Integrating Functions Over a Line (or Contour) Example: When dry, a climbing rope weighs 0.175 #/foot. The weight density for a wet rope hanging vertically is: wh()=+0.175 1 0.7 e−0.005h # per foot 0.30 0.28 0.26 Find the weight of a wet 2000’ 0.24 rope hanging vertically 0.22 0.20 Note: a dry rope 2000’ long would weigh 0.18 2000 feet x 0.175 #/foot = 350 # 0.16 Weight Density (pounds/foot)Weight Density 0 500 1000 1500 2000 Height Above Ground (feet) The weight of a wet rope is: h 2000 Total weight =∫∫whdh( )=+ 0.175() 1 0.7 e−.005h dh 00 2000 =−0.175he 140−.005h = 0.175() 2000 −+= 0.0064 140 374.5# ()0 Contour Integration with Vectors Example: if the force acting on a particle is Fa=−ˆy Newtons, how much work is required to move the particle from (0,0) to (1,1) along the path yx= ? F y W = work = force times distance, where (1,1) force is the force tangent to the path. Path, or Contour x (0,0) In this example, the angle between the path and the force is 45° over the entire path: FFTAN =°cos 45 The length of the path is 2 ⇒=°=work 2 cos 45 1 Nm To handle more general cases, where the vector or path varies, we will need to integrate. To do this, define a vector differential element, dl 1 Differential Length Vector dl is an infintesimally-short vector tangent to a curve or li ne In general, for Cartesian coordinates dl dl ddxadyadzal =++ˆˆˆx yz This can be put in terms of dx, dy, or dz only to facilitate integration. The length of the differential length vector is: 222 dll== d() dx + () dy + () dz Vector Contour Integration Needed to sum the tangential component of a vector function along a path. Generally of the form: ∫∫Fd⋅⋅AA or v Fd C Where the second form is for a closed contour. Note that any closed contour necessarily defines a surface. The dot product provides a scalar representing the component of the vector in the direction of the contour. Example ˆˆ If Faa=−2 xy + find the work required to go from (0,0) to (2,6) along the line yx=3 2 Vector Contour Integral Example Work = F d F = Tangential Force ∫ T l T C dW=⋅ F dll = Fcosθ d F ⋅=−+dl ( 22 aˆˆxy a) ⋅( dxa ˆ x + dya ˆ y + dza ˆ z) =−+ dx dy dy since y = 3x on the contour, = 3 ⇒ dy = 3dx y dx (2,6) F ⋅=−+ddxdxdxl 23 = (could have put in terms of dy also) so : ()2,6 2 y = 3x Work = ∫∫Fdl⋅= dx =2 ()0,0 0 x (0,0) Vector Contour Integral Example #2 A particle is attracted towards the origin with a force equal to kr, where r is the distance to the origin. How much work is required to move that particle from (0,1) to (1,2) along the path y = x2 + 1. y (1,2) Solution (x,y) First- find the force vector. (0,1) F y = x2 + 1 ˆ + ˆ Since xax yay points from the origin to the point (x, y), then x x ˆ − ˆ - ax yay points from (x, y) to the origin. 3 Vector Contour Integral Example #2 (2) The length of this vector is xyr2 +=2 , so F=− k() xaˆˆxy − ya (1,2) Work = Fd⋅⋅=−−⋅+ and Fd k()() xaˆˆ ya dxa ˆ dya ˆ ∫ llx yxy (0,1) (1,2) =−kxdxydy() − ⇒Work = ∫ kxdxydy () − − (0,1) 12 =−kxdxydyk ∫∫ − =−+()0.5 1.5 =− 2 k 01 Open and Closed Contours If a contour does not enclose a surface, it is an open contour: y Example: y = x2 x If a contour does enclose a surface, it is a closed contour: y Example: (x-3)2+(y-2)2 = 1 x 4 Tutorial on Double Integrals First consider a single integral f(x) arbitrary function b Range fxdx() = Total Area of x ∫ 123 a area of f(x) differential element x a dx b We can also express this as a double integral f(x) bbfx() fx() ∫∫dydx= ∫ y dx aa0 0 dy b fxdx( ) Same as before x ∫ a a dx Vector Surface Integration Example: For the fluid velocity given below, find the volume of fluid per second passing through the portion of the x=1 plane that extends from -1 to 1 on z and -2 to 2 on y. 32 Vxayaxza=++2 ˆˆxy ˆ z m/sec Only the x-component of velocity will pass z through the surface so we only need Vx. y Vx =2x =2 m/sec on the x = 1 plane Volume/sec = V(/)( m sdsm2 ) ∫∫ x x S Since Vx is constant on the surface we do not need to integrate Vds=× V area =2( m / s ) × 8( m23 ) = 16 m / sec ∫∫ xx S 5 Differential Surface Vector Needed when integrating a vector quantity passing through a surface ds is a vector that is normal to the surface. If the surface is closed, ds points out of the surface. In general, the field passing through a surface is given by: ∫∫V ⋅ds or ∫∫V ⋅ds if the surface is closed s Note: a closed surface defines a volume Surface Vector Integration Example #1 Find the rate of fluid flow through the surface on the z=0 plane defined by the region xy = 1 to 3 and = 2 to 4 if fluid flow 342 2 is defined by: Vyayz=+ˆˆˆxyz() +29 axa − m/sec z In this example, only the z component of fluid velocity y will pass through this surface. ds ds =± dsaˆˆzz =± dxdya 2 x so Vds⋅=± xdxdy 43 2 1 3 Flow rate = ∫∫Vds⋅= ∫ ∫ xdxdy =173 m / sec s 2 1 6 Cylindrical Coordinate System z r is the distance from the z-axis φ is the angle with respect to the x-axis z is z r z Any point in 3-D space can be defined by ( ) y r, φ, z φ Conversion to Cartesian Coordinates x = r cosφ y = r sin φ z = z x Cylindrical Coordinate System z arˆr is in the direction of increasing Top View ˆ y az aˆφ is in the direction of increasing φ aˆφ φ azˆ is in the direction of increasing r (r,φ, z) z aˆ aˆr φ y aˆ aaaˆˆˆ ⊥⊥ r φ rzφ x x Differential surface element vectors Curved surface ds = ±aˆr rdφdz Top & bottom ds = ±aˆz rdφdr Radial cut ds = ±aˆφ drdz Differential volume element dv= rdφ drdz 7 Comment on Radians Linear The linear distance traced out by φ is rφ if φ distance is in radians. distance traveled at a constant radius φ ≡ (radians) radius φ If a wheel rolls one revolution, it covers a r linear distance of 2π r Thus, there are 2π radians per revolution Coordinate Conversion: Cylindrical-Cartesian Top View Using trigonometry: y −1 x φ xr=⇒=cosφφ cos r r av φ y yr=⇒=sinφφ sin −1 av r r x zz= rxy=+22 Example If a point is defined by (r,φ, z) = (5, 30°, 8), that same point would be represented in Cartesian by (x, y, z) = (4.33, 2.5, 8) 8 Spherical Coordinates (R,θ,φ) z R = Distance from origin θ = Angle with z-axis φ = Angle with x-axis θ R y φ xR= sinθ cosφ x yR= sinθ sinφ zR= cosθ Spherical Differential Surface Element (dS) z Rsinθdφ Rdθ θ dS = R2sinθdθdφ y φ x 9 Another Way To Look At Rsinθ z r r = Rsinθ Cylindrical Spherical θ R y x So rdφ = Rsinθφ d Spherical Differential Volume Element (dV) z Rsinθdφ Rdθ dR θ dV = R2sinθdθdφdR y φ x 10 Spherical Examples What is the surface area of a sphere? 22ππ π π Area =∫∫ds= ∫ ∫R222sinθ ddθφ=− ∫ R cos θ d φ = 4 π R 00 0 0 What is the volume of a sphere? R 22ππR ππ R3 Volume =∫∫∫dV= ∫ ∫ ∫ R2 sinθ dRdθφ d= ∫ ∫ sin θθφ d d 000 00 3 0 π 2π RRR3332π 24π =−∫ cosθφd = φ = 3330 0 φ =0 Another Contour Integration Example In a previous example, we integrated a function over a straight line. In this example, we will integrate a function over a curved line. Calculate ∫? fxyd(, )A over the contour defined by the upper half of the circle x22+=yfxyxy4 in the direction shown, where (, ) = 2 y f o dℓ = rdφ = 2dφ n n io io t t a 2 c r dℓ is a differential length e g r = i e r d t in dφ element along the contour. x So the first step is to define dℓ (-2,0) (2,0) Because the geometry of the contour is circular, we will want to use polar coordinates ⇒ dℓ = rdφ = 2dφ on the contour, since r = 2 on the contour 11 Example (Continued) Since we are defining our variable of integration, dℓ in polar coordinates, we need to convert our function, f(x,y), into polar coordinates (i.e., f(r,φ) = f(2,φ) on the contour since r = 2 on the contour) xr==cosφ 2cosφφφ yr ==sin 2sin fxy(, )=⇒ xy2 fr (,φ ) = f (2,)φφφ =()() 2cos2 2sin 0 3 0 2 −16cosφ − 32 So ?∫∫fxyd(, )A ===()()2cosφφφ 2sin2 d π 33π n+1 n [ fx()] Note []fx() f'( xdx) = ∫ n +1 12.