6.4  AREA  OF  A SURFACE  OF REVOLUTION

When a curve in the plane is rotated about the x-or y-axis it forms a surface of

revolution, as in Figure 6.4.1.

 

 

The simplest surfaces of revolution are the right circular cylinders and cones. We can find their areas without calculus.

Figure 6.4.2 shows a right circular cylinder with height h and base of radius r. When the lateral surface is slit vertically and opened up it forms a rectangle with height h and base 2πr. Therefore its area is

 

                   lateral area of cylinder = 2πhr.

 

Figure 6.4.3 shows a right circular cone with slant height l and base of radius r.

When the cone is slit vertically and opened up, it forms a circular sector with radius l and arc length s=2πr. Using the formula A= __ sl  for the area of a sector, we see that the lateral surface of the cone has area

                            lateral area of cone =πrl 

 

 

 

 

 

 

 

 

 

 

Figure 6.4.4

 

Figure 6.4.4 shows the frustum of a cone with smaller radius r1, larger radius r2, and slant height l . The formula for the area of the lateral surface of a frustum of a cone is

                        

                        lateral area of frustum =π (r1 +r2 ) l.

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

The frustum therefore has lateral area

 

                                 A = πr2 l2  - πr1 l1.

The slant heights are proportional to the radii,

                            ____________________

 

The slant height lof the frustum is

                                l =  l2  －l1.

 

Using the last two equations,

                       

 

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

                              = π (r2 l2  + r1l2 - r2 l1 - r1 l1)

                              = π r2 l2  - π r1 l1 =A.

 

A Surface of revolution can be sliced into frustums in the same way that a solid of revolution can be sliced into discs or cylindrical shells. Consider a smooth curve segment

 

                                  y = f(x),    a≤  x ≤ b

 

in the first quadrant. When this curve segment is rotated about the y-axis it forms a surface of revolution ( Figure 6.4.5).

Here is the formula for the area.

 

AREA  OF SURFACE  OF  REVLUTION

                                     

                          __________________  (rotating about y-axis).

 

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

 

 

Figure 6.4.5

 

mal length Δs. When a piece Δs of the curve is rotated about 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, and bases of radius x and x + Δx. Thus compared to Δx,

 

                                 ______________________

 

                                 ___________________________

 

                                    ___________________________

 

 

Then by the Infinite Sum Theorem,

 

                        _____________________

 

 

 

EXAMPLE  1 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.

 

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

                            

 

 

 

SECOND  SOLUTION  This surface of revolution is a frustum of a cone, so the 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)

  = ________________________________________

 

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

 

EXAMPLE  2  The curve y= ___ x²,   0≤x ≤1, is rotated about the y-axis (Figure 6.4.8). Find the area of the surface of revolution.

 

 

 

 

 

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

 

                                 (2x + Δx) πΔs

 

Which is infinitely close to ΔA compared to Δx because it almost has the same shape as ΔA (Figure 6.4.9). The small cylinder has area 2xπ Δy. While this area is infinitesimal, it is not infinitely close to ΔA compared to Δx, because on dividing by Δx we get

 

 

 

 

 

 

 

 

 

 

Approximating the surface by small cylinders would give us the different and incorrect value____   for the surface area.

 

When a curve is given by parametric equations we get a formula for surface area of revolution analogous to the formula for lengths of parametric curves in Section 6.3

 

             Let     x=f(t),     y= g(t),    a ≤ t ≤ b

 

be a parametric curve in the first quadrant such that the derivatives are continuous and the curve does not retrace its path (Figure 6.4.10).

 

 

AREA  OF  SURFACE  OF  REVOLUTION

 

                         ______________________________    (rotating about y-axis).

 

To justify this new formula we observe that an infinitesimal piece of the surface is almost a cone frustum of radii x, x+Δx and slant height Δs. Thus compared to Δt,

 

                           Δ s≈___________________ Δt,

                           ΔA≈π(x + (x+ Δx)) Δs ≈ 2πxπΔs,

                           ΔA≈2π x _______________ Δt.

 

The Infinite Sum Theorem gives the desired formula for area.

 

This new formula reduces to our first formula when the curve has the simple form y=f(x). If y=f(x), a≤ x≤ b, take x=t and get

 

                      ____________________________   (about y-axis).

 

              Similarly, if x=g(y),  a≤ y≤ b,  we take y=t and get the formula

 

                       ________________________ (about y-axis).

 

EXAMPLE  3  The curve x=2t ², y=t3, 0 ≤ t≤ 1 is rotated about the y-axis.

               Find the area of the surface of revolution (Figure 6.4.11).

               We first find dx/dt and dy/dtand then apply the formula for area.

 

                      

 

 

 

 

 

 

 

 

 

 

 

Figure 6.4.11

 

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

 

 

 

EXAMPLE  4 

      Derive the formula A=4πr² for the area of the surface of a sphere of radius r.

      When the portion of the circle x²+ y²=r²in the first quadrant is rotated about the y-axis it will form a hemisphere of radius r(Figure 6.4.12). The surface of the sphere has twice the area of this hemisphere.

 

 

 

 

 

Figure 6.4.12

 

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

                   x = _____________________,  0 ≤ y≤ r.

 

Then ______________________

 

This derivative is undefined at y=0. To get around this difficulty we let 0< a < r and divide the surface into the two parts shown in Figure 6.4.13, the surface Bgenerated by the curve from y=0 to y=a and the surface Cgenerated by the curve from y=atoy=r.

 

The area of Cis

 

 

 

We could find the area of Bby taking xas the independent variable. However,

 

 

 

 

 

 

Figure 6.4.13

 

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

                  ___________________________

so               __________________________

and the sphere has area A = 4πr².

 

If a curve is rotated about the x-axis instead of the y-axis (Figure 6.4.14), we interchange xand y in the formulas for surface area,

 

                       _________________________  (about x-axis),

 

                        _________________________ (aboutx-axis),

 

                        _________________________  (about x-axis),

 

 

 

 

 

Figure 6.4.14

 

Most of the time the formula for surface area will give an integral which cannot be evaluated exactly but can only be approximated, for example by the Trapezoidal Rule.

 

EXAMPLE  5  Let Cbe the curve

                       y= x4,   0 ≤ x≤ 1.   (see Figure 6.4.15)

Set up 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

 

We cannot evaluate this integral, so we leave it in the above form. The Trapezoidal Rule can be used to get approximate values. When Δx=____the Trapezoidal Approximation is

                      A ~ 6.42,   error ≤ 0.26.

 

(b)             ________________dx

              =________________dx.

The Trapezoidal Approximation when Δx= ___ is

                      A ~ 3.582,   error ≤ 0.9.

 

PROBLEMS  FOR  SECTION  6.4

In Problem 1-12, find the area of the surface generated by rotating the given curve about the y-axis.

1   y=x²,   0≤ x≤ 2                   2  y=cx + d,  a ≤ x≤ b  

3   y=2x3/ ²,   0≤ x≤ 1                 4  y= _____( x²+ 2)3/2, 1 ≤ x≤ 2

5   y=_________,   1≤ x≤ 4

6   y=_________,   1≤ x≤ 2

7   y=_________,   1≤ x≤ 8

8   x=2t+ 1, y= 4-t,  0≤ t≤ 4

9   x=t+ 1, y=________,  0≤ t≤ 2

10  x=t ², y= __________,  0≤ t≤ 3

11  x=t 3, y=3t+1,        0≤ t≤ 1

12  x2/3+ y2/3 = 1 ,           first quadrant

 

In Problems 13-20, find the area of the surface generated by rotating the given curve about the x-axis.

13    y=________x 3,        0≤ x≤ 1

 

14   y=_________,        1 ≤ x≤ 2

15   y=_________,        1≤ x≤ 2

16   y=_________,        1≤ x≤ 2

17   y=_________,        3≤ x≤ 4

18   y=_________,        8≤ x≤ 27

19   x=2t + 1, y= 4- t,      0≤ t ≤ 4

20   x=t² + t, y= 2t + 1,     0≤ t ≤ 1

21   The part of the circle x²+ y²= r²between x=0 and x= a  in the first quadrant is rotated

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

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

 

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

23      y= x5,  0 ≤ x≤ 1

24      y= y + ___,  2 ≤ y≤ 3

25      x= t ² +t ,  y = t ²- 1, 1 ≤ t≤ 10

26      x= t4,  y= t3,       2 ≤ t≤ 4

27  Set up an integral for the area generated by rotating the curve y=_____ x², 0 ≤ x≤ 1

    about the x-axis and find the Trapezoidal Approximation with Δx=0.2

 

28  Set up an integral for the area generated by rotating the curve y=_____ x3, 0 ≤ x≤ 1

    about the y-axis and find the Trapezoidal Approximation with Δx=0.2

 

□29 show that the surface area of the torus generated by rotating the circle of radius rand center (c, 0) about the y-axis ( r < c ) is A = 4π²rc. Hint: Take y as the independent variable and use the formula ___ r dy/ ____ for the length of the arc of the circle from y=ato y=b.