第6.4节 旋转的表面积

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6.4AREA OF A SURFACE OF REVOLUTION

Whena curve in the plane is rotated about the x-or y-axisit forms a surface of

revolution,as in Figure 6.4.1.

Thesimplest surfaces of revolution are the right circular cylinders andcones. We can find their areas without calculus.

Figure6.4.2 shows a right circular cylinder with height h and baseof radius r. When the lateral surface is slit vertically andopened up it forms a rectangle with height h and base 2πr.Therefore its area is

lateralarea of cylinder = 2πhr.

Figure6.4.3 shows a right circular cone with slant height l and baseof radius r.

Whenthe cone is slit vertically and opened up, it forms a circular sectorwith radius l and arc length s=2πr. Using the formula A= __sl for the area of a sector, we see that the lateralsurface of the cone has area

lateralareaof cone =πrl

Figure6.4.4

Figure6.4.4 shows the frustum of a cone with smaller radiusr1,larger radius r2,and slant height l .The formula for the area of the lateral surface of a frustum of acone is

lateralarea of frustum =π (r1 +r2 ) l.

Thisformula is justified as follows. The frustum is formed by removing acone of radius r1 and slant height l1 from a cone ofradius r2 and slant height l2.

Thefrustum therefore has lateral area

A=πr2 l2 - πr1 l1.

Theslant heights are proportional to the radii,

____________________

Theslant height lofthe frustum is

l= l2 －l1.

Usingthe last two equations,

π(r1 + r2 ) l = π(r2 + r1) ( l2 - l1 )

=π(r2 l2 + r1l2 - r2 l1 - r1 l1)

=πr2 l2 - π r1 l1 =A.

ASurface of revolution can be sliced into frustums in the same waythat a solid of revolution can be sliced into discs or cylindricalshells. Consider a smooth curve segment

y= f(x), a≤ x ≤ b

inthe first quadrant. When this curve segment is rotated about they-axisit forms a surface of revolution ( Figure 6.4.5).

Hereis the formula for the area.

AREAOF SURFACE OF REVLUTION

__________________(rotatingabout y-axis).

Tojustify this formula we begin by dividing the interval [a,b]into infinitesimal subintervals of length Δx.This divides the curves into pieces of infinitesimal

Figure6.4.5

mallength Δs. When a piece Δs of the curve is rotatedabout the x-axis it sweeps out a piece of the surface,ΔA(Figure 6.4.6). Since Δs is almost a line segment,ΔA is almost a cone frustum of slant height Δs, andbases of radius x and x + Δx. Thus compared toΔx,

______________________

___________________________

___________________________

Thenby the Infinite Sum Theorem,

_____________________

EXAMPLE1 The line segment y=3x,from x=1 to x=4 , is rotated about the y-axis(Figure 6.4.7). Find the area of the surface of revolution.

FIRSTSOLUTION We use the integration formula. dy/dx =3, so

SECONDSOLUTION This surface of revolution is a frustum of a cone, sothe formula for the lateral area of a frustum can be used directly.From the diagram we see that the radii and slant height are:

r1= 1, r2 = 4,

l= distance from (1, 3) to (4, 12)

=________________________________________

ThenA = π(r1+ r2 ) l = π(1 + 4 )____= _____.

EXAMPLE2 The curve y= ___ x², 0≤x ≤1,is rotated about the y-axis (Figure 6.4.8). Find the area ofthe surface of revolution.

Infinding a formula for surface area, why did we divide the surfaceinto frustums of cones instead of into cylinders (as we did forvolumes) ? The reason is that to use the Infinite Sum Theorem we needsomething which is infinitely close to a small piece ΔA ofarea compared to Δx. The small frustum has area

(2x+ Δx) πΔs

Whichis infinitely close to ΔA compared to Δx because italmost has the same shape as ΔA (Figure 6.4.9). The smallcylinder has area 2xπ Δy. While this area isinfinitesimal, it is not infinitely close to ΔA compared toΔx, because on dividing by Δx we get

Approximatingthe surface by small cylinders would give us the different andincorrect value____ forthe surface area.

Whena curve is given by parametric equations we get a formula for surfacearea of revolution analogous to the formula for lengths of parametriccurves in Section 6.3

Letx=f(t), y= g(t),a ≤ t ≤ b

bea parametric curve in the first quadrant such that the derivativesare continuous and the curve does not retrace its path (Figure6.4.10).

AREAOF SURFACE OF REVOLUTION

______________________________ (rotating about y-axis).

Tojustify this new formula we observe that an infinitesimal piece ofthe surface is almost a cone frustum of radii x,x+Δxand slant height Δs.Thus compared to Δt,

Δs≈___________________ Δt,

ΔA≈π(x+ (x+Δx))Δs ≈2πxπΔs,

ΔA≈2πx _______________ Δt.

TheInfinite Sum Theorem gives the desired formula for area.

Thisnew formula reduces to our first formula when the curve has thesimple form y=f(x).If y=f(x), a≤ x≤ b, takex=t andget

____________________________(about y-axis).

Similarly,ifx=g(y), a≤ y≤ b,we take y=t andget the formula

________________________(about y-axis).

EXAMPLE3 The curve x=2t², y=t3,0 ≤ t≤ 1 is rotated about the y-axis.

Findthearea of the surface of revolution (Figure 6.4.11).

Wefirstfind dx/dt anddy/dtand then apply the formula for area.

Figure6.4.11

Letu=16+ 9t²,du=18tdt,dt=____du, t ²=______.Then u=16at t=0and u=25at t=1,so

EXAMPLE4

Derivetheformula A=4πr²for the area of the surface of a sphere of radius r.

Whenthe portion of the circle x²+ y²=r²inthe first quadrant is rotated about the y-axisit will form a hemisphere of radius r(Figure6.4.12). The surface of the sphere has twice the area of thishemisphere.

Figure6.4.12

Itis simpler to take yas the independent variable, so the curve has the equation

x= _____________________, 0≤ y≤ r.

Then______________________

Thisderivative is undefined at y=0.To get around this difficulty we let 0< a< r anddivide the surface into the two parts shown in Figure 6.4.13, thesurface Bgeneratedby the curve from y=0to y=aand the surface Cgeneratedby the curve from y=atoy=r.

Thearea of Cis

Wecould find the area of Bbytaking xasthe independent variable. However,

Figure6.4.13

Itis simpler to let a be an infinitesimal ε. Then B isan infinitely thin ring-shaped surface, so its area is infinitesimal.Therefore the hemisphere has area

___________________________

so__________________________

andthe sphere has area A = 4πr².

Ifa curve is rotated about the x-axisinstead of the y-axis(Figure 6.4.14), we interchange xandy inthe formulas for surface area,

_________________________(aboutx-axis),

_________________________(aboutx-axis),

_________________________(aboutx-axis),

Figure6.4.14

Mostof the time the formula for surface area will give an integral whichcannot be evaluated exactly but can only be approximated, for exampleby the Trapezoidal Rule.

EXAMPLE5 Let Cbethe curve

y=x4,0 ≤ x≤ 1.(see Figure 6.4.15)

Setup an integral for the surface area generated by rotating the curve Cabout(a) the y-axis,(b) the x-axis(see Figure 6.4.16).

(a)dy/dx = 4x3

A=____________ dx

=_____________dx

Wecannot evaluate this integral, so we leave it in the above form. TheTrapezoidal Rule can be used to get approximate values. When Δx=____theTrapezoidal Approximation is

A~6.42, error ≤0.26.

(b)________________dx

=________________dx.

TheTrapezoidal Approximation when Δx=___ is

A~3.582, error ≤0.9.

PROBLEMSFOR SECTION 6.4

InProblem 1-12, find the area of the surface generated by rotating thegiven curve about the y-axis.

1y=x²,0≤ x≤ 22 y=cx + d,a ≤ x≤ b

3y=2x3/ ²,0≤ x≤ 14 y=_____(x²+2)3/2,1 ≤ x≤ 2

5y=_________,1≤ x≤ 4

6y=_________,1≤ x≤ 2

7y=_________,1≤ x≤ 8

8x=2t+1, y=4-t,0≤ t≤ 4

9x=t+1, y=________,0≤ t≤ 2

10x=t²,y= __________,0≤ t≤ 3

11x=t3,y=3t+1,0≤ t≤ 1

12x2/3+y2/3 =1 , firstquadrant

InProblems 13-20, find the area of the surface generated by rotatingthe given curve about the x-axis.

13y=________x3,0≤ x≤ 1

14y=_________,1 ≤ x≤ 2

15y=_________,1≤ x≤ 2

16y=_________,1≤ x≤ 2

17y=_________,3≤ x≤ 4

18y=_________,8≤ x≤ 27

19x=2t+ 1,y= 4-t,0≤t ≤ 4

20x=t² + t, y= 2t+ 1,0≤t ≤ 1

21The part of the circle x²+y²=r²betweenx=0and x=a inthe first quadrant is rotated

aboutthex-axis.Find the area of the resulting zone of the sphere (0< a< r)

22Solve the above problem when the rotation is about the y-axis.

InProblems 23-26 set up integrals for the areas generated by rotatingthe given curve about (a) the y-axis,(b) the x-axis.

23y= x5,0 ≤ x≤ 1

24y= y + ___,2 ≤ y≤ 3

25x= t ² +t ,y = t ²-1, 1 ≤ t≤ 10

26x= t4,y=t3,2 ≤ t≤ 4

27Set up an integral for the area generated by rotating the curvey=_____x²,0 ≤ x≤ 1

aboutthex-axisand find the Trapezoidal Approximation with Δx=0.2

28Set up an integral for the area generated by rotating the curvey=_____x3,0 ≤ x≤ 1

aboutthey-axisand find the Trapezoidal Approximation with Δx=0.2

□29show that the surface area of the torus generated by rotating thecircle of radius randcenter (c,0) about the y-axis( r < c )is A = 4π²rc.Hint: Takey asthe independent variable and use the formula ___r dy/____forthe length of the arc of the circle from y=atoy=b.