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Course: Accelerated Engineering Calculus I Instructor: Michael Medvinsky 12.11.2 Volumes (6.2-6.3) Volume of a a simple body like circular\elliptical cylinder or rectangular parallelepiped (a box) is given by V = Ah, where A is the area of the basis and h is the height. For a more complicated body B we cut it into small pieces and approximate it by cylinders and adding the volumes. The volume of B is obtained by limiting process, when the pieces become infinitely small. x=b x=a a S ( xi ) S ( xi +1 ) xi xi +1 b We intersect\slicing B with a plane and obtaining a plane region that is called crosssections of B. Let A ( xi ) be the area of cross section in a plane Px . The volume of i cylinder is A ( xi* ) Δx and therefore the volume of the body is b ( ) n V = ∑ A x Δx = ∫ A ( x ) dx * j j=1 Ex 5. r x a Volume of sphere r ⎛ x3 ⎞ V = ∫ A ( x ) dx = ∫ π y dx = ∫ π ( r − x ) dx = 2π ∫ r − x dx = 2π ⎜ r 2 x − ⎟ = 3 ⎠0 ⎝ −r −r −r 0 r y r r r 2 2 2 2 2 ⎛ r3 ⎞ 2 4 ⎛ 1⎞ = 2π ⎜ r 3 − ⎟ = 2π r 3 ⎜ 1− ⎟ = 2π r 3 ⋅ = π r 3 ⎝ 3⎠ 3⎠ 3 3 ⎝ Ex 6. Find volume of body enclosed between the axes and 1− x 3 rotated around a) x-axis and b) y-axis. When we rotate around x-axis, we “slice” into discs and so the volume is given by b V = π ∫ f 2 (x) dx since the area is π r 2 = π f ( x ) , thus: 2 a 1 ( Vx = π ∫ 1− x 0 3 ) 2 1 1 ⎛ ⎛ 1 1⎞ 9 2 1 ⎞ dx = π ∫ 1− 2x + x dx = π ⎜ x − x 4 + x 7 ⎟ = π ⎜ 1− + ⎟ = π 4 7 ⎠0 ⎝ ⎝ 2 7 ⎠ 14 0 3 6 Course: Accelerated Engineering Calculus I Instructor: Michael Medvinsky In order to find the volume by on can “switch the axes” or in other words integrate by y, i.e. to find the area enclosed between axes and x = 3 1− y . Thus 0 Vy = π ∫ 1 ( 3 ) 0 0 1− y dy = π ∫ (1− y ) dy = −π ∫ t 2 2/3 1 1 1 2/3 1 dt = π ∫ t 2/3 0 t 5/3 3 dt = π = π 5/ 3 0 5 12.11.2.1 Volumes by Cylindrical Shells (6.3) The other method to solve the last volume can be used for more complicated shapes is called “Method of Cylindrical Shells”. xi xi +1 Consider cylindrical shell with inner radius x1 and outer radius x2 , the volume of the shell is the difference between the volumes of outer and inner cylinder. Thus V = V2 − V1 = π x22 h − π x12 h = π ( x22 − x12 ) h = π ( x2 + x1 ) ( x2 − x1 ) h = = 2π x2 + x1 ( x2 − x1 ) h = 2π ravg ( Δr ) h = 2π ravg ⋅ h ⋅ ( Δr ) = circumference ⋅ height ⋅ thikness 2 When we get body obtained by rotating around y-axis, instead of slicing cross sections we divide\peel the body into cylindrical shells, such that f(x) is a height, the volume of every shell is given by Vj = 2π x j ⋅ f ( x j ) ⋅ Δx where x j = ( ) n x j + x j+1 and therefore 2 V = lim ∑ 2π x j f x j Δx = ∫ 2π xf ( x ) dx n→∞ Ex 7. j=1 b a Find volume of body enclosed between the axes and 1− x 3 rotated around y-axis using method of cylindrical shells. 1 ⎛ x 2 x5 ⎞ ⎛ 1 1⎞ 3 V y = 2π ∫ x 1− x dx = 2π ∫ x − x dx = 2π ⎜ − ⎟ = 2π ⎜ − ⎟ = π ⎝ 2 5⎠ 5 ⎝ 2 5 ⎠0 0 0 1 ( 3 ) 1 4 12.11.3 Arc Length (6.4) When we speak about length of a curve we mean the distance between the end points of the graph after they are stretched like a graph was a string or thread. However we can’t really do it this way, therefore we need to a better method. Course: Accelerated Engineering Calculus I Instructor: Michael Medvinsky We approximate the curve by straight lines, i.e. we divide the curve into small pieces each one approximated by a line. The length of the curve y = f ( x ) is approximated by a sum of lengths of these lines. Such lines can be formulated as lj ( x) = ( ) ( ) ( ) ( ) f x j − f x j−1 x j f x j−1 − x j−1 f x j x − x j−1 x − xj f xj + f x j−1 = x+ x j − x j−1 x j−1 − x j x j − x j−1 x j − x j−1 ( ) ( ) The length of each line is given by (Pythagoras formula): (x lj = ) ( ( ) ( )) 2 j − x j−1 + l j x j − l j x j−1 ( = x j − x j−1 ) ( ) ( ) 2 (x = ) ( ( ) ( )) 2 j − x j−1 + f x j − f x j−1 2 = 2 ⎛ f x j − f x j−1 ⎞ 2 1+ ⎜ ⎟ ≈ 1+ ( f ' ( x )) Δx x j − x j−1 ⎝ ⎠ Thus the length of the curve is given by n n b L = lim ∑ l j = lim ∑ 1+ ( f ' ( x )) Δx = ∫ 1+ ( f ' ( x )) dx n→∞ n→∞ j=1 2 j=1 2 a b Similarly, when the curve is given by x = f ( y ) we have L = ∫ a 2 ⎛ dx ⎞ 1+ ⎜ ⎟ dy . ⎝ dy ⎠ When the curve is given by parameterization x = f ( t ), y = g ( t ) we have f ' ( t j ) ≈ Δx ≈ f ' ( t ) Δt and similarly Δy ≈ g' ( t ) Δt , thus lj = (x ) ( 2 j − x j−1 + y j − y j−1 2 b ) 2 ( f '(t ) Δt ) + ( g'(t ) Δt ) 2 ≈ 2 = Δx j , i.e. Δt ( f '(t )) + ( g'(t )) Δt 2 2 2 ⎛ dx ⎞ ⎛ dy ⎞ ⎜⎝ ⎟⎠ + ⎜⎝ ⎟⎠ dt dt dt and therefore L = ∫ a Consider a polar curve r = f (θ ) , in polar coordinates we have x = r cosθ = f (θ ) cosθ , y = r sin θ = f (θ ) sin θ and therefore dx = f ' (θ ) cosθ − f (θ ) sin θ and dθ dy = f ' (θ ) sin θ + f (θ ) cosθ . Thus dθ 2 b 2 b ⎛ dx ⎞ ⎛ dy ⎞ L = ∫ ⎜ ⎟ + ⎜ ⎟ dθ = ∫ ⎝ dθ ⎠ ⎝ dθ ⎠ a a b =∫ a ( f '(θ )) 2 b ( f '(θ ) cosθ − f (θ ) sinθ ) + ( f '(θ ) sinθ + f (θ ) cosθ ) 2 + f 2 (θ ) dθ = ∫ r 2 + ( f ' (θ )) dθ a 2 2 dθ Course: Accelerated Engineering Calculus I Instructor: Michael Medvinsky Compute arc length for curve described by ( x, y ) = ( t,t 2 ) on [ 0,1] Ex 8. 2 b 2 1 2 1 2 ⎛ dx ⎞ ⎛ dy ⎞ L = ∫ ⎜ ⎟ + ⎜ ⎟ dt = ∫ 1+ ( 2t ) dt = 1+ u 2 du = ∫ u=2t ⎝ ⎠ ⎝ ⎠ dt dt 20 a 0 du=2dt = u 1+ u Ex 9. 2 2 0 2 − ∫u 0 2u 2 1+ u 2 du = ... = ( ( 1 u 1+ u 2 + ln u + 1+ u 2 2 )) = 12 (2 2 0 ( 5 + ln 2 + 5 )) Compute arc length of r = 3cosθ + 7 sinθ b b L = ∫ r 2 + ( f ' (θ )) dθ = ∫ 2 a a 12.11.4 ( 3cosθ + 7 sin θ ) ( 2 + −3sin θ + 7 cosθ ) 2 b dθ = ∫ 9 + 7 dθ = 4 ( b − a ) a Average value of function Consider an average * * * y1 + y2 ++ yn f ( x1 ) + f ( x2 ) ++ f ( xn ) f ( x1* )+ f ( x2* )++ f ( xn* ) Avg = = = = b−a n n Δx = ( ) b 1 n 1 f x j Δx = f ( x ) dx ∑ b − a j=1 b − a ∫a Ex 1. An experiment measurement showed that the investigated phenomena could be described as f ( x ) = x 3 − 1 on time interval [1, 3] . Find the average value of this function. 3 3 ⎞ 1 1 ⎛ x4 1 ⎛ 81 1 ⎞ 81− 9 72 3 Avg = x − 1dx = − x ⎟ = ⎜ − 3 − + 1⎟ = = =9 ∫ ⎜ 3−1 1 2⎝ 4 4 ⎠ 8 8 ⎠1 2 ⎝ 4 Note that x 3 − 1 = 9 ⇒ 10 = x 3 ⇒ x = 3 10 , i.e. f ( −2 ) = favg . It is no coincidence. Theorem: If f(x) is continues on [a,b], the there is exists a c in [a,b] such that: f (c) = b 1 f ( x ) dx , in other words we have b − a ∫a b ∫ f ( x ) dx = f ( c )(b − a ) . a