10.15 极表和微积分:表层

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• 选择节极表和微积分:表层面积

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  极表和微积分:表层面积

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  On a weekend visit to her grandmother’s farm, Maia sees her grandmother peel an apple so that the peel remains a single, unbroken piece. Maia wonders if there’s a way to estimate the surface area of the apple’s peel. She’s been studying cardioid functions in calculus class, and thinks that a cardioid looks a bit like a cross section of an apple. How can she use to solve this problem and satisfy her curiosity?
  ::在周末访问祖母农场时,Maia看到她的祖母剥了苹果,这样皮就成了一个未碎的片子。 Maia想知道是否有办法估算苹果皮的表面面积。 她一直在研究微积分类的心血管组织功能,并且认为心血管组织看起来有点像苹果的交叉部分。 她如何用这个方法解决这个问题并满足她的好奇心?

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  Surface Area Using Polar Forms
  ::使用极地表表的地表面积

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  You can use integrals to find the surface area of solids formed by rotating a curve around the polar axis or $\begin{array}{r}\theta =\frac{\pi }{2}\end{array}$ . Why would an integral help you find the surface area of a solid? Imagine that you’re making a paper mache sculpture. In paper mache, you cover a form with thin strips of glue-soaked newspaper. Each piece of newspaper is a rectangle. This means that you could find the surface area of the solid by measuring the rectangles and then counting how many rectangles it takes to cover the form. Your answer wouldn’t be exact. Some strips of newspaper might overlap. You might need fractions of a strip to cover some odd corners. Still, adding up the areas of all the newspaper strips would give you a pretty good estimate of the amount of surface that you covered.
  ::您可以使用集成件来找到圆轴或2周围旋转曲线所形成的固体表面面积。 为何一个整体能帮助您找到固体表面面积? 想象一下您正在做一个纸大砍刀雕塑。 在纸大砍刀中, 您用胶水面报纸的薄条覆盖一种形态。 每一份报纸都是矩形。 这意味着您可以通过测量矩形并随后计出要覆盖表状的矩形面积来找到固体表面面积。 您的答案并不准确。 一些报纸的条纹可能会重叠。 您可能需要条纹以覆盖某些奇特的角。 但是, 加上所有报纸条纹的区块, 您就可以很好地估计覆盖的表面面积。

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  Using an integral to find surface area is a lot like using newspaper strips. Each strip is as long as the curve that you’ve rotated, but each strip is also infinitely thin. By adding up the areas of all the strips that cover the solid, you can find its surface area.
  ::使用一个构件查找表面区域与使用报纸条很相似。 每条条线都与您旋转的曲线一样长,但每条线也是无限薄的。 通过将覆盖固体的所有条条的面积加在一起,您可以找到其表面区域。

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  In polar form , the formula for the surface area of a curve revolved around the polar axis is
  ::以极形形式显示,围绕极轴旋转的曲线表面面积的公式是:

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  $\begin{array}{r}Are{a}_{surface}=2\pi \underset{a}{\overset{b}{\int }}r\sin \theta \sqrt{r^2+{\left(\frac{dr}{d\theta }\right)}^2}d\theta \end{array}$ .  The surface area for a curve revolved around $\begin{array}{r}\theta =\frac{\pi }{2}\end{array}$  is
  ::区域表面=2abrsinr2+(drd)2d。曲线的表面区域围绕%2

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  $\begin{array}{r}Are{a}_{surface}=2\pi \underset{a}{\overset{b}{\int }}r\cos \theta \sqrt{r^2+{\left(\frac{dr}{d\theta }\right)}^2}d\theta \end{array}$ . Sometimes, the surface area for a solid will give you an integral that you can solve symbolically. At other times, you may need to use a calculator or a web program to get a good numeric approximation for the integral.
  ::区域表面= 2 abrcosr2+( drd)2d。 有时, 固态的表面区域会为您提供一个可以象征性解析的内装件。 有时, 您可能需要使用计算器或网络程序来获得一个好的内装件数字近似值 。

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  Let's find the surface area of the curve $\begin{array}{r}r=\ln \theta \end{array}$  from $\begin{array}{r}\theta =1\end{array}$  to $\begin{array}{r}\theta =\pi \end{array}$  revolved around the line $\begin{array}{r}\theta =\frac{\pi }{2}\end{array}$ . 
  ::让我们找到曲线的表面积 r=ln \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ -\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

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  First, sketch the graph.
  ::首先,绘制图表。

  

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  You’ll be revolving the small area of the curve that is highlighted in red around the vertical line $\begin{array}{r}\theta =\frac{\pi }{2}\end{array}$ . Set up the formula for surface area of a revolution around $\begin{array}{r}\theta =\frac{\pi }{2}\end{array}$ .
  ::以红色标注的曲线小区域在垂直线 @ @#2. 设置革命地区2° 2 的公式。

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  $$\begin{array}{r}Are{a}_{surface}=2\pi \underset{\pi }{\overset{1}{\int }}\ln \theta \cos \theta \sqrt{(\ln \theta {)}^2+{\left(\frac{1}{\theta }\right)}^2}d\theta \end{array}$$

  
  ::面积表面=2 1 2 +(1)2 d

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  If you look at this integral, it’s pretty messy. There aren’t any obvious, elegant substitutions you can make to clean it up. For many curves, the integral for surface area can be extremely difficult to compute. In these cases, it’s fine to integrate numerically, especially since many of the applications for polar coordinates involve fields like engineering, where a good numeric approximation is the goal.
  ::如果看这个整体,它就会非常混乱。 没有任何明显的优雅的替代方法可以清理它。 对于许多曲线来说,表层的构件极难计算。 在这种情况下,用数字集成是件好事,特别是因为许多极地坐标应用都涉及工程学等领域,而工程学的目标是良好的数字近似值。

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  Use your calculator. Remember that surface area is always positive. You should get an answer close to 5.954.
  ::使用您的计算器。 记住表面积总是正数。 您应该得到接近5. 954 的答案 。

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  Examples
  ::实例

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  Example 1
  ::例1

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  Earlier, you were asked how  Maia can determine the surface area of an apple peel. Maia measures the apple and realizes that she can model its skin with a revolution of the equation  $\begin{array}{r}r=3+3\cos \theta \end{array}$ around the polar axis from 0 to $\begin{array}{r}\pi \end{array}$ , where all distances are in centimeters. 
  ::早些时候,有人问您Maia如何确定苹果皮的表面面积。 Maia测量了苹果,并意识到她可以通过在0到%的极轴周围的 r=3+3cos的方程式革命来模拟其皮肤,所有距离都以厘米计。

  

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  She substitutes her model into the surface area equation:
  ::她将模型换成表面积方程式:

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  $$\begin{array}{r}Are{a}_{surface}=2\pi \underset{0}{\overset{\pi }{\int }}(3+3\cos \theta )\sin \theta \sqrt{(3+3\cos \theta {)}^2+(-3\sin \theta {)}^2}d\theta \end{array}$$

  
  ::区域表面=20( 3+3cos)sin( 3+3cos)2+( - 3sin)2d( - 3sin)2d

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  Then, she factors and simplifies.
  ::然后,她就成为了各种因素,简化了。

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  $$\begin{array}{rl}& =2\pi \underset{0}{\overset{\pi }{\int }}3(1+\cos \theta )\sin \theta \sqrt{9(1+2\cos \theta +{\cos }^2\theta +{\sin }^2\theta )}d\theta \\ & =2\pi \underset{0}{\overset{\pi }{\int }}9(1+\cos \theta )\sin \theta \sqrt{1+2\cos \theta +1}d\theta \\ & =2\pi \underset{0}{\overset{\pi }{\int }}9(1+\cos \theta )\sin \theta \sqrt{2+2\cos \theta }d\theta \end{array}$$

  
  ::=203(1+222222241222222222222222222222422222224444222244422224424222442422242224424224222442222224

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  She multiplies by a form of one so that she can put the equation into a form that allows for $\begin{array}{r}u\end{array}$  substitution.
  ::将方程式乘以一种形式, 这样她就可以将方程式 变成一种允许 u 替代的方程式 。

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  $$\begin{array}{rl}& =2\pi \underset{0}{\overset{\pi }{\int }}9(1+\cos \theta )\sin \theta \sqrt{2+2\cos \theta }d\theta \cdot \frac{-4}{-4}\\ & =-\frac{1}{4}2\pi \underset{0}{\overset{\pi }{\int }}9\cdot 2\cdot (1+\cos \theta )\cdot -2\cdot \sin \theta \sqrt{2+2\cos \theta }d\theta \\ & =-\frac{9\pi }{2}\underset{0}{\overset{\pi }{\int }}(2+2\cos \theta )(-2\sin \theta )\sqrt{2+2\cos \theta }d\theta \\ u& =2+2\cos \theta \\ du& =-2\sin \theta d\theta \\ & =-\frac{9\pi }{2}\int u^{\frac{3}{2}}du\\ & =-\frac{9\pi }{2}\left[\frac{2}{5}{u}^{\frac{5}{2}}\right]\\ & =-\frac{9\pi }{2}{\left[\frac{2}{5}(2+2\cos \theta {)}^{\frac{5}{2}}\right]}_0^{\pi }\\ & =\frac{288\pi }{5}\\ & \approx 181c{m}^2\end{array}$$

  
  ::=20922222222441420092122222222222222222222222222222222581812

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  The surface area of the apple is approximately 181 square centimeters.
  ::苹果表面面积约为181平方厘米。

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  Example 2
  ::例2

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  Find the surface area of the solid created when you revolve $\begin{array}{r}r=\cos \theta \end{array}$  around the polar axis from 0 to $\begin{array}{r}\frac{\pi }{2}\end{array}$ . Use symbolic integration.
  ::查找当您将 r=cos 围绕极轴从 0 到 2 旋转时创建的固体表面区域。 使用符号集成 。

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  This revolution will create a sphere with a radius of 0.5.
  ::这场革命将创造一个半径为0.5的球体。

  

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  While it’s possible to solve this problem geometrically, it’s good practice to use an integral to find the surface area of this solid.
  ::虽然可以用几何方法解决这个问题, 但使用一个有机体来找到这一固体的表面面积是好的做法。

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  First, substitute $\begin{array}{r}r\end{array}$  into the surface area formula and simplify.
  ::首先,将 r 替换为表面积公式并简化。

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  $$\begin{array}{rl}Are{a}_{surface}& =2\pi \underset{a}{\overset{b}{\int }}r\sin \theta \sqrt{r^2+{\left(\frac{dr}{d\theta }\right)}^2}d\theta \\ & =2\pi \underset{0}{\overset{\frac{\pi }{2}}{\int }}\cos \theta \sin \theta \sqrt{{\cos }^2\theta +{\sin }^2\theta }d\theta \\ & =2\pi \underset{0}{\overset{\frac{\pi }{2}}{\int }}\cos \theta \sin \theta \sqrt{1}d\theta \\ & =2\pi \underset{0}{\overset{\frac{\pi }{2}}{\int }}\cos \theta \sin \theta d\theta \end{array}$$

  
  ::地表=2abrsinr2+(drd)220222222222222222222222222222222244444444444444444444444444422222244444444442222222222222444

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  You can use $\begin{array}{r}u\end{array}$  substitution to integrate at this point.
  ::您可以在此点使用 u 替换来整合 。

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  $$\begin{array}{rl}& 2\pi \underset{0}{\overset{\frac{\pi }{2}}{\int }}\cos \theta \sin \theta d\theta \\ u& =\sin \theta \\ du& =\cos \theta d\theta \\ & =2\pi \int udu=2\pi \left[\frac{1}{2}{u}^2\right]\\ & =2\pi {\left[\frac{1}{2}{\sin }^2\theta \right]}_0^{\frac{\pi }{2}}\\ & =\pi unit{s}^2\end{array}$$

  
  ::2 0222222222222222222222222

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  Example 3
  ::例3

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  Find the surface area of $\begin{array}{r}r=4+4\sin \theta \end{array}$  when the segment from $\begin{array}{r}\theta =-\frac{\pi }{2}\end{array}$  to  $\begin{array}{r}\frac{\pi }{2}\end{array}$  is revolved around the line  $\begin{array}{r}\theta =\frac{\pi }{2}\end{array}$ .
  ::查找 r= 4+4sin 的表面区域, 当% 2 到 2 的段围绕 2 线旋转时 。

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  First, identify the equation and sketch. This curve is a cardioid. From $\begin{array}{r}\theta =-\frac{\pi }{2}\end{array}$  to  $\begin{array}{r}\frac{\pi }{2}\end{array}$   is half the cardioid.
  ::首先,确定方程和草图。 这个曲线是一个心类。 从% 2 到% 2 是心类的一半 。

  

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  Now, use the formula for surface area.
  ::现在,对表面积使用公式。

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  $$\begin{array}{r}2\pi \underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}(4+4\sin \theta )\cos \theta \sqrt{(4+4\sin \theta {)}^2+(4\cos \theta {)}^2}d\theta \end{array}$$

  
  ::=============================================================================================================================================================================================================================================================================================================================================================================================================================================================================================================================

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  Factor to make the integral easier to simplify.
  ::使整体部分更容易简化的因子。

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  $$\begin{array}{rl}& =2\pi \underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}4(1+\sin \theta )\cos \theta \sqrt{16(1+\sin \theta {)}^2+16(\cos \theta {)}^2}d\theta \\ & =2\pi \underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}16(1+\sin \theta )\cos \theta \sqrt{(1+\sin \theta {)}^2+(\cos \theta {)}^2}d\theta \end{array}$$

  
  ::=2224(1+sin)cos16(1+sin)2+16(cos)2+16(cos)2d222222216(1+sin)cos(1+sin)2+(cos)2d2

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  Now, simplify.
  ::现在,简化。

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  $$\begin{array}{rl}& =2\pi \underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}16(1+\sin \theta )\cos \theta \sqrt{1+2\sin \theta +{\sin }^2\theta +{\cos }^2\theta }d\theta \end{array}$$

  
  ::=222216(1+sin1+2sin1+2sin22224d

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  Use trigonometric substitution to simplify the radical.
  ::使用三角替代来简化基体

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  $$\begin{array}{rl}& =2\pi \underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}16(1+\sin \theta )\cos \theta \sqrt{1+2\sin \theta +1}d\theta \\ & =2\pi \underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}16(1+\sin \theta )\cos \theta \sqrt{2+2\sin \theta }d\theta \end{array}$$

  
  ::=2222222222222222222222222

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  Now you’re ready to use $\begin{array}{r}u\end{array}$  substitution. If you split up the 16 into 4, 2, and 2, it becomes easier.
  ::现在,你准备使用U替代。 如果把16分成4、2和2, 就会容易一些。

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  $$\begin{array}{rl}& =2\pi \underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}(4)(2)(2)(1+\sin \theta )\cos \theta \sqrt{2+2\sin \theta }d\theta \\ & =8\pi \underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}(2+2\sin \theta )2\cos \theta \sqrt{2+2\sin \theta }d\theta \\ u& =2+2\sin \theta \\ du& =2\cos \theta \\ & =8\pi \int u^{\frac{3}{2}}du\\ & =8\pi \left[\frac{2}{5}{u}^{\frac{5}{2}}\right]\\ & =\frac{16\pi }{5}{\left[(2+2\sin \theta {)}^{\frac{5}{2}}\right]}_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\\ & =\frac{512\pi }{5}unit{s}^2\end{array}$$

  
  ::2222222222222222222222222222222228u32du=888288882888888888888882888288282888828888828882888=888888888888888888888[25=165[(2+225552}222=522222222222222222

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  Example 4
  ::例4

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  The curve $\begin{array}{r}r=5\cos \theta \end{array}$  is revolved around the polar axis from 0 to $\begin{array}{r}\frac{\pi }{3}\end{array}$ . Find the surface area of the revolution.
  ::曲线 r= 5cos 围绕极轴旋转 从 0 到 3. 寻找革命的表面区域 。

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  $$\begin{array}{rl}Are{a}_{surface}& =2\pi \underset{0}{\overset{\frac{\pi }{3}}{\int }}5\cos \theta \sin \theta \sqrt{(5\cos \theta {)}^2+(-5\sin \theta {)}^2}d\theta \\ & =2\pi \underset{0}{\overset{\frac{\pi }{3}}{\int }}5\cos \theta \sin \theta \sqrt{(25{\cos }^2\theta )+(25{\sin }^2\theta )}d\theta \\ & =2\pi \underset{0}{\overset{\frac{\pi }{3}}{\int }}5\cos \theta \sin \theta \sqrt{25({\cos }^2\theta +{\sin }^2\theta )}d\theta \\ & =2\pi \underset{0}{\overset{\frac{\pi }{3}}{\int }}5\cos \theta \sin \theta \sqrt{25(1)}d\theta \\ & =2\pi \underset{0}{\overset{\frac{\pi }{3}}{\int }}25\cos \theta \sin \theta d\theta \\ & =50\pi \underset{0}{\overset{\frac{\pi }{3}}{\int }}\cos \theta \sin \theta d\theta \\ & =-50\pi \underset{0}{\overset{\frac{\pi }{3}}{\int }}\cos \theta (-\sin \theta )d\theta \\ u& =\cos \theta \\ du& =-\sin \theta d\theta \\ & =-50\pi \int udu=-50\pi \left[\frac{1}{2}{u}^2\right]\\ & =-25\pi {\left[{\cos }^2\theta \right]}_0^{\frac{\pi }{3}}\\ & =\frac{75\pi }{4}\end{array}$$

  
  ::区域表面=20353555552203522203522522203552522220352220355251223350035003342035555055050505055[2203=754}2003=754

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  Example 5
  ::例5

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  Find the surface area of the curve $\begin{array}{r}r=-3-3\sin \theta \end{array}$  revolved around the line $\begin{array}{r}\theta =\frac{\pi }{2}\end{array}$  from $\begin{array}{r}-\frac{\pi }{2}\end{array}$  to $\begin{array}{r}\frac{\pi }{2}\end{array}$ . 
  

  $$\begin{array}{rl}{A}_{surface}& =2\pi \underset{\frac{\pi }{2}}{\overset{-\frac{\pi }{2}}{\int }}(-3-3\sin \theta )(\cos \theta )\sqrt{(-3-3\sin \theta {)}^2+(-3\cos \theta {)}^2}d\theta \\ & =2\pi \underset{\frac{\pi }{2}}{\overset{-\frac{\pi }{2}}{\int }}-3(1+\sin \theta )(\cos \theta )\sqrt{-3^2(1+3\sin \theta {)}^2+-3^2(\cos \theta {)}^2}d\theta \\ & =2\pi \underset{\frac{\pi }{2}}{\overset{-\frac{\pi }{2}}{\int }}-9(1+\sin \theta )(\cos \theta )\sqrt{(1+\sin \theta {)}^2+(\cos \theta {)}^2}d\theta \\ & =2\pi \underset{\frac{\pi }{2}}{\overset{-\frac{\pi }{2}}{\int }}-9(1+\sin \theta )(\cos \theta )\sqrt{(1+2\sin \theta +{\sin }^2\theta +{\cos }^2\theta )}d\theta \\ & =2\pi \underset{\frac{\pi }{2}}{\overset{-\frac{\pi }{2}}{\int }}-9(1+\sin \theta )(\cos \theta )\sqrt{(1+2\sin \theta +1)}d\theta \\ & =2\pi \underset{\frac{\pi }{2}}{\overset{-\frac{\pi }{2}}{\int }}-9(1+\sin \theta )(\cos \theta )\sqrt{(2+2\sin \theta )}d\theta \times \frac{4}{4}\\ & =-18\pi \underset{\frac{\pi }{2}}{\overset{-\frac{\pi }{2}}{\int }}\frac{1}{4}(2+2\sin \theta )(2\cos \theta )\sqrt{(2+2\sin \theta )}d\theta \\ u& =2+2\sin \theta \\ du& =2\cos \theta \\ & =\frac{-18\pi }{4}\int u^{\frac{3}{2}}du=\frac{-18\pi }{4}\left[\frac{2}{5}{u}^{\frac{5}{2}}\right]\\ & =\frac{-18\pi }{10}{\left[(2+2\sin \theta {)}^{\frac{5}{2}}\right]}_{\frac{\pi }{2}}^{-\frac{\pi }{2}}\\ & =57.6\pi unit{s}^2\end{array}$$

  
  ::找到曲线的表面面积 3 - 3sin} 3 - 2 - 2 - 2 -2 - 2 - 2 - 2 - 2 - 2 - 3 - 3 辛 (cos )(- 3 - 3 - 3 - 3 - ) 2+ (3 - 3 - 3 - 2 ) 2+ (3 - (3-3 - 3) 2 + (1+ 3- 3) 2 - 3 3 3 - 2 - 2 4 . 2 - 2 4, 2 3- 3 (cos _)( 3 - 3 - 3-3 - 3 - ) 2 + (3-1+ 3 3 3 3 3 - 3 3,2 4 (3 3 3 ) 2 - 2 - 2 - 2 3 3 - 3 3 4, 3 3 4, 2 - 3 3 - 3 - 3 3 3,2 3 3 3 4-2 5, 2 - 9 5 (1+2 5,2 5,2 5,2+2 5,2 5,2 5,2,2 5,2 5,2 5,2 5,2 5,2 5,2 5 5 5,2 5,2 5,2 5 5,2 5,2 5 5 5 5 5 5 5 5 5,8 5,5,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,

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  Your answer is $\begin{array}{r}57.6\pi unit{s}^2\end{array}$ .
  ::您的回答是 576.6 单位2。

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  Review
  ::回顾

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  For #1-10, set up, but do not evaluate, an integral that gives the surface area of the curve rotated about the given axis.
  ::对于 # 1- 10, 设置, 但不评价, 使曲线的表面区域旋转到给定轴上 。

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  1. $\begin{array}{r}r=\sin \theta \end{array}$  rotated about the line $\begin{array}{r}\theta =\frac{\pi }{2}\end{array}$  from $\begin{array}{r}\theta =0\end{array}$  to $\begin{array}{r}\theta =\frac{\pi }{2}\end{array}$ .
     ::r=sin 在2号线左右旋转,从0到2号线。
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  2. $\begin{array}{r}r=2+2\sin \theta \end{array}$  rotated about the polar axis from $\begin{array}{r}\theta =0\end{array}$  to $\begin{array}{r}\theta =\pi \end{array}$ .
     ::r=2+2sin 围绕极轴旋转,从0旋转到。
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  3. $\begin{array}{r}r=\cos \theta \sin \theta \end{array}$  rotated about the polar axis from $\begin{array}{r}\theta =0\end{array}$  to $\begin{array}{r}\theta =\frac{\pi }{2}\end{array}$ .
     ::r=cossin 沿极轴旋转,从0旋转到2。
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  4. $\begin{array}{r}r=\cos \theta \end{array}$  rotated about the line $\begin{array}{r}\theta =\frac{\pi }{2}\end{array}$  from $\begin{array}{r}\theta =0\end{array}$  to $\begin{array}{r}\theta =\pi \end{array}$ .
     ::r=cos在2号线上旋转 从0到 。
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  5. $\begin{array}{r}r={\sin }^2\theta \end{array}$  rotated about the line $\begin{array}{r}\theta =\frac{\pi }{2}\end{array}$  from $\begin{array}{r}\theta =0\end{array}$  to $\begin{array}{r}\theta =\frac{\pi }{3}\end{array}$ .
     ::r=sin2 2 2 2 0 3 3 2 2 2 2 2 2 2 2 2 2 2 3 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
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  6. $\begin{array}{r}r=\cos \theta \sin 2\theta \end{array}$  rotated about the line $\begin{array}{r}\theta =\frac{\pi }{2}\end{array}$  from $\begin{array}{r}\theta =0\end{array}$  to $\begin{array}{r}\theta =\frac{\pi }{3}\end{array}$ .
     ::r=cossin222222203333222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222
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  7. $\begin{array}{r}r=5-4\cos \theta \end{array}$  rotated about the line $\begin{array}{r}\theta =\frac{\pi }{2}\end{array}$  from $\begin{array}{r}\theta =0\end{array}$  to $\begin{array}{r}\theta =\frac{\pi }{2}\end{array}$ .
     ::r=5 - 4cos 2 2 0 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
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  8. $\begin{array}{r}r=\theta \cos \theta \end{array}$  rotated about the polar axis from $\begin{array}{r}\theta =\frac{\pi }{4}\end{array}$  to $\begin{array}{r}\theta =\frac{\pi }{2}\end{array}$ .
     ::rcos在极轴上旋转,从4旋转到2。
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  9. $\begin{array}{r}r=e^{\theta }\end{array}$  rotated about the line $\begin{array}{r}\theta =\frac{\pi }{2}\end{array}$  from $\begin{array}{r}\theta =0\end{array}$  to $\begin{array}{r}\theta =\frac{\pi }{2}\end{array}$ .
     ::\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\2\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\
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  10. $\begin{array}{r}r=1+\cos \theta \end{array}$  rotated about the polar axis from $\begin{array}{r}\theta =0\end{array}$  to $\begin{array}{r}\theta =\frac{\pi }{2}\end{array}$ .
      ::r=1+cos 沿极轴旋转,从0旋转到2。

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  For #11-15, find the surface area of the curve rotated about the given axis. Integrate symbolically or numerically as appropriate.
  ::对于 # 11-15, 找到在给定轴周围旋转的曲线的表面区域。 酌情以符号或数字方式整合 。

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  11. $\begin{array}{r}r=\sin \theta \end{array}$  rotated about the line $\begin{array}{r}\theta =\frac{\pi }{2}\end{array}$  from $\begin{array}{r}\theta =0\end{array}$  to $\begin{array}{r}\theta =\frac{\pi }{2}\end{array}$ .
      ::r=sin 在2号线左右旋转,从0到2号线。
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  12. $\begin{array}{r}r=2+2\sin \theta \end{array}$  rotated about the polar axis from $\begin{array}{r}\theta =0\end{array}$  to $\begin{array}{r}\theta =\pi \end{array}$ .
      ::r=2+2sin 围绕极轴旋转,从0旋转到。
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  13. $\begin{array}{r}r=\cos \theta \sin \theta \end{array}$  rotated about the polar axis from $\begin{array}{r}\theta =0\end{array}$  to $\begin{array}{r}\theta =\frac{\pi }{2}\end{array}$ .
      ::r=cossin 沿极轴旋转,从0旋转到2。
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  14. $\begin{array}{r}r=\cos \theta \end{array}$  rotated about the line $\begin{array}{r}\theta =\frac{\pi }{2}\end{array}$  from $\begin{array}{r}\theta =0\end{array}$  to $\begin{array}{r}\theta =\pi \end{array}$ .
      ::r=cos在2号线上旋转 从0到 。
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  15. $\begin{array}{r}r={\sin }^2\theta \end{array}$  rotated about the line $\begin{array}{r}\theta =\frac{\pi }{2}\end{array}$  from $\begin{array}{r}\theta =0\end{array}$  to $\begin{array}{r}\theta =\frac{\pi }{3}\end{array}$ .
      ::r=sin2 2 2 2 0 3 3 2 2 2 2 2 2 2 2 2 2 2 3 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2

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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 。