   Chapter 8.2, Problem 39E

Chapter
Section
Textbook Problem

Formula 4 is valid only when f(x) ≥ 0. Show that when f(x) is not necessarily positive, the formula for surface area becomes S = ∫ a b 2 π | f ( x ) | 1 + [ f ′ ( x ) ] 2   d x

To determine

To show: The formula for surface area is S=ab2π|f(x)|1+[f(x)]2dx , when f(x) is not necessarily positive.

Explanation

Given Information:

The function f(x) is not necessarily positive.

The area of the surface obtained by rotating the curve about x-axis is given by relation.

S=ab2πf(x)1+[f(x)]2dx

Here, the value of f(x)0 .

Calculation:

Consider a frustum of cone with top radius r1 and bottom radius r2 and slanting height l.

Show the frustum of cone as shown in Figure1.

Refer to Figure 1.

Show the surface area of the frustum of cone (shaded region) using the relation.

S=2πrl (1)

Here, the average radius of the shaded region, r=12(r1+r2) .

Consider the value of a curve y=f(x) and it is a continuous function.

Here, axb

The curve y=f(x) is rotated about x-axis as shown in Figure 2:

Show the surface obtained by rotating the curve y=f(x) about x-axis as shown in Figure 3:

Refer to Figure 3.

The function f is a continuous function.

The point Pi(xi,yi) lies on the curve.

The part of surface between xi and xi1 is approximated by rotating the line segment PiPi1 about x-axis.

The resulting surface is a band (shaded area) as shown in Figure 3.

The radius of the band, r=12(yi1+yi) .

The slant height of the band, l=|Pi1Pi| .

Calculate the slant height of the band using the relation:

l=|Pi1Pi|=1+[f(xi*)]2Δx

Here, xi* lies in the interval [xi1,xi] .

Consider the value of Δx to be very small and the function f(x) is not necessarily positive.

The value of yi=|f(xi)||f(xi*)| .

The value of yi1=|f(xi1)||f(xi*)| .

Modify Equation (1).

Substitute 12(yi1+yi) for r and |Pi1Pi| for l in Equation (1).

S=2πrl=2π12(yi1+yi)|Pi1Pi| (2)

Modify Equation (2)

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