AP Calculus BC 9.9 Finding the Area of the Region Bounded by Two Polar Curves Study Notes - New Syllabus
AP Calculus BC 9.9 Finding the Area of the Region Bounded by Two Polar Curves Study Notes- New syllabus
AP Calculus BC 9.9 Finding the Area of the Region Bounded by Two Polar Curves Study Notes – AP Calculus BC- per latest AP Calculus BC Syllabus.
LEARNING OBJECTIVE
- Definite integrals allow us to solve problems involving the accumulation of change in area or volume over an interval.
Key Concepts:
- Finding the Area of the Region Bounded by Two Polar Curves
Finding the Area of the Region Bounded by Two Polar Curves
Finding the Area of the Region Bounded by Two Polar Curves
In polar coordinates, a curve is represented as \( r = f(\theta) \), where \( r \) is the distance from the origin and \( \theta \) is the angle from the polar axis. When two polar curves are given, the goal is to find the area that lies inside one curve and outside the other over a specified interval of \( \theta \).
General Approach:
1. Identify the curves: Let the two curves be \( r = f(\theta) \) (outer curve) and \( r = g(\theta) \) (inner curve), with \( f(\theta) \geq g(\theta) \) on the interval of interest.
2. Determine the bounds of integration: Find the intersection points by solving \( f(\theta) = g(\theta) \) for \( \theta \) within the relevant range (often \( [0, 2\pi] \)).
3. Set up the area formula: The area between the curves is given by:
$ A = \dfrac{1}{2} \int_{\alpha}^{\beta} \left( [f(\theta)]^2 – [g(\theta)]^2 \right) \, d\theta $
Here, \( [f(\theta)]^2 \) is the squared distance of the outer curve from the origin, and \( [g(\theta)]^2 \) is that of the inner curve.
4. Integrate carefully: Simplify the integrand using trigonometric identities where needed, and evaluate the definite integral over the specified bounds.
5. Interpret the result: The result represents the area of the region bounded between the curves in square units.
Important Notes:
- If the curves intersect multiple times, the area may need to be computed over multiple intervals and summed.
- Symmetry (about the polar axis, vertical axis, or origin) can simplify calculations by allowing you to compute over part of the region and multiply the result.
- When deciding which curve is “outer” and which is “inner”, check values of \( r \) at different angles to ensure the ordering is correct over the integration range.
Example
The polar curves \( r = 2 – 2\cos\theta \) and \( r = 2 + 2\cos\theta \) are shown.
Which expression gives the total area of the shaded regions?
A. \( \displaystyle \int_{0}^{\pi} (2 + 2\cos\theta)^{2}\, d\theta \)\;
B. \( \displaystyle \int_{\pi/2}^{\pi} (2 + 2\cos\theta)^{2}\, d\theta \)\;
C. \( 8 \displaystyle \int_{0}^{\pi/2} (1 – \cos\theta)^{2}\, d\theta \)\;
D. \( \displaystyle \int_{0}^{\pi/2} \big((2 – 2\cos\theta)^{2} + (2 + 2\cos\theta)^{2}\big)\, d\theta \)
▶️ Answer/Explanation
Each shaded “cap” is the region common to the two curves near the \( y \)-axis. For \( 0 \le \theta \le \dfrac{\pi}{2} \)
we have \( \cos\theta \ge 0 \), so \( r_{\text{smaller}} = 2 – 2\cos\theta \) and the area element of the overlap is \( \dfrac{1}{2} r_{\text{smaller}}^{2}\, d\theta \).
By symmetry there are two caps (top and bottom) in the right half plane and two matching contributions from the left half plane, which together contribute a factor of \( 4 \). Hence
\( \text{Total Area} = 4 \cdot \dfrac{1}{2} \displaystyle \int_{0}^{\pi/2} (2 – 2\cos\theta)^{2}\, d\theta = 2 \displaystyle \int_{0}^{\pi/2} 4(1 – \cos\theta)^{2}\, d\theta = 8 \displaystyle \int_{0}^{\pi/2} (1 – \cos\theta)^{2}\, d\theta. \)
Correct choice: C.
Example
Find the area of the region inside \( r = 3 \) and outside \( r = 2\cos\theta \) over the range where they intersect.
▶️ Answer/Explanation
First, find intersection: \( 3 = 2\cos\theta \) gives \( \cos\theta = \dfrac{3}{2} \), which is not possible. This means the curves do not intersect; \( r=3 \) is always outside \( r = 2\cos\theta \).
Since they do not intersect, we find the area over \( 0 \leq \theta \leq 2\pi \):
$ A = \dfrac{1}{2} \int_{0}^{2\pi} \left( 3^2 – [2\cos\theta]^2 \right) \, d\theta $
$ A = \dfrac{1}{2} \int_{0}^{2\pi} \left( 9 – 4\cos^2\theta \right) \, d\theta $
Use \( \cos^2\theta = \dfrac{1 + \cos 2\theta}{2} \):
$ A = \dfrac{1}{2} \int_{0}^{2\pi} \left( 9 – 2 – 2\cos 2\theta \right) \, d\theta $
$ A = \dfrac{1}{2} \int_{0}^{2\pi} \left( 7 – 2\cos 2\theta \right) \, d\theta $
$ A = \dfrac{1}{2} \left[ 7\theta – \sin 2\theta \right]_{0}^{2\pi} $
Evaluate: \(\sin 4\pi = 0\), \(\sin 0 = 0\).
$ A = \dfrac{1}{2} \left[ 7(2\pi) \right] = 7\pi $
Final Answer: \( A = 7\pi \) square units.
Example
Find the area between \( r = 4\cos\theta \) and \( r = 2 \) from \( \theta = -\dfrac{\pi}{3} \) to \( \theta = \dfrac{\pi}{3} \).
▶️ Answer/Explanation
Outer curve: \( r = 4\cos\theta \), inner curve: \( r = 2 \).
$ A = \dfrac{1}{2} \int_{-\pi/3}^{\pi/3} \left( [4\cos\theta]^2 – [2]^2 \right) \, d\theta $
$ A = \dfrac{1}{2} \int_{-\pi/3}^{\pi/3} \left( 16\cos^2\theta – 4 \right) \, d\theta $
Use \( \cos^2\theta = \dfrac{1 + \cos 2\theta}{2} \):
$ A = \dfrac{1}{2} \int_{-\pi/3}^{\pi/3} \left( 8 + 8\cos 2\theta – 4 \right) \, d\theta $
$ A = \dfrac{1}{2} \int_{-\pi/3}^{\pi/3} \left( 4 + 8\cos 2\theta \right) \, d\theta $
Integrate: \( \int 4 \, d\theta = 4\theta \), \( \int 8\cos 2\theta \, d\theta = 4\sin 2\theta \)
$ A = \dfrac{1}{2} \left[ 4\theta + 4\sin 2\theta \right]_{-\pi/3}^{\pi/3} $
Since \( \sin 2\theta \) is odd, the sine term cancels.
$ A = \dfrac{1}{2} \times \left[ 4\left( \dfrac{\pi}{3} – \left( -\dfrac{\pi}{3} \right) \right) \right] = \dfrac{1}{2} \times \dfrac{8\pi}{3} = \dfrac{4\pi}{3} $
Final Answer: \( A = \dfrac{4\pi}{3} \) square units.
Example
The figure shows the polar curves \( r = 3\cos(3\theta) \) and \( r = 3 \).
What is the sum of the shaded areas?
▶️ Answer/Explanation
The shaded region is the area inside the circle \( r = 3 \) but outside the three-petaled rose \( r = 3\cos(3\theta) \). Since the rose lies entirely within the circle of radius 3, the shaded area equals \( \text{Area(circle)} – \text{Area(rose)} \).
Circle area: \( A_{\text{circle}} = \pi(3)^2 = 9\pi \).
Rose area: For \( r = a\cos(k\theta) \) with \( k \) odd, each petal area is \( \dfrac{a^2\pi}{4k} \).
Here \( a = 3 \) and \( k = 3 \), so the total area of all 3 petals is \( A_{\text{rose}} = \dfrac{a^2\pi}{4} = \dfrac{9\pi}{4} \).
(Derivation for one petal: \( A_{\text{petal}} = \dfrac{1}{2}\displaystyle \int_{-\pi/(2k)}^{\pi/(2k)} a^2\cos^2(k\theta)\, d\theta = \dfrac{a^2\pi}{4k} \).)
Shaded area: \( A_{\text{shaded}} = 9\pi – \dfrac{9\pi}{4} = \dfrac{27\pi}{4} \).
Final Answer: \( \dfrac{27\pi}{4} \).