The area bounded by a curve of polar equation \( r(\theta) \) and by the rays \( \theta = \theta_1\) and \( \theta = \theta_2\) is given by the formula [1] [2] [3]
\[ \dfrac{1}{2}\int_{\theta_1}^{\theta_2} \; r^2(\theta) \; d\theta \]
Fig.1 - Curve and Limits of Integration in Polar Coordinates
Examples and Solutions
Example 1
Use the formula given above to find the area of the circle enclosed by the curve \( r(\theta) = 2 \sin (\theta) \) whose graph is shown below and compare the result to the formula of the area of a circle given by \( \pi r^2 \) where \( r \) is the radius..
Fig.2 - Circle in Polar Coordinates \( r ( \theta) = 2 \sin \theta \)
Solution to Example 1
Note that the circle is swept by the rays \( \theta = \theta_1 \) and \( \theta = \theta_2 \) and we need to find \( \theta_1 \) and \( \theta_2 \).
Fig.3 - Circle Bounded by \( r(\theta) = 2 \sin (\theta) \) Swept by the Rays \( \theta = \theta_1 \) and \( \theta = \theta_2 \)
The circle may be assumed to start at the origin such that \( r(\theta) = 0 \) and finish at the origin \( r(\theta) =0 \). Hence the angles \( \theta_1 \) and \( \theta_2 \) are found by solving \( r(\theta) = 0 \). Which gives the equation
\( \sin ( \theta) = 0 \)
which gives the following solutions
The solutions are: \( \theta_1 = 0 \) , \( \theta_2 = \pi \)
The area \( A \) of the circle is given by
\[ A = \dfrac{1}{2}\int_{0}^{\pi} \; (2 \sin \theta )^2 \; d\theta \]
Simplify
Use the trigonometric identity \( \sin^2 \theta = \dfrac{1}{2} (1 - cos(2\theta) )\) and simplify to obtain
\[ A = \int_{0}^{\pi} \; (1 - cos(2\theta) ) \; d\theta \]
Evaluate the integral
\[ A = \int_{0}^{\pi} \; \left[\theta -\dfrac{1}{2}\sin (2 \theta ) \right]_0^{\pi} = \pi \]
The area of the given circle whose radius is \( 1 \) could have been calculated using the formula \( \pi \cdot r^2 = \pi \cdot 1^2 = \pi \)
Example 2
Find the area of the region enclosed by the curve \( r(\theta) = \sin ( 3 \theta) \) whose graph is shown below.
Fig.4 -Plot of Curve \( r (\theta) = \sin (3 \theta) \)
Solution to Example 2
Because of the symmetry of the graph of the curve, we need to find the area enclosed by one loop and then multiply the result by \( 3 \).
Fig.5 - Pplot of Curve \( r (\theta) = \sin (3 \theta) \) with Limits of Integration
The loop on the right (in blue) starts at the origin \( r (\theta) = 0\) and finish at the origin \( r (\theta) = 0\). Hence \( \theta_1 \) and \( \theta_2 \) are found by solving \( r(\theta) = 0 \). Which gives the equation
\( \sin (3 \theta) = 0 \)
which gives the general solutions: \( 3 \theta = n \pi \) with \( n = 0, \pm 1, \pm 2, ...\)
We need the solutions in quadrant (I) which are given by
\( \theta_1 = 0 \) and \( \theta_2 = \dfrac{\pi}{3} \)
The area of the shaded (in blue) loop is given by
\[ A = \dfrac{1}{2} \int_{0}^{\frac{\pi}{3}} \; (\sin 3 \theta )^2 \; d\theta \]
Use the trigonometric identity \( \sin^2 (3\theta) = \dfrac{1}{2} (1 - cos(6\theta) )\) and simplify to obtain
\[ A = \dfrac{1}{4} \int_{0}^{\frac{\pi}{3}} \; (1 - cos(6\theta)) \; d\theta \]
Evaluate the integral
\[ A = \dfrac{1}{4} \left[\theta - \frac{1}{6}\sin (6\theta) \right]_0^{\frac{\pi}{3}}= \dfrac{\pi}{12} \]
The total are of all three loops is given by \[ 3 A = \dfrac{ \pi}{4} \]
Example 3
Find the area of the region enclosed by the curves \( r_1(\theta) = \sin (\theta) \) and \( r_2(\theta) = 0.5 (1+\cos ( 2\theta)) \) as shown below.
Fig.6 -Plot of the Curves \( r_1 (\theta) = \sin (\theta) \) and \( r_2(\theta) = 0.5 (1+\cos ( 2\theta)) \)
Solution to Example 3
Because of the symmetry, we need to find the area of the surface enclosed by the two curves on the right, as shown below, and multiply the result by two.
Fig.7 - Plot of the Curves \( r_1 (\theta) = \sin (\theta) \) and \( r_3(\theta) = 0.5 (1+\cos ( 2\theta)) \) with Limits of Integration
The area to be found is made up of two parts: \( A_1 \) and \( A_2 \) given by
\[ A_1 = \dfrac{1}{2}\int_{\theta_1}^{\theta_2} \; r_1^2(\theta) \; d\theta \]
\[ A_2 = \dfrac{1}{2}\int_{\theta_2}^{\theta_3} \; r_2^2(\theta) \; d\theta \]
Find the limits of integration \( \theta_1 \), \( \theta_2 \) and \( \theta_3 \).
a) \( \theta_1 \) is found by solving the equation \( r_1 (\theta) = 0 \) , hence \( \sin (\theta) = 0 \) which gives the general solution \( \theta_1 = n \pi \) with \( n = 0, \pm 1, \pm 2, ...\)
The solution needed is \( \theta_1 = 0 \)
b) \( \theta_3 \) is found by solving the equation \( r_2 (\theta) = 0 \) which gives \( 0.5 (1+\cos ( 2\theta)) = 0 \)
\( \cos ( 2\theta) = -1 \) which gives the general solutions: \( \theta = n \dfrac{\pi}{2} \) with \( n = 0, \pm 1, \pm 2, ...\)
The needed solution \( \theta_3 = \dfrac{\pi}{2} \)
c) \( \theta_2 \) correspond to a ray that passes through the point of intersection of the curves \( r_1(\theta) \) and \( r_2(\theta) \) and is found by solving the equation \( r_1(\theta) = r_2(\theta) \) writen as
\( \sin (\theta) = 0.5 (1+\cos ( 2\theta)) \)
Use the identity \( \cos ( 2\theta) = 1 - 2 \sin^2 \theta \), and rewrite the equation as
\( \sin (\theta) = 0.5 (1+(1 - 2 \sin^2 \theta)) \)
Simplify
\( \sin^2 \theta + \sin (\theta) - 1 = 0 \)
Solve the above quadratic like equation to obtain the solutions
\( \sin (\theta')=\dfrac{-1+\sqrt{5}}{2} \approx 0.61803\) and \( \quad \sin (\theta")=\dfrac{-1-\sqrt{5}}{2} \approx -1.61803 \)
The quation \( \quad \sin (\theta") \approx -1.61803 \) has no solution.
Hence the general solution in quadrant (I) is given by \( \theta' = \arcsin (0.61803) + 2 n \pi \)
which gives \( \theta_2 = 0.61803 \)
\( A_1 \) and \( A_2 \) may be written as
\[ A_1 = \dfrac{1}{2}\int_{0}^{0.61803} \; \sin^2(\theta) \; d\theta \]
\[ A_2 = \dfrac{1}{2}\int_{0.61803}^{\frac{\pi}{2}} \; (0.5 (1+\cos ( 2\theta)))^2 \; d\theta \]
Use the identity \( \sin^2 \theta = \dfrac{1}{2} (1 - cos(2\theta) )\) in \( A_1 \) and simplify
\[ A_1 = \dfrac{1}{4}\int_{0}^{0.61803} \; (1 - cos(2\theta) ) \; d\theta \]
Evaluate \( A_1 \)
\[ A_1 = \dfrac{1}{4}[\theta -\dfrac{1}{2}\sin (2\theta) ]_{0}^{0.61803} \approx 0.03644625 \]
Expand the integrand in \( A_2 \)
\[ A_2 = \dfrac{1}{2}\int_{0.61803}^{\frac{\pi}{2}} \; 0.5^2 (1+\cos^2( 2\theta)+2 \cos (2\theta)) \; d\theta \]
Use the identity \( \cos^2 (2\theta) = \dfrac{1}{2} (1 + cos(4\theta) )\) in \( A_2 \) and simplify
\[ A_2 = \dfrac{0.5^2}{2}\int_{0.61803}^{\frac{\pi}{2}} \; (1+ \dfrac{1}{2} (1 + cos(4\theta) ) + 2 \cos (2\theta)) \; d\theta \]
Evaluate \( A_2 \)
\[ \dfrac{0.5^2}{2} \left[ \frac{3}{2}\theta+\frac{1}{8}\sin (4 \theta )+\sin (2 \theta) \right]_{0.61803}^{\frac{\pi}{2}} \approx 0.050885\]
The total area \( A \) enclosed by the two curves is \( 2 (A_1 + A_2) \)
\[ A \approx 0.1746625 \]
More References and Links
University Calculus - Early Transcendental - Joel Hass, Maurice D. Weir, George B. Thomas, Jr., Christopher Heil - ISBN-13 : 978-0134995540
Calculus - Gilbert Strang - MIT - ISBN-13 : 978-0961408824
Calculus - Early Transcendental - James Stewart - ISBN-13: 978-0-495-01166-8
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