. MA 105: Calculus Lecture 10 . Prof. B.V. Limaye IIT Bombay Friday, 3 February 2017 B.V. Limaye, IITB MA 105: Lec-10
. Area under a Curve . Let f : [ a , b ] → R be a bounded function. Suppose f ≥ 0 on [ a , b ], and let R f := { ( x , y ) ∈ R 2 : a ≤ x ≤ b and 0 ≤ y ≤ f ( x ) } . We say that R f has an area if f is Riemann integrable, and then we define ∫ b Area ( R f ) := f ( x ) dx . a If f : [ a , b ] → R is any function, then f = f + − f − , where f + = | f | + f f − = | f | − f and . 2 2 Note that f + ≥ 0 and f − ≥ 0. B.V. Limaye, IITB MA 105: Lec-10
. Positive and Negative Parts of a Function . In fact, for x ∈ [ a , b ], f + ( x ) = max { f ( x ) , 0 } and f − ( x ) = − min { f ( x ) , 0 } . The functions f + and f − are known as the positive part of f and the negative part of f , respectively. Clearly, f is bounded if and only if f + and f − are both bounded. Also, f is integrable if and only if f + and f − are both integrable, and then ∫ b ∫ b ∫ b f + ( x ) dx − f ( x ) dx = f − ( x ) dx a a a = Area ( R f + ) − Area ( R f − ) , which can be called the ‘signed area’ delineated by the curve y = f ( x ) , x ∈ [ a , b ]. B.V. Limaye, IITB MA 105: Lec-10
. Area between Curves . Let f 1 , f 2 : [ a , b ] → R be integrable functions such that f 1 ≤ f 2 . Let R := { ( x , y ) ∈ R 2 : a ≤ x ≤ b and f 1 ( x ) ≤ y ≤ f 2 ( x ) } be the region between the curves y = f 1 ( x ) and y = f 2 ( x ). Define ∫ b ( ) Area ( R ) := Area ( R f 2 − f 1 ) = f 2 ( x ) − f 1 ( x ) dx . a Let g 1 , g 2 : [ c , d ] → R be integrable functions such that g 1 ≤ g 2 . Let R := { ( x , y ) ∈ R 2 : c ≤ y ≤ d and g 1 ( y ) ≤ x ≤ g 2 ( y ) } be the region between the curves x = g 1 ( y ) and x = g 2 ( y ). Define ∫ d ( ) Area ( R ) := g 2 ( y ) − g 1 ( y ) dy . c If two curves cross each other, then we must find areas of several regions between them separately, and add them up. B.V. Limaye, IITB MA 105: Lec-10
y y y = f 2 ( x ) y = f ( x ) 1 y = f ( x ) 2 x a b y = f 1 ( x ) x a b y y d d b x = g 1 ( y ) x = g 2 ( y ) x = g 1 ( y ) x = g 2 ( y ) b c c x x B.V. Limaye, IITB MA 105: Lec-10
. Examples . (i) Let R denote the region enclosed by the loop of the curve y 2 = x (1 − x ) 2 , that is, the region bounded by the curves y = −√ x (1 − x ) and y = √ x (1 − x ). Now √ x (1 − x ) = −√ x (1 − x ) ⇐ ⇒ x = 0 or 1. Hence ∫ 1 ( √ x (1 − x ) − ( −√ x (1 − x )) dx = 8 ) Area ( R ) = 15 . 0 (ii) Let R denote the region bounded by the curves x = − 2 y 2 and x = 1 − 3 y 2 . Now − 2 y 2 = 1 − 3 y 2 ⇐ ⇒ y = ± 1, and − 2 y 2 ≤ 1 − 3 y 2 if y ∈ [ − 1 , 1]. Hence ∫ 1 ∫ 1 (1 − y 2 ) dy = 4 1 − 3 y 2 − ( − 2 y 2 ) ( ) Area ( R ) = dy = 3 . − 1 − 1 B.V. Limaye, IITB MA 105: Lec-10
. Polar coordinates . Review: The function cos − 1 : [ − 1 , 1] → [0 , π ] is one-one and onto. Let P := ( x , y ) ̸ = (0 , 0). There are unique r , θ ∈ R such that r > 0 , θ ∈ ( − π, π ] , x = r cos θ and y = r sin θ. x 2 + y 2 and √ In fact, r := { cos − 1 ( x / r ) if y ≥ 0 , θ := − cos − 1 ( x / r ) if y < 0 . (If y < 0, then | x / r | < 1, and − cos − 1 ( x / r ) ∈ ( − π ,0).) The pair ( r , θ ) is defined as the polar coordinates of P . B.V. Limaye, IITB MA 105: Lec-10
. Examples . Example: Let 0 < ϕ ≤ π/ 2, and let R denote the sector of a disc of radius a , marked by the points (0 , 0) , ( a , 0) and ( a cos ϕ, a sin ϕ ), that is, the region bounded by the curves a 2 − y 2 for y ∈ [0 , a sin ϕ ]. √ x = (cot ϕ ) y and x = y ( a os '; a sin ' ) ' b ot p y 2 2 = x = a � y x b b x 0 a Then ∫ a sin ϕ dy = a 2 ϕ (√ ) a 2 − y 2 − (cot ϕ ) y Area ( R ) = 2 . 0 a 2 θ. B.V. Limaye, IITB MA 105: Lec-10 =
. Curves given by Polar Equations . Let R denote the region bounded by the curve r = p ( θ ) and the rays θ = α, θ = β , where − π ≤ α < β ≤ π . Thus R := { ( r cos θ, r sin θ ) : α ≤ θ ≤ β and 0 ≤ r ≤ p ( θ ) } . Suppose p : [ α, β ] → R is integrable. (If α = − π and β = π , then we suppose p ( − π ) = p ( π ) . ) Partition [ α, β ] into α = θ 0 < θ 1 < · · · < θ n = β . Pick sample points γ i ∈ [ θ i − 1 , θ i ] for i = 1 , . . . , n . Area between the rays θ = θ i − 1 and θ = θ i is approximated by the area of a sector of a disc of radius r i := p ( γ i ), that is, by p ( γ i ) 2 ( θ i − θ i − 1 ) . 2 The sum of areas of these sectors is p ( γ i ) 2 ( θ i − θ i − 1 ) n ∑ . 2 i =1 B.V. Limaye, IITB MA 105: Lec-10
We define ∫ β ∫ β Area ( R ) := 1 p ( θ ) 2 d θ = 1 r 2 d θ. 2 2 α α Examples: (i) Let a > 0. Area of the disc enclosed by the ∫ π − π a 2 d θ = π a 2 . circle r = a is equal to 1 / 2 (ii) Let a > 0, and let R denote the region enclosed by the cardioid r = a (1 + cos θ ). Then ∫ π a 2 (1 + cos θ ) 2 d θ = 3 a 2 π Area ( R ) = 1 . 2 2 − π (iii) Let R denote the region that lies inside the circle r = 3 sin θ and outside the cardioid r = 1 + sin θ , where θ ∈ [0 , π ]. Now 3 sin θ = 1 + sin θ ⇐ ⇒ θ = π/ 6 or 5 π/ 6, and 1 + sin θ ≤ 3 sin θ if θ ∈ [ π/ 6 , 5 π/ 6]. Hence ∫ 5 π/ 6 Area ( R ) = 1 (3 sin θ ) 2 d θ − (1 + sin θ ) 2 ) ( d θ = π. 2 π/ 6 B.V. Limaye, IITB MA 105: Lec-10
. Volume of a solid . Let D be a bounded subset of R 3 . A cross-section of D by a plane in R 3 is called a slice of D . Let a < b , and suppose D lies between the planes x = a and x = b , which are perpendicular to the x -axis. For s ∈ [ a , b ], consider the slice of D by the plane x = s , namely { ( x , y , z ) ∈ D : x = s } , and suppose it has an area A ( s ). To find the volume of D , we proceed as follows. Partition [ a , b ] into a = x 0 < x 1 < · · · < x n = b . Pick sample points s i ∈ [ x i − 1 , x i ] for i = 1 , . . . , n . Volume between the planes x = x i − 1 and x = x i is approximated by the volume of a rectangular slab of width x i − x i − 1 and base area A ( s i ), that is, by A ( s i )( x i − x i − 1 ). n ∑ The sum of volumes of these slabs is A ( s i )( x i − x i − 1 ). i =1 B.V. Limaye, IITB MA 105: Lec-10
Slice Method: We define the volume of D by ∫ b Vol ( D ) := A ( x ) dx , a provided the ‘area function’ A : [ a , b ] → R is integrable. Examples (i) If D is a cylinder with cross-sectional area A and height h , then Vol ( D ) = Ah . (The ‘area function’ is the constant A .) (ii) Let a > 0, and let D denote the solid enclosed by the cylinders x 2 + y 2 = a 2 and x 2 + z 2 = a 2 . Then D lies between the planes x = − a and x = a . For s ∈ [ − a , a ], the slice { ( x , y , z ) ∈ D : x = s } is the square √ √ ( s , y , z ) ∈ R 3 : | y | ≤ a 2 − s 2 and | z | ≤ a 2 − s 2 } { , √ a 2 − s 2 ) 2 = 4 ( a 2 − s 2 ). Hence ( and so A ( s ) = 2 ∫ a ∫ a ds = 16 a 3 a 2 − s 2 ) ( Vol ( D ) = A ( s ) ds = 4 3 . − a − a B.V. Limaye, IITB MA 105: Lec-10
b x z √ a 2 − s 2 b y b b √ √ a 2 − s 2 a 2 − s 2 − b b y z b √ a 2 − s 2 − Figure : Solid enclosed by two cylinders and a slice resulting in a square region B.V. Limaye, IITB MA 105: Lec-10
. Solids of Revolution . If a subset D of R 3 is generated by revolving a planar region about an axis, then D known as a solid of revolution . Examples The spherical ball { ( x , y , z ) ∈ R 3 : x 2 + y 2 + z 2 ≤ a 2 } is obtained by revolving the semidisc { ( x , y ) ∈ R 2 : x 2 + y 2 ≤ a 2 and y ≥ 0 } about the x -axis. The cylindrical solid { ( x , y , z ) ∈ R 3 : x 2 + z 2 ≤ b 2 and 0 ≤ y ≤ h } is obtained revolving the rectangle [0 , b ] × [0 , h ] about the y -axis. B.V. Limaye, IITB MA 105: Lec-10
. Volume of a Solid of Revolution: Washer Method . Let D be the solid obtained by revolving the region between the curves y = f 1 ( x ) and y = f 2 ( x ), and the lines x = a and x = b , about the x -axis, where 0 ≤ f 1 ≤ f 2 . The slice of D at x ∈ [ a , b ] looks like a circular washer, that is, a disc of radius f 2 ( x ) from which a smaller disc of radius f 1 ( x ) has been removed, and so the area of the slice is A ( x ) := π ( f 2 ( x ) 2 − f 1 ( x ) 2 ). Suppose f 1 and f 2 are integrable on [ a , b ]. Then the area function A is integrable on [ a , b ], and by the slice method, ∫ b ∫ b f 2 ( x ) 2 − f 1 ( x ) 2 ) ( Vol ( D ) = A ( x ) dx = π dx . a a B.V. Limaye, IITB MA 105: Lec-10
If the inner radius of a washer is equal to 0, then the washer is in fact a disk. If this is the case for every x ∈ [ a , b ], then the washer method is called the disk method. y y = f ( x ) 2 y = f ( x ) 1 x a b z Figure : Illustrations of a disk and of a washer B.V. Limaye, IITB MA 105: Lec-10
Examples (i) Let D denote the solid obtained by rotating the region between the curves y = x and y = x 2 about the x -axis. Let f 1 ( x ) := x 2 and f 2 ( x ) := x for x ∈ [0 , 1]. The curves y = f 1 ( x ) and y = f 2 ( x ) intersect at x = 0 and x = 1, and 0 ≤ f 1 ≤ f 2 on [0 , 1]. By the washer method, ∫ 1 dx = 2 π ( x ) 2 − ( x 2 ) 2 ) ( Vol ( D ) = π 15 . 0 B.V. Limaye, IITB MA 105: Lec-10
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