8.1 Arc Length goo.gl/TuKjFG 1
Chapter 8
Further Applications of
Integra-tion
8.1
Arc Length (page 544)
W78DcwQWyfo 認識曲線長度的積 分公式, 在分割樣 本點取和求極限的 過 程 中, 所 有 分 割點都必須選在曲 線 上, 而 每 一 小 段的曲線弧長利用 線段長估計。 透過 均值定理, 線段長 可以和當中某一點 的切線斜率產生連 繫, 就把那一點當 成樣本點。 最後得 到的積分函數, 本 質上是切向量的長 度, 會在本單元的 最後補充說明。
The Arc Length Formula (page 544). If f′
is continuous on [a, b], then the length of the
curve y = f (x), a ≤ x ≤ b, is L= Z b a p1 + (f′(x))2dx = Z b a s 1 + dy dx 2 dx. (1)
If a curve has the equation x= g(y), c ≤ y ≤ d, and g′
(y) is continuous, then by interchanging
the roles of x and y, we obtain the following formula for its length:
L= Z d c p1 + (g′ (y))2dy = Z d c s 1 + dx dy 2 dy. (2)
Proof. Suppose that a curve C is defined by the equation y = f (x), where f (x) is continuous and a ≤ x ≤ b. We obtain a polygonal approximation to C by dividing the interval [a, b] into nsubintervals with endpoints x0, x1, . . . , xnand equal width ∆x. If yi= f (xi), then the point
Pi(xi, yi) lies on C and the polygon with vertices P0, P1, . . . , Pn is an approximation to C.
x y
Figure 1: We use the length of inscribed polygons to approximate the length of C.
We define the length L (弧長) of the curve C with equation y = f (x), a ≤ x ≤ b, as the
limit of the lengths of these inscribed polygons (if the limit exists): L = lim
n→∞ n P i=1|P i−1Pi|. When f′
is continuous on [a, b] (we say f ∈ C1([a, b])), then by the Mean Value Theorem, there is a number x∗
i ∈ (xi−1, xi) such that
|Pi−1Pi| =p(xi− xi−1)2+ (yi− yi−1)2 =p(∆x)2+ (f (xi) − f (xi−1))2
= q (∆x)2+ (f′(x∗ i)∆x)2 = q 1 + (f′(x∗ i))2∆x. Therefore, L= lim n→∞ n X i=1 |Pi−1Pi| =
2 8.1 Arc Length goo.gl/TuKjFG
To get the formula (2), by similar discussion, we divide c ≤ y ≤ d into n subinterval with endpoints y0, y1, . . . , yn and equal width ∆y, and rewrite |Pi−1Pi| as
|Pi−1Pi| =p(xi− xi−1)2+ (yi− yi−1)2=p(g(yi) − g(yi−1))2+ (∆y)2
= q (g′(y∗ i)∆x)2+ (∆y)2 = q 1 + (g′(y∗ i))2∆y, so L= lim n→∞ n X i=1 |Pi−1Pi| = 8VHJ7e3zRIw 若是把上半圓與下 半圓分別看成是函 數的圖形, 計算弧 長時要處理的積分 會是瑕積分, 這是 因為兩端點的切線 斜率會是正負無限 大的關係。
Example 1. Show that the circumference of a circle with radius r is 2πr.
Solution.
Example 2. Find the length of the curve x = 13√y(y − 3), 1 ≤ y ≤ 9.
這條曲線可以確實 算出弧長, 是因為 根號內部又可以整 理成完全平方的形 式, 就可以和根號 去掉。 各位平時應 該要一直記住: 平 方再開根號要加絕 對值, 所以平時應 養成習慣, 確定平 方與根號去掉後的 量是正量。 這個例題另一方面 示範以 y 為變數 的方式計算曲線弧 長。 Solution.
8.1 Arc Length goo.gl/TuKjFG 3
Example 3 (page 549). Find the length of the curve x23 + y 2 3 = a 2 3. YovNCXE2krk 各 位 在 看 影 片 前, 應該先猜一猜: 你 覺得星形線的長度 會比圓周長大還是 小? 然後再看影片 以確定你的直覺是 否正確。 Solution.
The Arc Length Function
弧長函數應細細體 會, 這個函數就是 度量從某個點為起 始點之下, 到對應 點之間的弧長。 弧長函數的變化率 是該點的切向量長 度, 這是微積分基 本定理順勢得出的 結果。
We will find it useful to have a function that measures the arc length of a curve from a particular starting point to any other point on the curve. If a smooth curve C has the equation y = f (x), a ≤ x ≤ b, let s(x) be the distance along C from the initial point P0(a, f (x)) to the
point Q(x, f (x)). Then
s(x) = Z x
a
p1 + (f′(t))2dt
is a function, called the arc length function (弧長函數). By the Fundamental Theorem of Calculus, we get
ds dx =p1 + (f ′(x))2 = s 1 + dy dx 2
In differential sense, we can view the arc length as the infinitesimal Pythagorean theorem: (ds)2= (dx)2+ (dy)2. Similarly, we have ds = s 1 + dx dy 2 dy.
Remark 4. In general, we can viewed the curve as r(x) = (x, f (x)). Then we have r′
(x) = (1, f′
(x)) and |r′
(x)| =p1 + (f′(x))2, so the arc-length formula becomes
s(x) = Z x
a |r ′
4 8.2 Arc Length goo.gl/7akKV6
8.2
Area of a Surface of Revolution (page 551)
In this section, we will derive the formula of the area of a surface of revolution.
KV6ZBsY74V0 理解 圓 柱 表 面 積、 圓錐表面積、 錐台 表面積的推導。 學 習這些幾何物件的 表面積之目的是在 下一個影片中要推 導旋轉體表面積公 式。
Example 1(page 551). Find the lateral surface area of a circular cylinder with radius r and height h.
Solution.
Example 2 (page 551). Find the lateral surface area of a circular cone with base radius r and slant height l.
Solution.
Example 3(page 551). Find the lateral surface area of a band, which is a portion of a circular cone with upper radius r1, lower radius r2, and slant height l.
8.2 Arc Length goo.gl/7akKV6 5
Example 4 (page 552). Find the surface area of the surface obtained by rotating the curve y = f (x), a ≤ x ≤ b, about the x-axis. (Assume that f ∈ C1([a, b]).)
Y6njzaRYuEE 各位應該花時間好 好體會旋轉體表面 積的公式推導, 給 定一條曲線, 利用 分割樣本點取和求 極限的過程, 所有 分割點都必須選在 曲線上, 在每個小 區間內用錐台的表 面積估計旋轉體表 面積, 最後得到的 公式在最下方。 這裡應強調的是證 明的難點, 在步驟 (c) 中, 對於曲線 弧長的部份, 利用 均值定理得到了樣 本點 x∗ i, 但是另 一方面錐台表面積 的公式中, 是選用 了中點 x∗∗ i ,這是 不同的點, 所以不 能直接寫出對應的 積 分 式(回想 定積 分的定義, 在黎曼 和的階段, 每一小 段當中必要選到一 模一樣的樣本點)。 為此, 我們在 (c) 的最後一步驟, 故 意做 f(x∗ i) 一加 一減, 則其中一部 份 可 以 湊 出積 分 式; 另一方面, 必 須再進一步估計剩 下的量是高階的無 窮小量, 這段的論 述就是 (d) 後面 的估計。 Solution.
(a) Partition: a = x0 < x1< x2<· · · < xn= b. We have ∆x = xi− xi−1= b−an .
(b) Approximate the curve (x, f (x)) by polygons ∪n
i=1Pi−1Pi, where Pi= (xi, f(xi)).
(c) By the mean value theorem, the surface area is approximated by
Sn= n X i=1 2π f (xi−1) + f (xi) 2 Pi−1Pi = n X i=1 2π f (xi−1) + f (xi) 2 p(f(xi) − f (xi−1))2+ (xi− xi−1)2 = where x∗∗ i ∈ [xi−1, xi] satisfies f (x∗∗i ) = f(xi−1)+f (xi) 2 and x ∗ i ∈ [xi−1, xi]. (d) When n → ∞, we have lim n→∞Sn= limn→∞ n X i=1 2π (f (x∗ i) + (f (x ∗∗ i ) − f (x ∗ i))) q 1 + (f′(x∗ i))2∆x = lim n→∞ n X i=1 2πf (x∗ i) q 1 + (f′ (x∗ i))2∆x + limn→∞ n X i=1 2π(f (x∗∗ i ) − f (x ∗ i)) q 1 + (f′ (x∗ i))2∆x = Z b a 2πf (x)p1 + (f′(x))2dx.
Here we need to estimate (use mean value theorem and extreme value theorem)
n X i=1 2π |f (x∗∗ i ) − f (x ∗ i)| q 1 + (f′(x∗ i))2∆x ≤ n X i=1 2π|f′ (x∗∗∗ i )∆x| q 1 + (f′(x∗ i))2∆x ≤ 2πM p 1 + M2 n X i=1 (∆x)2 = 2πMp1 + M2 n X i=1 b − a n2 2 = 2πMp1 + M2(b − a)2· 1 n → 0 as n → ∞. AIE-P3uG6cA 旋 轉 體 表 面 積 公 式: 圓周長乘以弧 長的積分。
Definition 5. Surface area (表面積) of the surface obtained by rotating the curve y = f(x), a ≤ x ≤ b, about x-axis is S = Z b a 2πf (x)p1 + (f′(x))2dx = Z b a 2πy s 1 + dy dx 2 dx.
6 8.2 Arc Length goo.gl/7akKV6
Recall that the differential of the arc length function is
(ds)2= (dx)2+ (dy)2 ⇒ ds = s 1 + dx dy 2 dy. (3)
If the curve is described as x = g(y), c ≤ y ≤ d, then the formula for surface area (rotating about x-axis) becomes
S= Z d c 2πy s 1 + dx dy 2 dy. (4)
Formula (3) and (4) can be formally written as
Z
2πy ds.
For rotation about the y-axis, the surface area formula (formally) becomes
S= Z 2πx ds, where ds = s 1 + dy dx 2 dx = s 1 + dx dy 2 dy
表面積公式: 圓周長乘以弧長的積分。 例題示範旋轉體表 面積的計算。Example 6. Find the area of surface obtained by rotating y = sin x, 0 ≤ x ≤ π, about the x-axis.
8.2 Arc Length goo.gl/7akKV6 7
Example 7 (page 556). Consider the region R = {(x, y)|x ≥ 1, 0 ≤ y ≤ 1x} rotating about the x-axis. 4qH7fgJuWyA 這個例子在數學上 具有非常深刻的意 義, 一個體積有限 的實體, 它的表面 積可以無限, 也就 是你無法拿油漆幫 這個實心物體完全 上色。 這個例子衝 擊到大家對於不同 維度之間的一些直 觀看法。
(a) Show that the volume of the resulting solid is finite.
(b) Show that the surface area is infinite. (The surface is called Gabriel’s horn.)
