Course: 10.8 参数形式和微积分:革命表面面积

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参数形式和微积分: 革命表面面积
Alicia is using duct tape to make a prom dress. How many square centimeters of hot pink-colored tape will she need to cover her skirt? Can she write a parametric equation to describe the shape of her gown and use that to figure out how much tape she needs?
::Alicia用胶带来做一个舞会礼服。她需要多少平方厘米的热粉色胶带来覆盖她的裙子?她能写一个参数方程式来描述她的礼服的形状,用它来弄清楚她需要多少胶带吗?

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

Once you’ve learned how to find the length of a curve in parametric form , you can find the surface area of a revolution. Why? Look at your sleeves. If you’re wearing a long sleeved shirt, you may notice that each sleeve is made up of many, tiny lines or thread. Together, they make up the whole piece of fabric. If you knew the length of each thread and the distance around the sleeve, you could calculate the amount of material covering your arm.
::一旦你学会了如何找到以参数形表示的曲线长度,你就可以找到革命的表面区域。为什么?看看袖子。如果你穿着长袖衬衫,你可能会注意到每个袖子都是由许多细细的线条或线条组成的。它们共同构成整个织物。如果你知道每条线的长度和袖子的距离,你就可以计算臂上材料的数量。

When you rotate a section of a curve around an axis, you’re basically making a sleeve for the axis. The ‘threads’ are infinitely thin, but you can measure their length, and you can calculate the circumference of the ‘sleeve.’ With this information, you can find the surface area of a rotation in parametric form.
::当您在轴周围旋转一条曲线时,您基本上是为轴做一个袖子。“线索”是无限细的,但可以测量它们的长度,也可以计算“袖子”的周长。 ”有了这些信息,您可以找到以参数形式旋转的表面区域。

Do you remember how to find the surface area of a cylinder? When you revolve a horizontal line segment around the $\begin{aligned} x \end{aligned}$ -axis or a vertical line segment around the $\begin{aligned} y \end{aligned}$ -axis, you get a cylinder.
::您还记得如何找到圆柱体的表面区域吗?当您绕 X 轴或 Y 轴周围的垂直线段旋转一个水平线段时,您可以得到一个圆柱体。

You may recall that the formula for the surface area of a cylinder, excluding the two ends, is $\begin{aligned} A = 2 π r h \end{aligned}$ , where $\begin{aligned} r \end{aligned}$ is the radius of the circle at the base of the cylinder and $\begin{aligned} h \end{aligned}$ is the height of the cylinder. When you create a surface by revolving a horizontal segment around the $\begin{aligned} x \end{aligned}$ -axis, the radius of the three dimensional solid is a constant – the distance of the line segment from the axis. So for example, if you create a solid by revolving a portion of the line $\begin{aligned} y = 5 \end{aligned}$ around the $\begin{aligned} x \end{aligned}$ -axis, the radius of your cylinder will be five.
::您可能记得, 圆柱体表面面积的公式, 不包括两端, 是 A=2rh , 其中 r 是圆柱体底部的圆圆半径, h 是圆柱体的高度。当您在 X 轴周围旋转一个水平段来创建表面时, 三维固体的半径是一个常数 — — 线段与轴的距离。因此, 例如, 如果您通过在 x 轴周围旋转一条线 y=5 来创建固体, 您圆柱体的半径将是 5 。

When you revolve a more complicated function around the $\begin{aligned} x \end{aligned}$ -axis, you get more complicated solids because the function value changes constantly, and the length of a curved solid is more difficult to measure than the height of a straight line. To find the surface area of complex solids, you can use an integral.
::当您围绕 x 轴旋转一个更复杂的函数时, 就会得到更复杂的固体, 因为函数值会不断变化, 而曲线的固体长度比直线的高度更难测量。要找到复杂固体的表面区域, 您可以使用一个组件。

The formula for the surface area of a parametric equation revolved around the $\begin{aligned} x \end{aligned}$ -axis takes the formula for the length of a parametric curve and multiplies it by the circumference of the solid at time $\begin{aligned} t \end{aligned}$ .
::围绕 x 轴旋转的参数方程式的表面积公式采用参数曲线长度的公式,乘以时间 t 的固体环形。

$\begin{aligned} A_{s u r f a c e} = \int_{a}^{b} 2 π y (t) \sqrt{(x^{'} (t))^{2} + (y^{'} (t))^{2}} d t \end{aligned}$

::平面ab2 y( t)( x_( t)) 2+(y_( t))2dt

Notice that this is a definite integral. You can’t find the surface area of a solid unless it has a finite height. If you revolve the solid around the $\begin{aligned} y \end{aligned}$ -axis instead, the equation changes slightly to:
::请注意这是一个明确的整体部分。您找不到固体的表面区域, 除非有一定的高度。如果您绕 Y 轴旋转固体, 方程式会稍有变化 :

$\begin{aligned} A_{s u r f a c e} = \int_{a}^{b} 2 π x (t) \sqrt{(x^{'} (t))^{2} + (y^{'} (t))^{2}} d t \end{aligned}$

::表面ab2 x(t)(x_(t))2+(y_(t))2dt

The equation is changed because the radius of a solid revolved around the $\begin{aligned} y \end{aligned}$ -axis is the distance between the axis and the $\begin{aligned} x \end{aligned}$ value of the equation at that point.
::由于围绕 Y 轴旋转的固体半径是轴与该点方程的 x 值之间的距离,所以方程被改变。

As with length of a curve formulas, many surface area formulas can be very difficult to integrate using normal methods. In these cases, it’s all right to use a calculator to find a numeric approximation for the integral. However, you should be able to complete the problems below in with the integration techniques you already know.
::与曲线公式的长度一样,许多表面积公式可能很难使用正常方法整合。在这种情况下,使用计算器为集成物找到数字近似值是完全正确的。但是,您应该能够用您已经知道的集成技术解决下面的问题。

For instance, find the surface area of the solid formed by rotating the following curve between $\begin{aligned} t = 0 \end{aligned}$ and $\begin{aligned} t = \frac{π}{2} \end{aligned}$ around the $\begin{aligned} x \end{aligned}$ -axis.
::例如,在 x 轴周围旋转 t=0 和 t2 之间的以下曲线,发现固体的表面区域。

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

::F(x(t),y(t))x(t)=5costy(t)=5sint

You are rotating a quarter circle around the $\begin{aligned} x \end{aligned}$ -axis. Star by finding the of $\begin{aligned} x (t) \end{aligned}$ and $\begin{aligned} y (t) \end{aligned}$ . Then use the surface area formula for revolutions around the $\begin{aligned} x \end{aligned}$ -axis.
::您正在 X 轴周围旋转四分之一圆。通过查找 x( t) 和 y( t) 的星。然后使用表面积公式进行 X 轴周围的革命。

$\begin{aligned} x^{'} (t) = - 5 \sin t \end{aligned}$
::{\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}

$\begin{aligned} y^{'} (t) = 5 \cos t \end{aligned}$
::y'(t)=5cost

$\begin{aligned} A_{s u r f a c e} & = \int_{0}^{\frac{π}{2}} 2 π y (t) \sqrt{(x^{'} (t))^{2} + (y^{'} (t))^{2}} d t \\ A_{s u r f a c e} & = \int_{0}^{\frac{π}{2}} 2 π (5 \sin t) \sqrt{(- 5 \sin t)^{2} + (5 \cos t)^{2}} d t \\ A_{s u r f a c e} & = \int_{0}^{\frac{π}{2}} 2 π (5 \sin t) \sqrt{(25 \sin^{2} t + 25 \cos^{2} t)} d t \\ A_{s u r f a c e} & = \int_{0}^{\frac{π}{2}} 2 π (5 \sin t) 5 \sqrt{\sin^{2} t + \cos^{2} t} d t \\ A_{s u r f a c e} & = 50 π \int_{0}^{\frac{π}{2}} \sin (t) \sqrt{1} d t \\ A_{s u r f a c e} & = 50 π \int_{0}^{\frac{π}{2}} \sin (t) d t \\ A_{s u r f a c e} & = 50 π {[- \cos (t)]}_{0}^{\frac{π}{2}} \\ A_{s u r f a c e} & = 50 π [- \cos (\frac{π}{2}) - (- \cos (0))] \\ A_{s u r f a c e} & = 50 π [0 + 1] = 50 π \end{aligned}$

::2+(y_(t))2dtApoloo=50°0°2sin(t)1dtApoloo=5002sin}(t)2+(5cos_t)2dtApoloo_02}(5sin}}}}(25sin2t+25cos2}}}(t)dtApoloo=02}(5sin}}(t)2+(y_(t))2+(y})2dtApoloool=5002sin(t)1dtApolooto=50002sin}(t)_(t)dt)_(dt)_(dt)ooooooooto=50[-(t)_(t)_02_A_50_(cos_(_)_(0+1)_=50__

The surface area of the revolution is $\begin{aligned} 50 π {units}^{2} \end{aligned}$ .
::革命的地表面积是50°2单位。

Above , you revolved a quarter-circle around the $\begin{aligned} x \end{aligned}$ -axis and found the surface area of half of a sphere. Now, revolve the same curve around the $\begin{aligned} y \end{aligned}$ -axis and find the area over the same range. What solid do you think you’ll make? What do you think the surface area of that solid will be?
::上面,你绕着 X 轴绕了四分之一个圆圈,发现了半个球体的表面区域。现在,在 Y 轴周围绕着同样的曲线转,并在相同的射程上找到这个区域。你认为你能够做哪些固体?你觉得这块固体的表面区域会是什么?

$\begin{aligned} x^{'} (t) = - 5 \sin t \end{aligned}$
::{\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}

$\begin{aligned} y^{'} (t) = 5 \cos t \end{aligned}$
::y'(t)=5cost

$\begin{aligned} A_{s u r f a c e} & = \int_{0}^{\frac{π}{2}} 2 π x (t) \sqrt{(x^{'} (t))^{2} + (y^{'} (t))^{2}} d t \\ A_{s u r f a c e} & = \int_{0}^{\frac{π}{2}} 2 π (5 \cos t) \sqrt{(- 5 \sin t)^{2} + (5 \cos t)^{2}} d t \\ A_{s u r f a c e} & = \int_{0}^{\frac{π}{2}} 2 π (5 \cos t) \sqrt{(25 \sin^{2} t + 25 \cos^{2} t)} d t \\ A_{s u r f a c e} & = \int_{0}^{\frac{π}{2}} 2 π (5 \cos t) 5 \sqrt{\sin^{2} t + \cos^{2} t} d t \\ A_{s u r f a c e} & = 50 π \int_{0}^{\frac{π}{2}} \cos t \sqrt{1} d t \\ A_{s u r f a c e} & = 50 π \int_{0}^{\frac{π}{2}} \cos (t) d t \\ A_{s u r f a c e} & = 50 π {[\sin (t)]}_{0}^{\frac{π}{2}} \\ A_{s u r f a c e} & = 50 π [\sin (\frac{π}{2}) - (\sin (0))] \\ A_{s u r f a c e} & = 50 π [1 - 0] = 50 π \end{aligned}$

::022 x (t)(x(t)) 2+(y(t) ) 2dtA forsploo022(5cos) (-5sint) 2+(5cost)2+(5cost)2&(5cost)2_(25sin2t+25cos2t)2}(5cost)2+(5cos%2t)2+(y}})2+(y})2dtAground=5002}(t) 地表=50}(sin)02A 地表=50}(sin(t)]0)-(sin2)-(0)]Aboomloo=50[1-0]=50(7)

The quarter circle happens to create the same solid and have the same surface area no matter which axis you revolve it around.
::四分之一圆碰巧形成同样的固体并且有着相同的表面积不管你绕着哪个轴环绕

Examples
::实例

Example 1
::例1

Earlier, you were asked about making a duct tape skirt. Alicia can use to model the shape of her skirt and to compute her duct tape needs. Assume that her skirt can be modeled by the following parametric equation when the segment from $\begin{aligned} t = 1 \end{aligned}$ to $\begin{aligned} t = 8 \end{aligned}$ is revolved around the $\begin{aligned} y \end{aligned}$ -axis.
::Alicia可以用她的裙子的形状来模拟和计算她的胶带需要。假设从t=1到t=8的线段围绕Y轴旋转时,她的裙子可以用以下的参数方程来模拟。

$\begin{aligned} F (t) & = (x (t), y (t)) \\ x (t) & = \frac{1}{2} t \\ y (t) & = \frac{1}{3} t^{2} \end{aligned}$

::F(t) = (x(t) y(t) ) x(t) = 12ty(t) = 13t2

To find the amount of hot pink duct tape that she’ll need for her skirt, Alicia can use the equation for the surface area of a revolution around the $\begin{aligned} y \end{aligned}$ -axis. She can use substitution to solve the problem.
::为了找到她裙子需要的热粉色胶带数量,Alicia可以使用Y轴周围革命表面的方程式。她可以使用替代方程式解决问题。

$\begin{aligned} F (t) & = (x (t), y (t)) \\ x (t) & = \frac{1}{2} t \\ y (t) & = \frac{1}{3} t^{2} \\ x^{'} (t) & = \frac{1}{2}, y^{'} (t) = \frac{2}{3} t \end{aligned}$

::F(t) = (x(t) y(t) y(t) x(t) = 12ty(t) = 13t2x_(t) = 12,y_(t) = 23t

$\begin{aligned} A_{s u r f a c e} & = \int_{a}^{b} 2 π (x (t)) \sqrt{(x^{'} (t))^{2} + (y^{'} (t))^{2}} d t \\ A_{s u r f a c e} & = \int_{1}^{8} 2 π (\frac{1}{2} t) \sqrt{{(\frac{1}{2})}^{2} + {(\frac{2}{3} t)}^{2}} d t \\ A_{s u r f a c e} & = π \int_{1}^{8} t \sqrt{\frac{1}{4} + \frac{4}{9} t^{2}} d t \end{aligned}$

:x(t)x_(t)x_(t))2+(y_(t))2+(y_(t))2dtAoum_182}(12)2(12)2+(23t)2dt_(23t)2dtAoum_}18t14+49t2dt

$\begin{aligned} u = \frac{1}{4} + \frac{4}{9} t^{2} \end{aligned}$
::u=14+49t2

$\begin{aligned} d u = \frac{8}{9} t \end{aligned}$
::=89吨

$\begin{aligned} A_{S u r f a c e} & = \frac{9}{8} π \int u^{\frac{1}{2}} d u \\ A_{S u r f a c e} & = \frac{9}{8} π [\frac{2}{3} u^{\frac{3}{2}}] \\ A_{S u r f a c e} & = \frac{9}{8} π {[\frac{2}{3} {(\frac{1}{4} + \frac{4}{9} t)}^{\frac{3}{2}}]}_{1}^{8} \\ A_{S u r f a c e} & = 360. 803 \end{aligned}$

::[23u32] [23(14+49t)32] [18(15)=360.803]

Example 2
::例2

Revolve the following function around the $\begin{aligned} x \end{aligned}$ -axis, then find its surface area between $\begin{aligned} t = 0 \end{aligned}$ and $\begin{aligned} t = 4 \end{aligned}$ :
::在 x 轴周围循环以下函数,然后在 t=0 和 t=4 之间发现其表面面积:

$\begin{aligned} F (x) & = (x (t), y (t)) \\ x (t) & = t^{2} \\ y (t) & = t \end{aligned}$

::F(x)=(x(t),y(t))x(t)=t2y(t)=t

Use the surface area formula.
::使用表面积公式。

$\begin{aligned} F (x) & = (x (t), y (t)) \\ x (t) & = t^{2} \\ y (t) & = t \\ x^{'} (t) & = 2 t \\ y^{'} (t) & = 1 \end{aligned}$

::F(x) = (x) = (x(t) y(t) ) x(t) = t2y(t) = tx_(t) = 2ty_(t) =1

$\begin{aligned} A_{s u r f a c e} & = \int_{0}^{4} 2 π y (t) \sqrt{(x^{'} (t))^{2} + (y^{'} (t))^{2}} d t \\ A_{s u r f a c e} & = \int_{0}^{4} 2 π t \sqrt{(2 t)^{2} + (1)^{2}} d t \\ A_{s u r f a c e} & = \int_{0}^{4} 2 π t \sqrt{4 t^{2} + 1} d t \\ A_{s u r f a c e} & = 2 π \int_{0}^{4} t \sqrt{4 t^{2} + 1} d t \end{aligned}$

::地表@042}y(t)(x_(t)) 2+(y_(t)) 2dtA 地表 @042}t(t)(2t) 2+(1)2dtA 地表 @042}t4}t2+1dtA 地表=2}044}t2+1dtA 地表=2}t2}t2+1dt

To integrate, use substitution.
::集成,使用替代。

$\begin{aligned} A_{s u r f a c e} & = 2 π \int_{0}^{4} t \sqrt{4 t^{2} + 1} d t \\ u & = 4 t^{2} + 1 \\ d u & = 8 t \\ A_{s u r f a c e} & = 2 π \int_{0}^{4} \frac{8}{8} t \sqrt{4 t^{2} + 1} d t \\ A_{s u r f a c e} & = \frac{π}{4} \int \sqrt{u} d u = \frac{π}{4} [\frac{2}{3} (u)^{\frac{3}{2}}] = \frac{π}{4} {[\frac{2}{3} (4 t^{2} + 1)^{\frac{3}{2}}]}_{0}^{4} = \frac{π}{6} [(4 (16) + 1)^{\frac{3}{2}}] - \frac{π}{6} (1) \\ = \frac{π}{6} (65 \sqrt{65} - 1) \approx 273.8666 \end{aligned}$

::地表=204t4t2+1dtu=4t2+1du=8tA地表=20488t4t2+1dtA地表=4udu4[23(u)32]4[23(4t2+132)]04}6[4(4(16)+132)]6(1)(6)(6565-1)273.8666]

Example 3
::例3

Set up the following surface area problems. You do not need to solve them.
::设置以下的表面区域问题。您不需要解决这些问题。

From 1 to 3, revolved around the $\begin{aligned} x \end{aligned}$ -axis:
::1到3,围绕X轴旋转:

$\begin{aligned} F (t) & = (x (t), y (t)) \\ x (t) & = 3 t^{3} - 7 t^{2} \\ y (t) & = t^{4} + 10 \end{aligned}$

::F(t) = (x(t) = (x(t) y(t) ) x(t) = 3t3- 7t2y(t) = t4+10

From $\begin{aligned} \frac{π}{6} \end{aligned}$ to $\begin{aligned} \frac{π}{4} \end{aligned}$ , revolved around the $\begin{aligned} y \end{aligned}$ -axis:
::从6点到4点,围绕Y轴旋转:

$\begin{aligned} F (t) & = (x (t), y (t)) \\ x (t) & = \tan t \\ y (t) & = t^{\frac{- 4}{3}} \end{aligned}$

::F(t) = (x(t) y(t) ) x(t) = tanty(t) = t-43

First find $\begin{aligned} x^{'} (t) \end{aligned}$ and $\begin{aligned} y^{'} (t) \end{aligned}$ , then substitute into the formula for surface area.
::首先找到 x_(t) 和 y_(t) , 然后替换为表面积的公式。

$\begin{aligned} F (t) & = (x (t), y (t)) \\ x (t) & = 3 t^{3} - 7 t^{2} \\ y (t) & = t^{4} + 10 \\ x^{'} (t) & = 9 t^{2} - 14 t \\ y^{'} (t) & = 4 t^{3} \\ A_{s u r f a c e} & = \int_{1}^{3} 2 π (t^{4} + 10) \sqrt{(9 t^{2} - 14 t)^{2} + (4 t^{3})^{2}} d t \end{aligned}$

::F( t) = (x( t) = (x( t) ,y( t) ) x( t) = 3t3- 7t2y( t) = t4+10x*( t) = 9t2- 14ty_ ( t) = 4t3A 地表= 132&( t) (t4+10) (9t2- 14t) 2+( 4t3) 2dt

$\begin{aligned} F (t) & = (x (t), y (t)) \\ x (t) & = \tan (t) \\ y (t) & = t^{\frac{- 4}{3}} \\ x^{'} (t) & = \sec^{2} (t) \\ y^{'} (t) & = \frac{- 4}{3} (t)^{\frac{- 7}{3}} \\ A_{s u r f a c e} & = \int_{\frac{π}{6}}^{\frac{π}{4}} 2 π \tan (t) \sqrt{(\sec^{2} (t))^{2} + {(\frac{- 4}{3} (t)^{\frac{- 7}{3}})}^{2}} d t \end{aligned}$

::F( t) = (x( t) = (x( t) ,y( t) x( t) = tan( t) y( t) = t- 43x( t) = sec2 - ( t) y( t) = ( t) = = (t) = = (t) = (t) = = = (t) = (t) = (y2) = (t) = (t) = (yc2) = (t) = (y2) = (t) = (t) = sec2 (t) 2+ (dt) +( 43( t)- 732dt)

Example 4
::例4

Find the area from $\begin{aligned} t = 0 \end{aligned}$ to $\begin{aligned} t = 7 \end{aligned}$ when you revolve the following curve around the $\begin{aligned} y \end{aligned}$ -axis.
::当您在 Y 轴周围旋转下一条曲线时, 从 t=0 到 t=7 查找区域。

$\begin{aligned} F (t) & = (x (t), y (t)) \\ x (t) & = t \\ y (t) & = \frac{4}{3} t + 7 \end{aligned}$

::F(t) = (x(t) y(t) ) x(t) = ty(t) = 43t+7

$\begin{aligned} F (t) & = (x (t), y (t)) \\ x (t) & = t \\ y (t) & = \frac{4}{3} t + 7 \\ x^{'} (t) & = 1 \\ y^{'} (t) & = \frac{4}{3} \end{aligned}$

::F(t) = (x(t) y(t) ) x(t) = ty(t) = 43t+7x_(t) = 1y}(t) = 43

$\begin{aligned} A_{s u r f a c e} & = \int_{0}^{7} 2 π t \sqrt{1^{2} + {(\frac{4}{3})}^{2}} d t \\ = 2 π \int_{0}^{7} t \sqrt{\frac{25}{9}} = \frac{10}{3} π \int_{0}^{7} t \\ = \frac{10}{3} π {[\frac{1}{2} t^{2}]}_{0}^{7} \\ = \frac{5}{3} π (49) - \frac{5}{3} π (0) \\ = \frac{245}{3} π \end{aligned}$

::地表072t12+(432dt=207t259=103*07t=103}[12t2]07=53}(49)-53(0)=2453

Notice that the ‘curve’ that you revolved here was actually an oblique line, and that when you revolved it, it made a cone.
::注意你在这里转动的“曲线”其实是一条斜线, 当你转动时,它就形成了锥形。

Example 5
::例5

Revolve the following parametric function around the $\begin{aligned} x \end{aligned}$ -axis and find the surface area of the figure from $\begin{aligned} t = 0 \end{aligned}$ to $\begin{aligned} t = π \end{aligned}$ .
::在x轴周围循环以下参数函数,并找到图的表面积,从t=0到t。

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

::F(t) = (x(t) y(t) ) x(t) = 3cos(t) y(t) = 3sin(t)

$\begin{aligned} x^{'} (t) = - 3 \sin t \end{aligned}$
::{\fn方正黑体简体\fs18\b1\bord1\shad1\3cH2F2F2F

$\begin{aligned} y^{'} (t) = 3 \cos t \end{aligned}$
::y'(t)=3cost

$\begin{aligned} A_{s u r f a c e} & = \int_{0}^{π} 2 π (3 \sin (t)) \sqrt{(- 3 \sin t)^{2} + (3 \cos t)^{2}} d t \\ = \int_{0}^{π} 6 π (3 \sin (t)) \sqrt{\sin^{2} t + \cos^{2} t} d t \\ = \int_{0}^{π} 6 π (3 \sin (t)) \sqrt{1} d t \\ = 18 π [- \cos (π) - (- \cos (0))] \\ = 18 π (1 + 1) \\ = 18 π (2) \\ = 36 π \end{aligned}$

::3sin(t) (-3sin(t)) 2+(3cos(t) 2dt) @06(3sin(t))sin2t+cos2t}tdt06(t)1dt=18[-cos()-(-cos(0)]=18(1+1)=18(2)=36

Notice that by going from 0 to $\begin{aligned} π \end{aligned}$ , you revolved a semi-circle around the $\begin{aligned} x \end{aligned}$ -axis. You created a sphere. The geometric formula for the surface area of a sphere is $\begin{aligned} 4 π r^{2} \end{aligned}$ , and the radius of the revolution was 3.
::注意从 0 到 , 你绕着 X 轴旋转了半圆圈。你创造了一个球体。球体表面的几何公式是 4°r2, 革命半径是 3 。

Review
::回顾

For #1-5, set up but do not solve the surface area problems.
::为1 -5设立,但不能解决地表问题。

From $\begin{aligned} t = 1 \end{aligned}$ to $\begin{aligned} t = 5 \end{aligned}$ , revolved around the $\begin{aligned} x \end{aligned}$ -axis:
$\begin{aligned} F (t) & = (x (t), y (t)) \\ x (t) & = 2 t^{2} + 1 \\ y (t) & = 4 t \end{aligned}$

::从t=1到t=5,围绕x轴:F(t)=(x(t),y(t))x(t)=2t2+1y(t)=4t

From $\begin{aligned} t = \frac{π}{6} \end{aligned}$ to $\begin{aligned} t = \frac{π}{3} \end{aligned}$ , revolved around the $\begin{aligned} y \end{aligned}$ -axis:
$\begin{aligned} G (t) & = (x (t), y (t)) \\ x (t) & = \tan (t) \\ y (t) & = t^{2} \end{aligned}$

::从 t6 到 t3, 围绕 Y 轴 : G( t) = (x( t),y( t) ) x( t) =tan( t) y( t) = t2

From $\begin{aligned} t = 0 \end{aligned}$ to $\begin{aligned} t = 1.5 \end{aligned}$ , revolved around the $\begin{aligned} y \end{aligned}$ -axis:
$\begin{aligned} H (t) & = (x (t), y (t)) \\ x (t) & = t + 7 \\ y (t) & = 2 t^{3} \end{aligned}$

::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

From $\begin{aligned} t = 0 \end{aligned}$ to $\begin{aligned} t = \frac{π}{2} \end{aligned}$ , revolved around the $\begin{aligned} x \end{aligned}$ -axis:
$\begin{aligned} J (t) & = (x (t), y (t)) \\ x (t) & = 3 \cos (t) \\ y (t) & = 2 \sin (t) \end{aligned}$

::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)

From $\begin{aligned} t = 0 \end{aligned}$ to $\begin{aligned} t = \frac{π}{2} \end{aligned}$ , revolved around the $\begin{aligned} y \end{aligned}$ -axis:
$\begin{aligned} J (t) & = (x (t), y (t)) \\ x (t) & = 3 \cos (t) \\ y (t) & = 2 \sin (t) \end{aligned}$

::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)

Use a calculator to evaluate the integrals from #4 and #5 and determine the approximate surface areas of the solids. Earlier, you saw that when a quarter circle was revolved around the $\begin{aligned} x \end{aligned}$ -axis its surface area was the same as its surface area when it was revolved around the $\begin{aligned} y \end{aligned}$ -axis. Why is that not true for $\begin{aligned} J (t) \end{aligned}$ ?
::使用计算器来评估来自 # 4 和 # 5 的构件, 并确定固体的大致表面积。早些时候, 您看到, 当 X 轴周围旋转四分之一圆时, 其表面积与 y 轴周围旋转时的表面积相同。为什么J (t) 不这样呢 ?

Find the area from $\begin{aligned} t = 0 \end{aligned}$ to $\begin{aligned} t = 2 \end{aligned}$ when you revolve the following curve around the $\begin{aligned} x \end{aligned}$ -axis.
$\begin{aligned} K (t) & = (x (t), y (t)) \\ x (t) & = 3 t - 4 \\ y (t) & = t \end{aligned}$

::当您在 x- 轴. K( t) = (x( t) ,y( t) ) x( t) = 3t- 4y( t) = t 周围旋转以下曲线时, 从 t=0 到 t=2 查找区域

Revolve the following parametric function around the $\begin{aligned} x \end{aligned}$ -axis and find the surface area of the figure from $\begin{aligned} t = 0 \end{aligned}$ to $\begin{aligned} t = \frac{π}{2} \end{aligned}$ .
$\begin{aligned} M (t) & = (x (t), y (t)) \\ x (t) & = 3 \cos (2 t) \\ y (t) & = 3 \sin (2 t) \end{aligned}$

::在 x 轴周围循环以下的参数函数, 并查找图的表面面积, 从 t=0 到 t2. M( t) = (x( t),y( t) ) x( t) = 3cos ( 2t)y( t) = 3sin ( 2t) = 3sin @ ( 2t)

What type of solid was created by revolving the function in #8 around the $\begin{aligned} x \end{aligned}$ -axis?
::将8号的函数旋转到 X 轴周围, 创造了何种类型的固态 ?

Revolve the following parametric function around the $\begin{aligned} y \end{aligned}$ -axis and find the surface area of the figure from $\begin{aligned} t = 0 \end{aligned}$ to $\begin{aligned} t = 2 \end{aligned}$ .
$\begin{aligned} N (t) & = (x (t), y (t)) \\ x (t) & = t \\ y (t) & = - 2 t \end{aligned}$

::围绕 Y 轴循环以下的参数函数, 并查找图的表面区域, 从 t=0 到 t=2. N( t) = (x( t),y( t)) x( t) =ty( t) @ @ @% 2t

What type of solid was created by revolving the function in #10 around the $\begin{aligned} y \end{aligned}$ -axis?
::将第10号中的函数围绕 Y 轴旋转, 创造了何种类型的固态 ?

How would the solid created by $\begin{aligned} N (t) \end{aligned}$ change if it was instead revolved around the $\begin{aligned} x \end{aligned}$ -axis?
::如果N(t)创造的固体转而围绕X轴,那么它会如何改变呢?

Revolve the following parametric function around the $\begin{aligned} x \end{aligned}$ -axis and find the surface area of the figure from $\begin{aligned} t = 0 \end{aligned}$ to $\begin{aligned} t = π \end{aligned}$ . Use a calculator to help.
$\begin{aligned} P (t) & = (x (t), y (t)) \\ x (t) & = 2 \cos (t) \\ y (t) & = 5 \sin (t) \end{aligned}$

::在 x 轴周围循环以下的参数函数, 并找到图的表面积, 从 t=0 到 t 。使用计算器来帮助。 P( t) = (x( t), y( t) ) x( t) = 2cos( t) y( t) = 5sin( t)

What type of solid was created by revolving the function in #13 around the $\begin{aligned} x \end{aligned}$ -axis?
::将第13号中的函数旋转到 X 轴周围, 创造了何种类型的固态 ?

Revolve the following parametric function around the $\begin{aligned} x \end{aligned}$ -axis and find the surface area of the figure from $\begin{aligned} t = 0 \end{aligned}$ to $\begin{aligned} t = 2 \end{aligned}$ . Use a calculator to help.
$\begin{aligned} Q (t) & = (x (t), y (t)) \\ x (t) & = t^{3} \\ y (t) & = t \end{aligned}$

::在 x 轴周围循环以下的参数函数, 并找到图的表面积, 从 t=0 到 t=2 。使用计算器来帮助。 Q( t) = (x( t),y( t) ) x( t) = t3y( t) = t

Review (Answers)
::回顾(答复)

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。

Section outline

General

参数形式和微积分: 革命表面面积

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

Examples ::实例

Example 1 ::例1

Example 2 ::例2

Example 3 ::例3

Example 4 ::例4

Example 5 ::例5

Review ::回顾

Review (Answers) ::回顾(答复)

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

Examples
::实例

Example 1
::例1

Example 2
::例2

Example 3
::例3

Example 4
::例4

Example 5
::例5

Review
::回顾

Review (Answers)
::回顾(答复)