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Theorem 2: Area of Region bounded by a Curve

THEOREM 2

Let r = f(θ) be continuous and r ≥ 0 for a ≤ θ ≤ b, where b ≤ a + 2π. Then the region R bounded by the curve r = f(θ) and the lines 0 = a and 0 - b has area

07_trigonometric_functions-521.gif

Discussion

Imagine a point P moving along the curve r = f(θ) from θ = a to θ = b. The line OP will sweep out the region R in Figure 7.9.2. Since b ≤ a + 2π, the line will complete at most one revolution, so no point of R will be counted more than once.

07_trigonometric_functions-522.gif07_trigonometric_functions-523.gif

Figure 7.9.2                                                         Figure 7.9.3

The formula for area can be seen intuitively by considering an infinitely small wedge ΔA of R between θ and θ + Δθ. (Figure 7.9.3). The wedge is almost a sector

of a circle of radius j(θ) with central angle Δθ, so

ΔA ≈ ½ f(θ)2 Aθ (compared to Δθ). By the Infinite Sum Theorem,

07_trigonometric_functions-524.gif

The actual proof follows this intuitive idea but the area of ΔA must be computed more carefully.

07_trigonometric_functions-525.gif

Figure 7.9.4

PROOF

Let Δθ be positive infinitesimal and let θ be a hyperreal number between a and b - Aθ. Consider the wedge of R with area ΔA between θ and θ + Aθ. Since f(θ) is continuous, it has a minimum value m and maximum value M between θ and θ + Aθ, and furthermore,

m ≈ f(θ), M ≈ f(θ).

The sector between θ and Aθ of radius m is inscribed in ΔA while the sector of radius M is circumscribed about ΔA. (Figure 7.9.4 shows the inscribed and circumscribed sectors for real Aθ and infinitesimal Δθ.) By Theorem 1, the two sectors have areas ½m2 Δθ and ½M2 Δθ. Moreover, ΔA is between those two areas,

½m2 Δθ ≤ ΔA ≤ ½M2 Δθ, ½W ≤ ΔA/Aθ ≤ ½M2

Taking standard parts,

½f (θ)2 ≤ st(ΔA/Δθ) ≤ ½f (θ)2.

Therefore

ΔA/Aθ x ½f(θ)2,

and by the Infinite Sum Theorem,

07_trigonometric_functions-526.gif

Theorem 1 is also true in the case that r = f(θ) is continuous and r ≤ 0. Since

07_trigonometric_functions-527.gif

the region R bounded by the curve r = f(θ) has the same area as the region S bounded by the curve r = -f(θ). Both areas are positive. As we can see from Figure 7.9.5, S looks exactly like R but is on the opposite side of the origin.

07_trigonometric_functions-528.gif

Figure 7.9.5

Example 1: Area of a "Four-leaf Clover"

Our next example shows why the hypothesis that r has the same sign for a ≤ θ ≤ b is needed in Theorem 2r.


Last Update: 2006-11-08