“JUST THE MATHS” SLIDES NUMBER 13.9 INTEGRATION APPLICATIONS 9 (First moments of a surface of revolution) by A.J.Hobson 13.9.1 Introduction 13.9.2 Integration formulae for first moments 13.9.3 The centroid of a surface of revolution
UNIT 13.9 - INTEGRATION APPLICATIONS 9 FIRST MOMENTS OF A SURFACE OF REVOLUTION 13.9.1 INTRODUCTION Let C denote an arc (with length s ) in the xy -plane of cartesian co-ordinates. Let δs denote the length of a small element of this arc. Then, for the surface obtained when the arc is rotated through 2 π radians about the x -axis, the “first mo- ment” about a plane through the origin, perpendicular to the x -axis, is given by lim C 2 πxyδs, � δs → 0 where x is the perpendicular distance, from the plane of moments, of the thin band, with surface area 2 πyδs , so generated. 1
13.9.2 INTEGRATION FORMULAE FOR FIRST MOMENTS (a) Consider an arc of the curve whose equation is y = f ( x ) , joining two points, P and Q, at x = a and x = b , respec- tively. Q y ✻ δs δy r r P ✲ x O a δx b The arc may divided up into small elements of typical length, δs , by using neighbouring points along the arc, separated by typical distances of δx (parallel to the x - axis) and δy (parallel to the y -axis). From Pythagoras’ Theorem � 2 � δy � ( δx ) 2 + ( δy ) 2 = � � � 1 + δs ≃ δx. � � δx 2
For the surface of revolution of the arc about the x -axis, the first moment becomes � 2 � δy x = b � � lim x = a 2 πxy � 1 + δx � � � δx δx → 0 � 2 � d y � b � � = a 2 πxy � 1 + d x � � d x Note: If the curve is given parametrically by x = x ( t ) , y = y ( t ) then, d y d y d t d x = . d x d t Hence, � 2 + � � 2 � � � d y � d x 2 � � d y � � d t d t � � 1 + = , � � d x d x d t provided that d x d t is positive on the arc being considered. 3
If d x d t is negative on the arc, then the previous line needs to be prefixed by a negative sign. Using integration by substitution, � � 2 2 � � d y d y . d x � b � t 2 � � � � a 2 πxy � 1 + d x = t 1 2 πxy � 1 + d t d t, � � � � d x d x where t = t 1 when x = a and t = t 2 when x = b . The first moment about the plane through the origin, perpendicular to the x -axis is given by � 2 2 � d x d y � t 2 � � First moment = ± t 1 2 πxy + d t, � � � d t d t according as d x d t is positive or negative. (b) For an arc whose equation is x = g ( y ) , contained between y = c and y = d , we may reverse the roles of x and y in the previous section so that the first moment about a plane through the origin, perpendicular to the y -axis is as follows: 4
� 2 � d x � � d � First moment = c 2 πyx � � 1 + d y. � � d y y ✻ S d δy δs R c ✲ x O rr δx Note: If the curve is given parametrically by x = x ( t ) , y = y ( t ) , where t = t 1 when y = c and t = t 2 when y = d , then the first moment about a plane through the origin, perpendicular to the y -axis is given by � 2 2 � d x d y � t 2 � � First moment = ± t 1 2 πyx + d t, � � d t d t � according as d y d t is positive or negative. 5
EXAMPLES 1. Determine the first moment about a plane through the origin, perpendicular to the x -axis, for the hemispher- ical surface of revolution (about the x -axis) of the arc of the circle whose equation is x 2 + y 2 = a 2 , lying in the first quadrant Solution y ✻ ✡ ✡ a ✡ ✡ ✡ ✲ x O 2 x + 2 y d y d x = 0 . Hence, d y d x = − x y. 6
The first moment about the specified plane is therefore given by � x 2 + y 2 � � � 1 + x 2 � � � a � a � � � � 0 2 πxy y 2 d x = 0 2 πxy d x. � � � � y 2 But x 2 + y 2 = a 2 . Thus, the first moment becomes � a 0 2 πax d x = [ πax 2 ] a 0 = πa 3 . 2. Determine the first moments about planes through the origin, (a) perpendicular to the x -axis and (b) perpen- dicular to the y -axis, of the first quadrant arc of the curve with parametric equations x = a cos 3 θ, y = a sin 3 θ. Solution y ✻ ✲ x O d x d θ = − 3 a cos 2 θ sin θ and d y d θ = 3 a sin 2 θ cos θ. 7
Hence, the first moment about the x -axis is given by � 0 � 9 a 2 cos 4 θ sin 2 θ + 9 a 2 sin 4 θ cos 2 θ d θ. 2 2 πxy − π On using cos 2 θ + sin 2 θ ≡ 1, this becomes � π 2 πa 2 cos 3 θ sin 3 θ. 3 a cos θ sin θ d θ 2 0 � π 6 πa 3 cos 4 θ sin 4 θ d θ. = 2 0 Using 2 sin θ cos θ ≡ sin 2 θ , the integral reduces to 3 πa 3 � π sin 4 2 θ d θ 2 0 8 = 3 πa 3 1 − 2 cos 4 θ + 1 + cos 8 θ � π d θ 2 0 32 2 π = 3 πa 3 0 = 9 πa 3 3 θ 2 − sin 4 θ + sin 8 θ 2 128 . 32 2 16 By symmetry, or by direct integration, the first mo- ment about a plane through the origin, perpendicular to the y -axis is also 9 πa 3 128 . 8
13.9.3 THE CENTROID OF A SURFACE OF REVOLUTION Having calculated the first moment of a surface of rev- olution about a plane through the origin, perpendicular to the x -axis, it is possible to determine a point, ( x, 0), on the x -axis with the property that the first moment is given by Sx , where S is the total surface area. The point is called the “centroid” or the “geometric centre” of the surface of revolution and, for the surface of revolution of the arc of the curve whose equation is y = f ( x ), between x = a and x = b , the value of x is given by � 2 d x � � d y � � b a 2 πxy � 1 + � d x x = � � 2 d x � � d y � b a 2 πy � 1 + � d x � � 2 d x � � d y � b a xy � 1 + � d x = . � 2 d x � � � d y � b � 1 + a y � d x Note: The centroid effectively tries to concentrate the whole surface at a single point for the purposes of consid- ering first moments. 9
In practice, the centroid of a surface corresponds to the position of the centre of mass of a thin sheet, for example, with uniform density. EXAMPLES 1. Determine the position of the centroid of the surface of revolution (about the x -axis) of the arc of the circle whose equation is x 2 + y 2 = a 2 , lying in the first quadrant. Solution y ✻ ✡ ✡ a ✡ ✡ ✡ ✲ x O From Example 1 in Section 13.9.2, the first moment of the surface about a plane through the origin, perpen- dicular to the the x -axis is equal to πa 3 . 10
The total surface area is � � 1 + x 2 � � a � � 0 2 πy y 2 d x. � � Using x 2 + y 2 = a 2 , � a 0 2 πa d x = 2 πa 2 . surface area = Hence, x = πa 3 = a 2 . 2 πa 2 2. Determine the position of the centroid of the surface of revolution (about the x -axis) of the first quadrant arc of the curve with parametric equations x = a cos 3 θ, y = a sin 3 θ. Solution y ✻ ✲ x O 11
From Example 2 in Section 13.9.2, the first moment of the surface about a plane through the origin, perpen- dicular to the x -axis is equal to 9 πa 3 128 . The total surface area is given by � 0 2 2 πa sin 3 θ. 3 a cos θ sin θ d θ − π � π 3 a 2 sin 4 θ cos θ d θ = 2 0 π sin 5 θ = 3 πa 2 2 = 3 πa 2 5 . 5 0 Thus, x = 9 πa 3 128 ÷ 3 πa 2 5 or x = 15 a 128 . 12
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