AP Calculus BC 9.8 Find the Area of a Polar Region or the Area Bounded by a Single Polar Curve Study Notes - New Syllabus
AP Calculus BC 9.8 Find the Area of a Polar Region or the Area Bounded by a Single Polar Curve Study Notes- New syllabus
AP Calculus BC 9.8 Find the Area of a Polar Region or the Area Bounded by a Single Polar Curve 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:
- Find the Area of a Polar Region or the Area Bounded by a Single Polar Curve
Find the Area of a Polar Region or the Area Bounded by a Single Polar Curve
Find the Area of a Polar Region or the Area Bounded by a Single Polar Curve
In polar coordinates, the location of a point is determined by a radius \( r \) (distance from the origin) and an angle \( \theta \) (measured from the positive \( x \)-axis). The area of a region bounded by a polar curve can be found using integration techniques adapted to polar geometry.
For a polar curve \( r = f(\theta) \) between angles \( \theta = \alpha \) and \( \theta = \beta \), the area \( A \) is given by:
$ A = \dfrac{1}{2} \int_{\alpha}^{\beta} \left[ f(\theta) \right]^2 \, d\theta $
- This formula comes from summing up infinitely many sectors of radius \( r \) and infinitesimal angle \( d\theta \).
- The factor \( \dfrac{1}{2} \) arises from the formula for the area of a sector: \( \dfrac{1}{2} r^2 \theta \).
Derivation of the Formula
- The area of a small sector with radius \( r \) and angle \( \Delta \theta \) is: $ \Delta A \approx \dfrac{1}{2} r^2 \, \Delta \theta $
- Replacing \( r \) with \( f(\theta) \) and letting \( \Delta \theta \to 0 \): $ dA = \dfrac{1}{2} \left[ f(\theta) \right]^2 \, d\theta $
- Integrating from \( \alpha \) to \( \beta \) gives the total area: $ A = \dfrac{1}{2} \int_{\alpha}^{\beta} \left[ f(\theta) \right]^2 \, d\theta $
Choosing Limits of Integration
- Identify the range of \( \theta \) that sweeps out the desired region.
- For a complete curve, the limits often correspond to one full petal or symmetric section (which can be multiplied for total area).
- When curves intersect, solve \( f_1(\theta) = f_2(\theta) \) to find the intersection angles for the limits.
Symmetry Considerations
If the curve has symmetry, you can compute the area of one symmetric section and multiply by the number of such sections:
- Symmetry about the polar axis (horizontal line): replace \( \theta \) with \( -\theta \).
- Symmetry about the vertical line \( \theta = \dfrac{\pi}{2} \): replace \( \theta \) with \( \pi – \theta \).
- Symmetry about the pole (origin): replace \( r \) with \( -r \) or \( \theta \) with \( \theta + \pi \).
Area Between Two Polar Curves
If two curves \( r = f(\theta) \) and \( r = g(\theta) \) share the same center and satisfy \( f(\theta) \geq g(\theta) \) for all \( \theta \) in the interval, then the area between them is:
$ A = \dfrac{1}{2} \int_{\alpha}^{\beta} \left( \left[ f(\theta) \right]^2 – \left[ g(\theta) \right]^2 \right) \, d\theta $
Units
- If \( r \) is in meters and \( \theta \) is in radians, the resulting area will be in square meters (\( m^2 \)).
- It is essential that \( \theta \) be in radians for the formula to work correctly.
Example
Find the area enclosed by one petal of the polar curve \( r = 4\sin(2\theta) \).
▶️ Answer/Explanation
One petal occurs when \( \theta \) runs from \( 0 \) to \( \dfrac{\pi}{2} \).
Using the polar area formula:
$ A = \dfrac{1}{2} \int_{0}^{\pi/2} \left[ 4\sin(2\theta) \right]^2 \, d\theta $
$ A = \dfrac{1}{2} \int_{0}^{\pi/2} 16\sin^2(2\theta) \, d\theta $
Use the identity \( \sin^2(2\theta) = \dfrac{1 – \cos(4\theta)}{2} \):
$ A = 8 \int_{0}^{\pi/2} \dfrac{1 – \cos(4\theta)}{2} \, d\theta $
$ A = 4 \int_{0}^{\pi/2} \left( 1 – \cos(4\theta) \right) \, d\theta $
$ A = 4\left[ \theta – \dfrac{\sin(4\theta)}{4} \right]_{0}^{\pi/2} $
$ A = 4\left[ \dfrac{\pi}{2} – 0 \right] = 2\pi $
Final Answer: \( A = 2\pi \) square units.
Example
Find the total area enclosed by the cardioid \( r = 3 + 3\cos\theta \).
▶️ Answer/Explanation
The curve is symmetric about the polar axis, so we compute the area from \( 0 \) to \( \pi \) and double it.
$ A = 2 \times \dfrac{1}{2} \int_{0}^{\pi} \left[ 3 + 3\cos\theta \right]^2 \, d\theta $
$ A = \int_{0}^{\pi} 9(1 + \cos\theta)^2 \, d\theta $
Expand: \( (1 + \cos\theta)^2 = 1 + 2\cos\theta + \cos^2\theta \)
Replace \( \cos^2\theta \) with \( \dfrac{1 + \cos 2\theta}{2} \):
$ A = 9 \int_{0}^{\pi} \left( \dfrac{3}{2} + 2\cos\theta + \dfrac{\cos 2\theta}{2} \right) \, d\theta $
$ A = 9\left[ \dfrac{3}{2}\theta + 2\sin\theta + \dfrac{\sin 2\theta}{4} \right]_{0}^{\pi} $
Evaluate: \( \sin\pi = 0, \ \sin 2\pi = 0 \)
$ A = 9\left[ \dfrac{3}{2} \pi + 0 + 0 \right] = \dfrac{27\pi}{2} $
Final Answer: \( A = \dfrac{27\pi}{2} \) square units.
Example
Find the area between the curves \( r = 2 \) and \( r = 4\cos\theta \) 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} $
Evaluate: symmetry makes \( \sin 2\theta \) term vanish at \( \pm\pi/3 \).
$ A = \dfrac{1}{2} \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.