SalasSV_09_09 - 9.9 THE AREA OF A SURFACE OF REVOLUTION;...

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9.9 THE AREA OF A SURFACE OF REVOLUTION; CENTROID OF A CURVE ± 575 ± 9.9 THE AREA OF A SURFACE OF REVOLUTION; THE CENTROID OF A CURVE; PAPPUS’S THEOREM ON SURFACE AREA The Area of a Surface of Revolution In Figure 9.9.1 you can see the frustum of a cone; one radius marked r , the other R . The slant height is marked s . An interesting elementary calculation that we leave to you shows that the area of this slanted surface is given by the formula (9.9.1) A = π ( r + R ) s . (Exercise 21) This formula forms the basis for all that follows. s R r Figure 9.9.1 Let C be a curve in the upper half-plane (Figure 9.9.2). The curve can meet the x -axis, but only at a ±nite number of points. We will assume that C is parametrized by a pair of continuously differentiable functions x = x ( t ), y = y ( t ), t [ c , d ]. Furthermore, we will assume that C is simple : no two values of t between c and d give rise to the same point of C ; that is, the curve does not intersect itself. y x y x C Figure 9.9.2 If we revolve C about the x -axis, we obtain a surface of revolution. The area of that surface is given by the formula (9.9.2) A = ± d c 2 π y ( t ) ² [ x ± ( t )] 2 + [ y ± ( t )] 2 dt . We will try to outline how this formula comes about. The argument is similar to the one given in Section 9.8 for the length of a curve. Each partition P ={ c = t 0 < t 1 < ··· < t n = d } of [ c , d ] generates a polygonal approximation to C (Figure 9.9.3). Call this polygonal approximation C p . By revolving C p about the x -axis, we get a surface made up of n conical frustums. The i th frustum (Figure 9.9.4) has slant height y x y ( t i ) x ( t i ) x ( t i – 1 ) y ( t i – 1 ) Figure 9.9.3 s i = ² [ x ( t i ) x ( t i 1 )] 2 + [ y ( t i ) y ( t i 1 )] 2 = ³ ´ x ( t i ) x ( t i 1 ) t i t i 1 µ 2 + ´ y ( t i ) y ( t i 1 ) t i t i 1 µ 2 ( t i t i 1 ).
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576 ± CHAPTER 9 THE CONIC SECTIONS; POLAR COORDINATES; PARAMETRIC EQUATIONS y x y ( t i – 1 ) s i y ( t i ) Figure 9.9.4 The lateral area π [ y ( t i 1 ) + y ( t i )] s i [see Formula (9.9.1)] can be written π [ y ( t i 1 ) + y ( t i )] ± ² x ( t i ) x ( t i 1 ) t i t i 1 ³ 2 + ² y ( t i ) y ( t i 1 ) t i t i 1 ³ 2 ( t i t i 1 ). There exist points t i , t ∗∗ i , t i all in [ t i 1 , t i ] such that y ( t i ) + y ( t i 1 ) = 2 y ( t i ), x ( t i ) x ( t i 1 ) t i t i 1 = x ± ( t ∗∗ i ), y ( t i ) y ( t i 1 ) t i t i 1 = y ± ( t i ). intermediate-value theorem ´ mean-value theorem ´ Let ± t i = t i t i 1 . We can now write the lateral area of the i th frustum as 2 π y ( t i ) µ [ x ± ( t ∗∗ i )] 2 + [ y ± ( t i )] 2 ± t i . The area generated by revolving all of C p is the sum of these terms: 2 π y ( t 1 ) µ [ x ± ( t ∗∗ 1 )] 2 + [ y ± ( t 1 )] 2 ± t 1 + ···+ 2 π y ( t n ) µ [ x ± ( t ∗∗ n )] 2 + [ y ± ( t n )] 2 ± t n . This is not a Riemann sum: we don’t know that t i = t ∗∗ i = t i . But it is “close” to a Riemann sum. Close enough that, as || P || → 0, this “almost” Riemann sum tends to the integral.
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This note was uploaded on 10/12/2010 for the course MATH 12345 taught by Professor Smith during the Spring '10 term at University of Houston - Downtown.

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SalasSV_09_09 - 9.9 THE AREA OF A SURFACE OF REVOLUTION;...

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