第6.4节 旋转体的表面积

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 i and arc length s=2πr. Using the formula A= __ si for the area of a sector, we see that the lateral surface of the cone has area

Lateral area of cone =π rl

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 l of 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 A compared to x because it almost has the same shape as A (Figure 6.4.9). The small cylinder has area 2xπ Δy. While this area is infinitesimal, it is not infinitely close to A compared to x, because on dividing by x 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 ≈2π xπΔ s,

ΔA≈2π _______________ 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=2t², 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/dt and then apply the formula for area.

Figure 6.4.11

Let u=16 + 9t², du=18tdt, dt=____ du, t²=______. 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 =r2 in 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 y as 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 B generated by the curve from y=0 to y=a and the surface C generated by the curve from y=a to y=r.

The area of C is

We could find the area of B by taking x as 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 x and y in the formulas for surface area,

_________________________ (about x-axis),

_________________________ (about x-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 C be 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 C about (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.

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

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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.

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□29 show that the surface area of the torus generated by rotating the circle of radius r and 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=a to y=b.