10.9 参数形式和微积分:革命量

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  参数形式和微积分:革命量

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  Martin is a potter who displays his mugs, jugs, and vases at juried exhibitions around the country. Every piece he creates is totally unique. At a craft show in Dubuque, Iowa, a customer wants to know exactly how many cubic centimeters of clay it took to make a particular vase. How can Martin calculate the volume of the clay used to make the vase? 
  ::马丁是个陶匠,他在全国各地的陪审团展览上展示自己的杯子、杯子和花瓶。他创造的每一件作品都是独一无二的。在爱荷华州的杜布克(Dubuque)的一个工艺展上,一位顾客想知道究竟用了多少立方厘米的粘土来制作一个花瓶。马丁如何计算用来制作花瓶的粘土量?

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  Volume of Revolution with Parametric Equations
  ::具有参数等量的革命量

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  Have you ever made cookies using a roll of pre-made, refrigerated cookie dough? To make the cookies, you must slice the dough into a group of thin cylinders. If you put the cylinders back together, you can recreate the roll of cookie dough.
  ::您有没有做过饼干, 使用一卷预制、 冷冻的饼干面团? 要做饼干, 您必须将饼干切成一组薄筒。 如果您把圆筒重新组合起来, 您可以重新创建饼干面团卷 。

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  When you find the volume of a revolution of a parametric curve, you use a process a lot like the one you use to make slice and bake cookies. When a curve is revolved around an axis, every cross section is a circle . The entire shape is made of very, very thin cylinders, just like the roll of cookie dough. The volume of the entire shape will be the sum of the volumes of the cylinders.
  ::当您发现参数曲线革命的体积时,您会使用类似于您用来做切片和烤饼干的过程。当曲线围绕轴旋转时,每个横截段都是圆形。整个形状由非常非常薄的圆柱组成,就像饼干卷。整个形状的体积将是圆柱体的体积总和。

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  The geometric formula for the volume of a cylinder is $\begin{array}{r}A=\pi r^{2}h\end{array}$ , where  $\begin{array}{r}r\end{array}$ is the radius and  $\begin{array}{r}h\end{array}$ is the height. The formula for a parametric solid is similar. The radius is the distance from the axis of rotation to the value of the function at a time, $\begin{array}{r}t\end{array}$ . The height of the cylinder is defined as the change in distance over time- a derivative. You can use an integral to add all of the cylinders together to find the volume of the solid.
  ::圆柱体体积的几何公式是 Ar2h, 其中 r 是半径, h 是高度。 参数固体的公式相似。 半径是旋转轴与函数值之间的距离, t。 圆柱体的高度被定义为时间距离的变化- 衍生物。 您可以使用一个元件将所有圆柱体相加, 以查找固体的体积 。

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  The formula to find the volume of a curve revolved around the $\begin{array}{r}x\end{array}$ -axis is $\begin{array}{r}\pi \underset{a}{\overset{b}{\int }}y(t{)}^2(x^{\mathrm{\prime }}(t))dt\end{array}$ . If the curve revolves around the $\begin{array}{r}y\end{array}$ -axis, the formula is $\begin{array}{r}\pi \underset{a}{\overset{b}{\int }}x(t{)}^2(y^{\mathrm{\prime }}(t))dt\end{array}$ .
  ::查找围绕 x 轴旋转的曲线体积的公式为 aby(t)2(x(t))dt。如果曲线围绕 y 轴旋转,则公式为 abx(t)2(yä(t))dt。

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  Let's revolve the curve described by the parametric equation below around the $\begin{array}{r}x\end{array}$ -axis from  $\begin{array}{r}t=0\end{array}$ to  $\begin{array}{r}t=\frac{\pi }{2}\end{array}$ and find the volume of the resulting three dimensional shape.
  ::让我们绕过下方X轴周围的参数方程式所描述的曲线 从t=0到t=2, 并找到由此产生的三维形状的体积。

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  $$\begin{array}{rl}F(t)& =(x(t),y(t))\\ x(t)& =3\cos t\\ y(t)& =3\sin t\end{array}$$

  
  ::F(t) = (x(t) y(t) ) x(t) = 3costy(t) = 3sint

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  To find the volume of this revolution, use the formula $\begin{array}{r}\pi \underset{a}{\overset{b}{\int }}y(t{)}^2(x^{\mathrm{\prime }}(t))dt\end{array}$ .
  ::要查找此革命的量, 请使用 {aby( t) 2( x( t)) dt 公式 。

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  $$\begin{array}{rl}\pi \underset{a}{\overset{b}{\int }}y(t{)}^2(x^{\mathrm{\prime }}(t))dt& =\pi \underset{0}{\overset{\frac{\pi }{2}}{\int }}(3\sin t{)}^2(-3\sin (t))dt\\ & =\pi \underset{0}{\overset{\frac{\pi }{2}}{\int }}-27{\sin }^{2}t\sin tdt\\ & =-27\pi \underset{0}{\overset{\frac{\pi }{2}}{\int }}{\sin }^{3}tdt\end{array}$$

  
  ::{\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2款

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  From a table of integrals, you can find that the integral for  $\begin{array}{r}{\sin }^{3}tdt\end{array}$ is $\begin{array}{r}\frac{1}{3}{\cos }^{3}t-\cos t\end{array}$ , so
  ::从一个整体体的表格中,你可以发现,sin3的有机体是13cos3_t-cos_t,所以

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  $$\begin{array}{rl}-27\pi \underset{0}{\overset{\frac{\pi }{2}}{\int }}{\sin }^{3}tdt& =-27\pi {\left[\frac{1}{3}{\cos }^{3}t-\cos t\right]}_0^{\frac{\pi }{2}}\\ & =-27\pi \left[\left[\frac{1}{3}{\cos }^3\left(\frac{\pi }{2}\right)-\cos \left(\frac{\pi }{2}\right)\right]-\left[\frac{1}{3}{\cos }^3(0)-\cos (0)\right]\right]\\ & =-27\pi \left[\left[\frac{1}{3}(0)-(0)\right]-\left[\frac{1}{3}(1)-1\right]\right]\\ & =-27\pi \left(\frac{2}{3}\right)\\ & =-18\pi \end{array}$$

  
  ::-=YTET -=YTET -伊甸园字幕组=-伊甸园字幕组=- 翻译:

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  Remember, the quantity is negative because travel the curve in a counter clockwise direction, and thus the results of the integration are negative. The volume is actually $\begin{array}{r}18\pi {\text{units}}^3\end{array}$ .
  ::记住, 数量是负数, 因为曲线向反时钟方向移动, 因此整合结果为负数 。 体积实际上是 18Q 单位 3 。

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  This revolution made half a sphere. Recall that the geometric formula for the volume of a sphere is $\begin{array}{r}\frac{4}{3}\pi r^3\end{array}$ . So, using geometry, this semi-sphere would have the area  $\begin{array}{r}\frac{1}{2}\left(\frac{4}{3}\pi r^3\right)=\frac{2}{3}(27\pi )=18\pi {\text{units}}^3\end{array}$ . The integral method for finding the volume of a solid gives the same result that the geometric method yields. However, you can also use the integral method to find the volume of irregular or unusual shapes.
  ::这场革命产生了半个球体。 回顾一个球体体积的几何公式是43°r3。 因此,使用几何,这个半球体的面积为12(43°r3)=23(27°)=18°3。 查找固体体积的综合方法得出的结果与几何方法的产值相同。 但是,您也可以使用整体方法来查找异常或异常形状的体积。

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  Now, let's find the volume of the solid formed when the following parametric curve is rotated around the $\begin{array}{r}y\end{array}$ -axis from  $\begin{array}{r}t=1\end{array}$ to $\begin{array}{r}t=5\end{array}$ .
  ::现在,让我们在以下的参数曲线在y轴周围从t=1到t=5旋转时,找到固体的体积。

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  $$\begin{array}{rl}F(t)& =(x(t),y(t))\\ x(t)& =5t^2\\ y(t)& =6t\end{array}$$

  
  ::F(t) = (x(t) y(t) ) x(t) = 5t2y(t) = 6t

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  Use the volume equation for a curve revolved around the $\begin{array}{r}y\end{array}$ -axis.
  ::使用音量方程式来绘制围绕 Y 轴的曲线 。

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  $$\begin{array}{rl}\pi \underset{a}{\overset{b}{\int }}x(t{)}^2(y^{\mathrm{\prime }}(t))dt& =\pi \underset{1}{\overset{5}{\int }}(5t^2{)}^2(6)dt\\ & =6\pi \underset{1}{\overset{5}{\int }}25t^{4}dt\\ & =150\pi \underset{1}{\overset{5}{\int }}{t}^4\\ & =150\pi {\left[\frac{1}{5}{t}^5\right]}_1^5\\ & =93750\pi -30\pi \\ & =93720\pi \end{array}$$

  
  ::@abx(t)2(y_(t))dt @%15(5t2)2(6)t=61525t4dt=150}15t4=150}150}[15t5]15=93750}30{93720}

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  The volume of the solid is $\begin{array}{r}93720\pi {\text{units}}^3\end{array}$ .
  ::固体体积为93720立方体3。

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  Examples
  ::实例

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  Example 1
  ::例1

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  Earlier, you were asked about how Martin can model the volume of a particular vase. Martin can model the vase by revolving two parametric curves around the $\begin{array}{r}y\end{array}$ -axis from .25 to 1.5. The first curve, $\begin{array}{r}F(t)\end{array}$ , will model the outside of the vase. The second curve, $\begin{array}{r}G(t)\end{array}$ , will model the inner wall of the vase. If Martin finds the volume of the solid formed by the outer curve and subtracts the volume of the solid formed by the inner curve, he can find out approximately how many cubic centimeters of clay make up the vase.
  ::早些时候,有人问马丁如何模拟某个花瓶的体积。 Martin可以通过在 Y 轴周围旋转两个参数曲线,从 25 到 1.5 来模拟花瓶。 第一个曲线, F (t) 将模拟花瓶的外部。 第二个曲线, G (t) 将模拟花瓶的内壁。 如果马丁发现外曲线形成的固体体积, 并减去内曲线形成的固体体积, 他就能发现有多少立方厘米的粘土组成花瓶。

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  $$\begin{array}{rl}F(t)& =(x(t),y(t))\\ x(t)& =8t^2\\ y(t)& =6t^5\\ \\ G(t)& =(x(t),y(t))\\ x(t)& =7t^2\\ y(t)& =6t^5\end{array}$$

  
  ::F( t) = (x( t) = 8t2y( t) = 6t5G( t) = (x( t) ,y( t) ) x( t) = 7t2y( t) = 6t5

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  First, find the volume of the revolution for $\begin{array}{r}F(t)\end{array}$ .
  ::首先,找到F(t)的革命量。

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  $$\begin{array}{rl}x(t)& =8t^2\\ y(t)& =6t^5\\ y^{\mathrm{\prime }}(t)& =30t^4\\ \\ \pi \underset{.25}{\overset{1.5}{\int }}x(t{)}^2{y}^{\mathrm{\prime }}(t)dt& =\pi \underset{.25}{\overset{1.5}{\int }}(8t^2{)}^2(30t^4)dt\\ & =\pi \underset{.25}{\overset{1.5}{\int }}64t^4(30t^4)dt\\ & =1920\pi \underset{.25}{\overset{1.5}{\int }}{t}^{8}dt\\ & =1920\pi {\left[\frac{1}{9}{t}^9\right]}_{.25}^{1.5}\\ & =640\frac{\pi }{3}((1.5{)}^9-(.25{)}^9)\\ & =8201.249\pi \end{array}$$

  
  :t)=8t2y(t)=6t5y_(t)=30t4251.5x(t)=30t4255}(t)251.5(t)8t2,2(30t4)dt251.564t4(30t4)dt=1920251.5t8dt=1920}[19t9]251.5=640}3((1.5)9-(25)9)=8201.249__

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  The volume of the solid created by $\begin{array}{r}F(t)\end{array}$  is $\begin{array}{r}8201.249\pi c{m}^3\end{array}$ . Now, compute the volume of the inner solid, $\begin{array}{r}G(t)\end{array}$ .
  ::F( t) 创建的固体体积为 8201.249 cm3。 现在, 计算内固体的体积, G( t) 。

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  $$\begin{array}{rl}G(t)& =(x(t),y(t))\\ x(t)& =7t^2\\ y(t)& =6t^5\\ y^{\mathrm{\prime }}(t)& =30t^4\\ \\ \pi \underset{.25}{\overset{1.5}{\int }}x(t{)}^2{y}^{\mathrm{\prime }}(t)dt& =\pi \underset{.25}{\overset{1.5}{\int }}(7t^2{)}^2(30t^4)dt\\ & =\pi \underset{.25}{\overset{1.5}{\int }}49t^4(30t^4)dt\\ & =1470\pi \underset{.25}{\overset{1.5}{\int }}{t}^{8}dt\\ & =490\frac{\pi }{3}{\left[t^9\right]}_{.25}^{1.5}\\ & =6279.081\pi \end{array}$$

  
  ::G( t) = (x( t) ,y( t) x( t) = 7t2y( t) = 6t5y_ (t) = 6t5y}(t) = 30t4} (t) = 30t4) 2.549t4 (30t4) 251.549t4 (30t4) dt= 1470} 251.5t8dt= 490_3[t9] 251.5= 6279.081}

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  The volume of the solid created by $\begin{array}{r}G(t)\end{array}$  is $\begin{array}{r}6279.081\pi c{m}^3\end{array}$ .
  ::G(t) 创造的固体体积为6279.081立方厘米。

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  Subtract to find the volume of the clay, and you’ll see that Martin used approximately  $\begin{array}{r}1922.168\pi c{m}^3\end{array}$ of clay to make the vase.
  ::马丁用大约1922.168立方厘米的粘土制造花瓶。

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  Example 2
  ::例2

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  A certain glass bead can be described as the parametric equation
  ::某种玻璃珠可以称为参数方程

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  $$\begin{array}{rl}F(t)& =(x(t),y(t))\\ x(t)& =5\cos (t)\\ y(t)& =5\sin (t)\end{array}$$

  
  ::F(t) = (x(t) y(t) ) x(t) = 5cos(t) y(t) = 5sin(t)

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  Rotated around the $\begin{array}{r}x\end{array}$ -axis from $\begin{array}{r}t=0\end{array}$  to $\begin{array}{r}t=\pi \end{array}$  with a core described by the rotation of
  ::围绕 x 轴旋转, 从 t=0 旋转到 t 。

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  $$\begin{array}{rl}G(t)& =(x(t),y(t))\\ x(t)& =4\cos (t)\\ y(t)& =4\sin (t)\end{array}$$

  
  ::G(t) = (x(t) y(t) ) x(t) = 4cos(t) y(t) = 4sin(t)

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  around the $\begin{array}{r}x\end{array}$ -axis from  $\begin{array}{r}t=0\end{array}$ to $\begin{array}{r}t=\pi \end{array}$ . Find the volume of the material that makes up the bead.
  ::从 t=0 到 t\\\\\\\\\\\\ x轴周围的 X 轴。 查找组成珠子的材料的体积 。

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  First, find the volume of the whole bead.
  ::首先,找到整个珠子的体积

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  $$\begin{array}{rl}\pi \underset{a}{\overset{b}{\int }}y(t{)}^2(x^{\mathrm{\prime }}(t))dt& =\pi \underset{0}{\overset{\pi }{\int }}(5\sin (t){)}^2(-5\sin t)dt\\ & =-125\pi \underset{0}{\overset{\pi }{\int }}{\sin }^3(t)dt\\ & =-125\pi {\left[\frac{1}{3}{\cos }^{3}t-\cos t\right]}_0^{\pi }\\ & =-125\pi \left[\frac{2}{3}-\frac{-2}{3}\right]\\ & =\frac{-500}{3}\pi \end{array}$$

  
  ::-=YTET -伊甸园字幕组=-伊甸园字幕组=- 翻译:

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  The formula produces a negative volume because  $\begin{array}{r}t\end{array}$ travels the curve in a counter-clockwise direction. The actual volume is $\begin{array}{r}\frac{500}{3}\pi \end{array}$ .
  ::公式产生的负体积为负体积,因为 t 向反时针方向移动曲线。 实际体积为 5003。

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  Now, find the volume of the hollow core.
  ::现在,找到空心核心的体积

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  $$\begin{array}{rl}\pi \underset{a}{\overset{b}{\int }}y(t{)}^2(x^{\mathrm{\prime }}(t))dt& =\pi \underset{0}{\overset{\pi }{\int }}(4\sin (t){)}^2(-4\sin t)dt\\ & =-64\pi \underset{0}{\overset{\pi }{\int }}{\sin }^3(t)dt\\ & =-64\pi {\left[\frac{1}{3}{\cos }^{3}t-\cos t\right]}_0^{\pi }\\ & =-64\pi \left[\frac{2}{3}-\frac{-2}{3}\right]\\ & =\frac{-256}{3}\pi \end{array}$$

  
  ::\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\2563\\\\\\\\\\\\\\\\\\\\\\\\\\\

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  The actual volume of the hollow core is $\begin{array}{r}\frac{256}{3}\pi \end{array}$ . Subtract to find the volume of the material that makes up the bead.
  ::空心核心的实际体积为2563。减号以找到构成珠子的材料体积。

  $$\begin{array}{r}\frac{500}{3}\pi -\frac{256}{3}\pi =\frac{244}{3}\pi \end{array}$$

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  The material that makes up the bead has a volume of  $\begin{array}{r}\frac{244}{3}\pi {\text{units}}^3\end{array}$ .
  ::构成珠子的材料数量为2443瓦3单位。

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  Example 3
  ::例3

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  Find the volume of the solid created when the following parametric curve is revolved around the $\begin{array}{r}x\end{array}$ -axis from  $\begin{array}{r}t=5\end{array}$ to $\begin{array}{r}t=10\end{array}$ .
  ::查找以下参数曲线围绕 x 轴从 t=5 到 t=10 围绕 x 轴旋转时所创建的固体体积。

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  $$\begin{array}{rl}F(t)& =(x(t),y(t))\\ x(t)& =t\\ y(t)& =\frac{3}{4}t+5\end{array}$$

  
  ::F(t) = (x(t) y(t) ) x(t) = ty(t) = 34t+5

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  $$\begin{array}{rl}y(t)& =\frac{3}{4}t+5\\ x^{\mathrm{\prime }}(t)& =1\\ \pi \underset{a}{\overset{b}{\int }}y(t{)}^2(x^{\mathrm{\prime }}(t))dt& =\pi \underset{5}{\overset{10}{\int }}{\left(\frac{3}{4}t+5\right)}^{2}dt\\ & =\pi \underset{5}{\overset{10}{\int }}\left(\frac{9}{16}{t}^2+\frac{15}{2}t+25\right)dt\\ & =\pi {\left[\frac{9}{48}{t}^3+\frac{15}{4}{t}^2+25t\right]}_5^{10}\\ & =\pi \left[\left[\frac{9}{48}(10{)}^3+\frac{15}{4}(10{)}^2+25(10)\right]-\left[\frac{9}{48}(5{)}^3+\frac{15}{4}(5{)}^2+25(5)\right]\right]\\ & =\pi (812.5)-\pi (242.1875)\\ & =570.3125\pi \end{array}$$

  
  ::y( t) = 34t+5x}( t) = 1 aby( t) 2( x} (t) d510( 34t+5) 510( 34t+5) 2dt510( 916t2+152t+25) d[ 948t3+154t2+25t] 510}[948( 1010) 3+1544( 10) 2+25( 10)]] - [948(5)3+154(5)2+25(5)] ] ( 812.5) ( ( 242. 1875)=570. 3125]]

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  The volume is $\begin{array}{r}570.3125\pi {\text{units}}^3\end{array}$ . Notice that this revolution created a cone.
  ::体积是570.3125立方单位3. 注意这场革命创造了锥体。

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  Example 4
  ::例4

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  Find the volume of the solid created when the following curve is revolved around the $\begin{array}{r}y\end{array}$ -axis from 0 to $\begin{array}{r}\frac{\pi }{2}\end{array}$ .
  ::查找以下曲线围绕 Y 轴从 0 到 2 旋转时所创建的固体体积。

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  $$\begin{array}{rl}G(t)& =(x(t),y(t))\\ x(t)& =5\cos (t)\\ y(t)& =5\sin (t)\end{array}$$

  
  ::G(t) = (x(t) y(t) ) x(t) = 5cos(t) y(t) = 5sin(t)

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  $$\begin{array}{rl}x(t)& =5\cos (t)\\ y(t)& =5\sin (t)\\ y^{\mathrm{\prime }}(t)& =5\cos (t)\\ \\ \pi \underset{a}{\overset{b}{\int }}x(t{)}^2(y^{\mathrm{\prime }}(t))dt& =\pi {\int }_0^{\frac{\pi }{2}}(5\cos t{)}^2(5\cos t)dt\\ & =125\pi \underset{0}{\overset{\frac{\pi }{2}}{\int }}(\cos t{)}^{3}dt\\ & =125\pi {\left[\sin t-\frac{1}{3}{\sin }^{3}t\right]}_0^{\frac{\pi }{2}}\\ & =\frac{250\pi }{3}\end{array}$$

  
  :t)=5cos}(t)=5sin}(t)=5cos}(t)=5cos}(t)=5cos}(t)=5cos}(t)=(t){abx(t)2(y_(t))}dt=02(5cos}}(t)=2(5cos}}}(t)dt=12502(cos}}}}(t)3dt=125}[sin}(t)-13sin3}}02=250}3

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  Example 5
  ::例5

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  Find the volume of plastic needed to make an object described by the revolution of  $\begin{array}{r}H(t)\end{array}$ from  $\begin{array}{r}\frac{\pi }{6}\end{array}$ to  $\begin{array}{r}\frac{5\pi }{6}\end{array}$ around the $\begin{array}{r}x\end{array}$ -axis, with a core described by  $\begin{array}{r}K(t)\end{array}$ drilled out.
  ::找到制造H(t)革命所描述的X轴周围6至5°6的物体所需的塑料体积,K(t)所描述的芯被钻出。

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  $$\begin{array}{rl}H(t)& =(x(t),y(t))\\ x(t)& =10\cos t\\ y(t)& =4\sin t\\ \\ K(t)& =(x(t),y(t))\\ x(t)& =t\\ y(t)& =.5\end{array}$$

  
  :t)=(x(t),y(t))x(t)=10costy(t)=4sin}}tK(t)=(x(t),y(t))x(t)=ty(t)=5。

  .

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  $$\begin{array}{rl}H(t)& =(x(t),y(t))\\ x(t)& =10\cos t\\ y(t)& =4\sin t\\ x^{\mathrm{\prime }}(t)& =-10\sin t\\ \\ K(t)& =(x(t),y(t))\\ x(t)& =t\\ y(t)& =.5\\ x^{\mathrm{\prime }}(t)& =1\end{array}$$

  $$\begin{array}{rl}\text{Volume}& =\pi \underset{\frac{\pi }{6}}{\overset{\frac{5\pi }{6}}{\int }}(4\sin t{)}^2(-10\sin t)dt-\pi \underset{\frac{\pi }{6}}{\overset{\frac{5\pi }{6}}{\int }}(.5{)}^2(1)dt\\ & =\pi \underset{\frac{\pi }{6}}{\overset{\frac{5\pi }{6}}{\int }}-160{\sin }^{3}tdt-\pi \underset{\frac{\pi }{6}}{\overset{\frac{5\pi }{6}}{\int }}.25dt\\ & =-160\pi {\left[\frac{1}{3}{\cos }^{3}t-\cos t\right]}_{\frac{\pi }{6}}^{\frac{5\pi }{6}}-\pi [.25t{]}_{\frac{\pi }{6}}^{\frac{5\pi }{6}}\\ & \approx -654.613\end{array}$$

  
  ::H( t) = (x( t) ,y( t) x( t) x( t) = 10cos( t) ty( t) = 4sin( t) }( t) { @ 10sin} (t) = (x( t) = (x( t) ) 5x* (t) = (t) = 5x* (t) = 1 Volume* 65* 6( 4sin})( 2- 10sin} (t) dt) = 65}6}6(6) 2 (1) dt 65_ 6_ 6- 160sin3} (t) @ 65_ 6. 25dt}}} [13 coscos3, } - cos3} - cos}}}} __ 656[ [ 6} [ 6_ 65_64.613

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  The volume of the plastic needed is approximately $\begin{array}{r}654.613unit{s}^3\end{array}$ .
  ::所需的塑料量约为654.613个单位3。

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  Review
  ::回顾

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  For #1-7, set up but do not solve the following volume problems. 
  ::对于#1-7, 设置但不解决以下的量问题 。

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  1. From  $\begin{array}{r}t=1\end{array}$ to $\begin{array}{r}t=5\end{array}$ , revolved around the $\begin{array}{r}x\end{array}$ -axis:
     $$\begin{array}{rl}F(t)& =(x(t),y(t))\\ x(t)& =2t^2+1\\ y(t)& =4t\end{array}$$
     
     ::从t=1到t=5,围绕x轴:F(t)=(x(t),y(t))x(t)=2t2+1y(t)=4t
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  2. From $\begin{array}{r}t=1\end{array}$ to $\begin{array}{r}t=5\end{array}$ , revolved around the $\begin{array}{r}y\end{array}$ -axis:
     $$\begin{array}{rl}F(t)& =(x(t),y(t))\\ x(t)& =2t^2+1\\ y(t)& =4t\end{array}$$
     
     ::从t=1到t=5,围绕 Y 轴: F(t) = (x(t),y(t) ) x(t) = 2t2+1y(t) =4t
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  3. From $\begin{array}{r}t=\frac{\pi }{6}\end{array}$  to $\begin{array}{r}t=\frac{\pi }{3}\end{array}$ , revolved around the $\begin{array}{r}y\end{array}$ -axis:
     $$\begin{array}{rl}G(t)& =(x(t),y(t))\\ x(t)& =\tan (t)\\ y(t)& =t^2\end{array}$$
     
     ::从 t6 到 t3, 围绕 Y 轴 : G( t) = (x( t),y( t) ) x( t) =tan( t) y( t) = t2
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  4. From $\begin{array}{r}t=\frac{\pi }{6}\end{array}$  to $\begin{array}{r}t=\frac{\pi }{3}\end{array}$ , revolved around the $\begin{array}{r}x\end{array}$ -axis:
     $$\begin{array}{rl}G(t)& =(x(t),y(t))\\ x(t)& =\tan (t)\\ y(t)& =t^2\end{array}$$
     
     ::从t6到t3,围绕 x 轴: G( t) = (x( t),y( t) ) x( t) =tan( t) y( t) = t2
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  5. From  $\begin{array}{r}t=0\end{array}$ to $\begin{array}{r}t=1.5\end{array}$ , revolved around the $\begin{array}{r}y\end{array}$ -axis:
     $$\begin{array}{rl}H(t)& =(x(t),y(t))\\ x(t)& =t+7\\ y(t)& =2t^3\end{array}$$
     
     ::从t=0到t=1.5,围绕 Y 轴旋转: H( t) = (x( t),y( t) ) x( t) = t+7y( t) =2t3
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  6. From  $\begin{array}{r}t=0\end{array}$ to  $\begin{array}{r}t=\frac{\pi }{2}\end{array}$ , revolved around the $\begin{array}{r}x\end{array}$ -axis:
     $$\begin{array}{rl}J(t)& =(x(t),y(t))\\ x(t)& =3\cos (t)\\ y(t)& =2\sin (t)\end{array}$$
     
     ::从t=0到t2,围绕x轴旋转:J(t)=(x(t),y(t))x(t)=3cos(t)y(t)=2sin(t)
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  7. From $\begin{array}{r}t=0\end{array}$ to  $\begin{array}{r}t=\frac{\pi }{2}\end{array}$ , revolved around the $\begin{array}{r}y\end{array}$ -axis:
     $$\begin{array}{rl}J(t)& =(x(t),y(t))\\ x(t)& =3\cos (t)\\ y(t)& =2\sin (t)\end{array}$$
     
     ::从 t=0 到 t2, 围绕 Y 轴旋转: J( t) = (x( t),y( t) ) x( t) = 3cos *( t) y( t) = 2sin ( t)
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  8. Revolve the following parametric function around the $\begin{array}{r}x\end{array}$ -axis and find the volume of the solid from $\begin{array}{r}t=0\end{array}$ to  $\begin{array}{r}t=\frac{\pi }{2}\end{array}$ .
     $$\begin{array}{rl}M(t)& =(x(t),y(t))\\ x(t)& =3\cos (2t)\\ y(t)& =3\sin (2t)\end{array}$$
     
     ::在 x 轴周围循环以下的参数函数, 并查找从 t=0 到 t2. M( t) = (x( t),y( t)) x( t) = 3cos ( 2t)y( t) = 3sin( 2t) = 3sin}( 2t) 的固体体积
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  9. Revolve the following parametric function around the $\begin{array}{r}y\end{array}$ -axis and find the volume of the solid from  $\begin{array}{r}t=0\end{array}$ to $\begin{array}{r}t=2\end{array}$ .
     $$\begin{array}{rl}N(t)& =(x(t),y(t))\\ x(t)& =t\\ y(t)& =-2t\end{array}$$
     
     ::围绕 Y 轴循环以下的参数函数, 并查找从 t=0 到 t=2. N( t) = (x( t),y( t)) x( t) =ty( t)\\\\\\\ 2t 的固体体积
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  10. Revolve the following parametric function around the $\begin{array}{r}x\end{array}$ -axis and find volume of the solid from  $\begin{array}{r}t=0\end{array}$ to $\begin{array}{r}t=\pi \end{array}$ . Use a calculator to help.
      $$\begin{array}{rl}p(t)& =(x(t),y(t))\\ x(t)& =2\cos (t)\\ y(t)& =5\sin (t)\end{array}$$
      
      ::在 x 轴周围循环以下的参数函数, 并查找从 t= 0 到 t 的固体体积 。 使用计算器帮助. p( t) = (x( t), y( t) ) x( t) = 2cos *( t) = 5sin ( t)
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  11. Revolve the following parametric function around the $\begin{array}{r}x\end{array}$ -axis and find the volume of the solid from  $\begin{array}{r}t=0\end{array}$ to $\begin{array}{r}t=2\end{array}$ .
      $$\begin{array}{rl}Q(t)& =(x(t),y(t))\\ x(t)& =t^3\\ y(t)& =t\end{array}$$
      
      ::在 x 轴周围循环以下的参数函数, 并查找从 t=0 到 t=2. Q( t) = (x( t),y( t)) x( t) =t3y( t) =t
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  12. Find the volume of the solid created when the following parametric curve is revolved around the $\begin{array}{r}x\end{array}$ -axis from  $\begin{array}{r}t=5\end{array}$ to $\begin{array}{r}t=10\end{array}$ .
      $$\begin{array}{rl}R(t)& =(x(t),y(t))\\ x(t)& =\frac{2}{3}t\\ y(t)& =t+5\end{array}$$
      
      ::查找以下参数曲线围绕 x 轴从 t=5 到 t=10.R(t) = (x(t),y(t)) x(t) = 23ty(t) = t+5 围绕 x 轴旋转时所创建的固体体积
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  13. Find the volume of the solid created when the following curve is revolved around the $\begin{array}{r}x\end{array}$ -axis from 0 to $\begin{array}{r}\pi \end{array}$ .
      $$\begin{array}{rl}A(t)& =(x(t),y(t))\\ x(t)& =3\cos (t)\\ y(t)& =3\sin (t)\end{array}$$
      
      ::查找以下曲线围绕 X 轴从 0 到 . A( t) = (x( t) ,y( t) ) x( t) = 3cos ( t) y( t) = 3sin ( t) = 3sin ( t) 时创建的固体的体积
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  14. Find the volume of the solid created when the following curve is revolved around the $\begin{array}{r}x\end{array}$ -axis from -2.9896 to 2.9896.
      $$\begin{array}{rl}B(t)& =(x(t),y(t))\\ x(t)& =t\\ y(t)& =\frac{1}{4}\end{array}$$
      
      ::查找以下曲线围绕 x 轴旋转时所创建的固体的体积, 该曲线由 -2.9896 到 2.9896.B(t) = (x(t),y(t)) x(t) =ty(t)=14
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  15. Find the volume of plastic needed to make an object defined by the revolution of  $\begin{array}{r}A(t)\end{array}$ (as described in #13) with a core of  $\begin{array}{r}B(t)\end{array}$ (as described in #14) drilled out.
      ::找到制造A(t)革命(如第13节所述)所定义的、B(t)核心(如第14节所述)被钻出的物体所需的塑料体积。

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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 。