革命表面面积

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Previous concepts developed methods for determining the volume of a solid of revolution , using cross-sectional areas (slices, disks, washers). Associated with the volume is the surface area generated by the length of the function curve as it is revolved around the axis-of-revolution. We know how to compute the differential and total arc length of a function curve. How do we formulate the rotation of this length about the axis-of-rotation to determine area of a surface of revolution ?
::先前的概念开发了用来确定革命固体体积的方法,使用跨部门区域(切片、磁盘、洗衣机)确定革命固体体积。与体积相关的是围绕革命轴旋转的函数曲线长度产生的表面积。我们知道如何计算一个函数曲线的差异和总弧长度。我们如何在旋转轴上制定这一长度的旋转,以确定革命表面的面积?

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Area of a Surface of Revolution
::革命表面面积

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How do we find the area of a surface that is generated by revolving a curve about an axis or a line. For example, a circular cylinder can be generated by revolving a line segment about any axis that is parallel to it.
::我们如何在轴或线上找到旋转曲线所生成的表面区域。例如,圆圆圆圆柱体可以通过在与之平行的任何轴上旋转一条线段而生成。

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The surface area, $\begin{align*}S\end{align*}$ , of that revolution can be fairly easily determined to be $\begin{align*}S=2 \pi r h\end{align*}$ , where $\begin{align*}r\end{align*}$  is the radius of revolution, and $\begin{align*}h\end{align*}$  is the length (height) of the line that is being revolved.
::这场革命的表面积S可以很容易地确定为S=2rh,r是革命的半径,h是正在旋转的线的长度(高度)。

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The following definition and formulation of the area of a surface of revolution is based on revolving a differential arc length about an axis and integrating over the length of the revolution.
::以下对革命表面区域的定义和表述是基于将一个轴面的差弧长度旋转,并结合革命的长度。

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The area of a surface of revolution  is i f  $\begin{align*}f(x)\end{align*}$  is a smooth and non-negative function in the interval $\begin{align*}[a,b]\end{align*}$ , then the surface area  $\begin{align*}S\end{align*}$ generated by revolving the curve  $\begin{align*}y=f(x)\end{align*}$  about the $\begin{align*}x\end{align*}$ -axis is defined by
::革命表面的面积是如果 f(x) 在间隔[a,b] 中是一个平滑的非负负函数,那么通过旋转关于 x轴的曲线y=f(x) y=f(x) 所生成的表面积S的定义是

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$$\begin{align*}S=\int\limits^b_a 2 \pi f(x) \sqrt{1+[f^\prime (x)]^2} dx=\int\limits^b_a 2 \pi f (x) \sqrt{1+\left(\frac{dy}{dx}\right)^2}dx.\end{align*}$$
::S=ba2f(x)1+[f_(x)]2dx=ba2f(x)1+(uddx)2dx。

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Similarly: If  "> $\begin{align*}g(y)\end{align*}$  is a smooth and non-negative function in the interval $\begin{align*}[c,d]\end{align*}$ , then the surface area  $\begin{align*}S\end{align*}$ generated by revolving the curve  "> $\begin{align*}x=g(y)\end{align*}$  about the $\begin{align*}y\end{align*}$ -axis is defined by
::同样:如果g在[c,d]间隔内是一个平滑的非负负函数,那么通过旋转关于y轴的曲线 x=g所生成的表面积S的定义是:

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$$\begin{align*}S=\int\limits^d_c 2 \pi g(y) \sqrt{1+[g^\prime (y)]^2} dy=\int\limits^d_c 2 \pi g(y) \sqrt{1+\left(\frac{dx}{dy}\right)^2}dy.\end{align*}$$
::S=dc2g1+[g]2dy=dc2g1+(dxdy)2dy。

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Note that in each integrand the terms $\begin{align*}\sqrt{1+[f^\prime (x)]^2} dx\end{align*}$  and %5D%5E2%7D%20dy"> $\begin{align*}\sqrt{1+[g^\prime (y)]^2} dy\end{align*}$  represent differential arc length along $\begin{align*}f(x)\end{align*}$  and  "> $\begin{align*}g(y)\end{align*}$ introduced in a previous concept.
::请注意,在每整数中,1+[f_(x)]2dx和1+[g_]2dy的术语代表前一个概念中采用的f(x)和g的弧长度差。

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Show that the formula  $\begin{align*}S=4 \pi r^2\end{align*}$ for the surface area of a sphere with radius $\begin{align*}r\end{align*}$ , can be derived by rotating a semi- circle about the $\begin{align*}x\end{align*}$ -axis.
::显示一个半径为 r 的球的表面积的公式S=4r2, 可以通过在 X 轴上旋转半圆心来得出。

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The equation of a semi-circle in Quadrants I and II is given by $\begin{align*}f(x)=\sqrt{r^2-x^2}\end{align*}$ . Because this function is symmetric about the $\begin{align*}y\end{align*}$ -axis, consider rotating only the Quadrant I portion of the semi-circle about the $\begin{align*}x\end{align*}$ -axis to generate half a sphere, then multiplying the area result by two.
::F(x)r2-x2给出了二次半圆轴I和二次半圆轴的方程式。 由于此函数对准 Y 轴, 考虑仅旋转 X 轴半圆轴I 的 Quadrant I 部分来生成半球, 然后将区域结果乘以 2 。

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S = b ∫ a 2 π f ( x ) √ 1 + [ f ′ ( x ) ] 2 d x = 2 r ∫ 0 2 π √ r 2 − x 2 √ 1 + [ − x √ r 2 − x 2 ] 2 d x = 4 π r ∫ 0 √ r 2 − x 2 + x 2 d x = 4 π r [ x ] r 0 = 4 π r 2

$$\begin{align*}S &= \int\limits^b_a 2 \pi f (x) \sqrt{1+[f^\prime (x)]^2} dx\\ &= 2 \int\limits^r_0 2 \pi \sqrt{r^2-x^2} \sqrt{1+\left[\frac{-x}{\sqrt{r^2-x^2}}\right]^2}dx\\ &= 4 \pi \int\limits^r_0 \sqrt{r^2-x^2+x^2} dx\\ &= 4 \pi r [x]^r_0\\ &= 4 \pi r^2\end{align*}$$
::S=ba2f(x)1+[f_(x)]2dx=2r}022r2_x21+[-xr2-x2]2dx=4}0r2_r2,2dx=4}x2+x2&x2dx=4r[x]r0=4r2]

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Examples
::实例

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Example 1
::例1

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Earlier, you were asked how to formulate the rotation of function curve length about the axis-of-rotation to determine area of a surface of revolution. This concept presented the formulation as the integral of the circular area segment  $\begin{align*}2 \pi r \cdot d s\end{align*}$ along the appropriate coordinate axis.
::早些时候,有人问您如何对旋转轴的函数曲线长度进行旋转,以确定革命表面的面积。这个概念作为圆形区域部分 2rdsa 和适当的坐标轴的有机组成部分提出。

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Example 2
::例2

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Find the surface area that is generated by revolving  $\begin{align*}y=x^3\end{align*}$  on [0, 2] about the $\begin{align*}x\end{align*}$ -axis.
::查找 X 轴 [0, 2] 上的旋转 y=x3 生成的表面积。

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The surface area  $\begin{align*}S\end{align*}$  is
::S表层面积为

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$$\begin{align*}S &= \int\limits^b_a 2 \pi y \sqrt{1+\left(\frac{dy}{dx}\right)^2}dx\\ &= \int\limits^2_0 2 \pi x^3 \sqrt{1+(3x^2)^2}dx\\ &= 2 \pi \int\limits^2_0 x^3 (1+9x^4)^{\frac{1}{2}}dx.\end{align*}$$
::S=ba2y1+(uddx)2dx=202x31+(3x2)2x2=220x3(1+9x4)12dx。

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Using  $\begin{align*}u\end{align*}$ -substitution by letting  $\begin{align*}u=1+9x^4\end{align*}$ (and $\begin{align*}du=36x^3 dx\end{align*}$ ) 
::用 u =1+9x4 (和 = 36x3dx) 使用 u 替换

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$$\begin{align*}S &= 2 \pi \int\limits^{145}_1 u^{\frac{1}{2}} \frac{du}{36}\\ &= \frac{2 \pi}{36} \left[\frac{2}{3}u^{\frac{3}{2}}\right]^{145}_1\\ &= \frac{2 \pi}{36} \cdot \frac{2}{3} [(145)^{\frac{3}{2}}-1]\\ & \approx \frac{4 \pi}{108} [1745]\\ & \approx 203\end{align*}$$
::S=21451u12du36=236[23u32]1451=23623[(145)32-1]4108[1745]203

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Example 3
::例3

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Find the area of the surface generated by revolving the graph of  $\begin{align*}f(x)=x^2\end{align*}$ on the interval  $\begin{align*}[0, \sqrt{3}]\end{align*}$  about the $\begin{align*}y\end{align*}$ -axis.
::在y轴的间隔[0]-3]中查找通过旋转 f(x) =x2 图形生成的表面面积。

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Since the curve is revolved about the $\begin{align*}y\end{align*}$ -axis, we apply
::由于曲线是围绕 Y 轴旋转的, 我们申请

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$\begin{align*}S=\int\limits^d_c 2 \pi x \sqrt{1+\left(\frac{dx}{dy}\right)^2}dy\end{align*}$ .
::S=d=c2x1+(dxdy)2dy。

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So we write  $\begin{align*}y=x^2\end{align*}$  as  $\begin{align*}x=\sqrt{y}\end{align*}$ .
::因此,我们将y=x2写成 xy。

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In addition, the interval on the $\begin{align*}x\end{align*}$ -axis  $\begin{align*}[0, \sqrt{3}]\end{align*}$ becomes [0, 3]. Thus
::此外,X轴[0]-3的间隔也成为[0,3]。

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$$\begin{align*}S=\int\limits^3_0 2 \pi \sqrt{y} \sqrt{1+\left(\frac{1}{2\sqrt{y}}\right)^2}dy.\end{align*}$$
::S=302y1+(12y)2dy。

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Simplifying,
::简化,

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$\begin{align*}S=\pi \int\limits^3_0 \sqrt{4 y+1}dy\end{align*}$ .
::304y+1dy。

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With the aid of $\begin{align*}u\end{align*}$ -substitution, let  $\begin{align*}u=4y+1\end{align*}$ ,
::在u替代帮助下,让u=4y+1,

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$$\begin{align*}S &= \frac{\pi}{4} \int\limits^{13}_1 u^{\frac{1}{2}}du\\ &= \frac{\pi}{6} [(13)^{\frac{3}{2}}-1]\\ &= \frac{\pi}{6} [46.88-1]\\ & \approx 24\end{align*}$$
::S4131u12du6[(13)32-1]6[46.88-1]24]

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Example 4
::例4

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Find the surface area generated by revolving the function  $\begin{align*}f(x)=e^{0.1x}\end{align*}$  about the $\begin{align*}x\end{align*}$ -axis over the interval [0, 4].
::查找在 [0, 4] 间隔范围内旋转函数 f(x) =e0. 1x 生成的 x 轴的表面面积。

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Since the revolution is about the $\begin{align*}x\end{align*}$ -axis, the surface area formulation to use is $\begin{align*}S=\int\limits^b_a 2 \pi f(x) \sqrt{1+[f^\prime (x)]^2}dx\end{align*}$ .
::由于革命是关于x轴的,需要使用的表面积配方是S=ba2f(x)1+[f*(x)]2dx。

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With $\begin{align*}f^\prime (x)=0.1 e^{0.1x}\end{align*}$ , the surface area is computed as follows:
::F_(x)=0.1e0.1x,表面积计算如下:

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$$\begin{align*}S &= \int\limits^b_a 2 \pi f(x) \sqrt{1+[f^\prime (x)]^2}dx\\ &= \int\limits^4_0 2 \pi e^{0.1x} \sqrt{1+[0.1e^{0.1x}]^2}dx\\ &= 2 \pi \int\limits^4_0 e^{0.1x} \sqrt{1+[0.1e^{0.1x}]^2}dx\\ &= 31.14 && \ldots \text{Numberically evaluated using calculator.}\end{align*}$$
::S=ba2f(x)1+[f_(x)]2dx=402e0.1x1+[0.1e0.1x]2dx=240e0.1x}1+[0.1e0.1x]2dx=31.14...使用计算器进行数字评估。

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The area of the surface of revolution is 31.14 square units.
::革命表面面积为31.14平方单位。

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Review
::回顾

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For  #1-3 find the area of the surface generated by revolving the curve about the $\begin{align*}x\end{align*}$ -axis.
::对于 # 1-3 , 找到在 X 轴周围旋转曲线所生成的表面区域 。

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1. $\begin{align*}y=3x, 0 \le x \le 1\end{align*}$
   ::y=3x,0xx1
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2. $\begin{align*}y=\sqrt{x}, 1 \le x \le 9\end{align*}$
   ::yx,1xx%9 yx,1xx%9
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3. $\begin{align*}y=\sqrt{4-x^2,}-1 \le x \le 1\end{align*}$
   ::y4-x2, - 1x1

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For  #4-6 find the area of the surface generated by revolving the curve about the $\begin{align*}y\end{align*}$ -axis.
::对于#4-6, 找到在Y轴周围旋转曲线所生成的表面区域。

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4. $\begin{align*}x=7y+2, 0 \le y \le 3\end{align*}$
   ::x=7y+2,0y3
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5. $\begin{align*}x=y^3, 0 \le y \le 8\end{align*}$
   ::x=y3,0y8
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6. $\begin{align*}x=\sqrt{9-y^2}, -2 \le y \le 2\end{align*}$
   ::x9-y2,-22
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7. Show that the surface area of a sphere of radius  $\begin{align*}r\end{align*}$ is $\begin{align*}4 \pi r^2\end{align*}$ .
   ::显示半径 r 的表面区域为 4°r2。
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8. Show that the lateral area  $\begin{align*}S\end{align*}$ of a right circular cone of height  $\begin{align*}h\end{align*}$ and base radius  $\begin{align*}r\end{align*}$ is $\begin{align*}S=\pi r \sqrt{r^2+h^2}\end{align*}$ .
   ::显示右圆锥体高度 h 和基半径 r 的平面区域 S 是 Srr2+h2。
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9. Find the surface area generated by rotating the line  $\begin{align*}y=x\end{align*}$ about the $\begin{align*}x\end{align*}$ -axis on the interval $\begin{align*}0 < x < 5\end{align*}$ .
   ::查找在 0 < x < 5 间距的 X 轴上旋转线 y=x 所生成的表面面积。
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10. Set up, but do not solve, an integral to calculate the surface area created by revolving  $\begin{align*}y=\cos x, \frac{\pi}{4} < x < \frac{\pi}{2}\end{align*}$ about the $\begin{align*}x\end{align*}$ -axis.
    ::设置, 但不解析, 用于计算 y=cosx, 4 <x2 所创建的关于 x 轴的表面面积 。
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11. Find the surface area generated by rotating the curve  $\begin{align*}y=\sqrt{1-x^2}, 0 < x < 0.5\end{align*}$ about the $\begin{align*}x\end{align*}$ -axis.
    ::查找旋转曲线 y1- x2,0 < x < 0.5 所生成的关于 x 轴的表面积。
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12. Find the surface area generated by rotating the curve $\begin{align*}y=\sqrt{x}, 1 < x < 4\end{align*}$ , about the $\begin{align*}x\end{align*}$ -axis.
    ::查找旋转 yx,1 < x<4 曲线所生成的关于 x 轴的表面区域。
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13. Find the surface area generated by rotating the line  $\begin{align*}y=x\end{align*}$ about the $\begin{align*}y\end{align*}$ -axis on the interval $\begin{align*}0 < x < 5\end{align*}$ .
    ::查找在0 < x < 5 间距上的 Y 轴上旋转 y =x 线所生成的表面区域。
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14. Set up, but do not solve, an integral to calculate the surface area created by revolving  $\begin{align*}y=\cos x, \frac{\pi}{4}, < x < \frac{\pi}{2}\end{align*}$ about the $\begin{align*}y\end{align*}$ -axis.
    ::设置, 但不解析, 以计算 y=cosx, 4, <x2 所创建的 Y 轴的表面面积 。
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15. Find the surface area generated by rotating the curve $\begin{align*}y=\sqrt{1-x^2}, 0 < x < 0.5\end{align*}$ , about the $\begin{align*}y\end{align*}$ -axis.
    ::查找旋转 y1- x2, 0 < x < 0.5, 所生成的 Y 轴的表面积 。
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16. Set up, but do not solve, two integrals to calculate the surface area generated by rotating the curve $\begin{align*}y=\sqrt{x}, 1 < x < 4\end{align*}$ , about the $\begin{align*}y\end{align*}$ -axis. One should be in terms of $\begin{align*}x\end{align*}$ , one in terms of $\begin{align*}y\end{align*}$ .
    ::设置, 但不解析, 两个集成件来计算旋转 y 轴的 y 轴 y x, 1 < x < 4 所生成的表面积。 一个应该是 x, 一个是 y 。
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17. Find the surface area generated by rotating the curve $\begin{align*}y=x^2, 1 < x < 3\end{align*}$ , about the $\begin{align*}y\end{align*}$ -axis.
    ::查找旋转 y =x2, 1 < x < 3 曲线生成的 Y 轴的表面积 。

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Review (Answers)
::回顾(答复)

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Click to see the answer key or go to the Table of Contents and click on the Answer Key under the 'Other Versions' option.
::单击可查看答题键, 或转到目录中, 单击“ 其他版本” 选项下的答题键 。