate the area enclosed between these curves.'

The curves intersect when 3\sin\theta=3\cos\theta \\ \frac{\sin\theta}{\cos\theta}=\frac{3}{3} \\ \tan\theta=1

so , \theta=\frac{\pi}{4}

The common region is symmetric about the line (\theta=\frac{\pi}{4}), so we first find half of the area and then double it.

For (0\le\theta\le\frac{\pi}{4}), the smaller radius is (r=3\sin\theta).

Thus, A_{\text{half}} =\frac{1}{2}\int_0^{\pi/4}(3\sin\theta)^2,d\theta =\frac{9}{2}\int_0^{\pi/4}\sin^2\theta,d\theta

Using \sin^2\theta=\frac{1-\cos2\theta}{2},
A_{\text{half}}=\frac{9}{4}\int_0^{\pi/4}(1-\cos2\theta)d\theta =\frac{9}{4}\left[\theta-\frac{\sin2\theta}{2}\right]_0^{\pi/4}\\[3pts]\frac{9}{4}\left(\frac{\pi}{4}-\frac{1}{2}\right).
Therefore, A=2A_{\text{half}} =\frac{9}{2}\left(\frac{\pi}{4}-\frac{1}{2}\right) =\frac{9\pi}{8}-\frac{9}{4} square units.

Sample Problems

Question 1: Find the area enclosed by the curve r = 3 in the first quadrant. 

The area can be calculated using the formula studied above, 

A = \frac{1}{2}\int^{\frac{\pi}{2}}_{0}r^2d\theta

A = \frac{1}{2}\int^{\frac{\pi}{2}}_{0}3^2d\theta

A = \frac{1}{2}\times9[\theta]^{\frac{\pi}{2}}_{0}

A = \frac{9\pi}{4}

Question 2: Find the area enclosed by the curve r = 5 + θ in the first two quadrants. 

The area can be calculated using the formula studied above, 

A = \frac{1}{2}\int^{\frac{\pi}{2}}_{0}r^2d\theta

A = \frac{1}{2}\int^{\pi}_{0}(5 + \theta)^2d\theta

A = \frac{1}{2}\int^{\pi}_{0}(25 + 10.\theta + \theta^2)d\theta

A = \frac{1}{2}[25\theta + 5.\theta^2 + \frac{\theta^3}{3})]^{\pi}_{0}

A = [\frac{25\pi}{2}+ \frac{5\pi^2}{2} + \frac{\pi^3}{6}]

Question 3: Find the area enclosed by the curve r = 5 + e^θ in the first two quadrants. 

The area can be calculated using the formula studied above, 

A = \frac{1}{2}\int^{\frac{\pi}{2}}_{0}r^2d\theta

A = \frac{1}{2}\int^{\pi}_{0}(5 + e^\theta)^2d\theta

A = \frac{1}{2}\int^{\pi}_{0}(25 + 10e^\theta + e^{2\theta})d\theta

A = \frac{1}{2}[25\theta + 10e^\theta + \frac{e^{2\theta}}{2})]^{\pi}_{0}

A = \frac{1}{2}(25\pi + 10e^\pi + \frac{e^{2\pi}}{2}-10-\frac{1}{2})

A = \frac{25\pi}{2} + 5e^\pi + \frac{e^{2\pi}}{4}-\frac{21}{4})

Question 4: Find the area under this curve from θ = 0 to θ = 180°: r = 1 + cos(θ)

A = \int^{b}_{a} \frac{1}{2}r^2d\theta

In this case, a = 0 and b = 180° and r = 1 + cos(θ)

Plugging these values into the formula, 

A = \int^{\pi}_{0} \frac{1}{2}r^2d\theta \\ = A = \int^{\pi}_{0} \frac{1}{2}(1  + cos(\theta))^2d\theta \\ = A = \frac{1}{2}\int^{\pi}_{0}(1 + cos^2(\theta) +2cos(\theta))d\theta

Rearranging the equation with the identity cos(2θ) = 1 - 2cos²(θ)

A =  \frac{1}{2}\int^{\pi}_{0}(1 + (\frac{1 + cos(2\theta)}{2}) +2cos(\theta))d\theta

A =  \frac{1}{2}\int^{\pi}_{0}(\frac{3}{2}  + \frac{cos(2\theta)}{2} +2cos(\theta))d\theta

A =  \frac{1}{2}[\frac{3}{2}\theta +2sin(\theta) + \frac{1}{4}sin(2\theta)]^{\pi}_{0} 

A =  \frac{3}{4}\pi

Practice Problem

Question 1: Find the area enclosed by the polar curve r = 4 for 0 ≤ θ ≤ 2π.

Question 2: Find the area enclosed by the cardioid r = 1+cos⁡θ for 0 ≤ θ ≤ 2π.

Question 3: Find the area of one petal of the rose curve r = 2sin⁡(2θ).

Question 4: Find the area between the curves r = 3 and r = 3sin⁡θ for 0 ≤ θ ≤ π.

Question 5: Find the area common to the curves r = 2sin⁡θ for r = 2cos⁡θ.