Line Integrals - Vector Calculus - PDF

L INE INTEGRALS - AV ECTOR NTON,B C ALCULUS - EVINS BASED ON L INE INTEGRAL AND S URFACE INTEGRALS V ECTOR VALUED FUNCTION OF A REAL VARIABLE...

L INE INTEGRALS - AV ECTOR NTON,B C ALCULUS - EVINS BASED ON L INE INTEGRAL AND S URFACE INTEGRALS V ECTOR VALUED FUNCTION OF A REAL VARIABLE f (t) = t 2 is a real valued function of one independent variable. f : R → R2 has vectors in the codomain as values. f : R → R3 also. Hence domain is 1-dimensional represented by a parameter. Codomain is a 2D/3D space, consisting of vectors. Eg: f (t) = t 2 i + sin t j + et k is a vector valued function (vvf) Image of f (t) = x(t) i + y (t) j + z(t) k is a 3-D curve. f (t) = (x(t), y (t), z(t)) is also the same function. L INE INTEGRAL AND S URFACE INTEGRALS V ECTOR VALUED FUNCTION OF MANY VARIABLES Eg: f (x, y ) = x 2 i + sin y j + exy k is a vector valued function of two variables x, y. The domain is a part of the plane. At each point of the domain, a vector is assigned. See f (x, y ) = (x, 2y , x) = x i + 2y j + x k gives f (a, b) = a i + 2b j + a k at each point (a, b) of R Plot f (x, y ) = i + j + k L INE INTEGRAL AND S URFACE INTEGRALS V ECTOR FIELDS In some cases, at all points in the region under observation, there are vectors acting upon. Such situation wherein at all the points of the domain, a vector valued function is defined is called a vector field. The gravitational field acts at every particle/objects till a distance. The magnetic field around a magnet Velocity vector field of a flowing liquid It is actually a vvf considered at each point in a domain, mathematically. Equation of a vector field is the equation of the vvf creating the field: F(x, y , z) = f1 (x, y , z)i + f2 (x, y , z)j + f3 (x, y , z)k L INE INTEGRAL AND S URFACE INTEGRALS G RAPHICAL REPRESENTATION OF A VECTOR FIELD It is a representation of the behaviour of vectors, given at a collecton of points in the domain. The length of the vectors may not be actual! Since the (x, y , z) represents the position of the point, we may take the position vector r as the input. That is F(x, y , z) = F(r). L INE INTEGRAL AND S URFACE INTEGRALS I NVERSE SQUARE FIELDS Some vector fields are of special interest. They are very much depending on the distance from the soarce creating the field. The Gravitational vector field gives the force of attraction exerted on each point away from the center of earth. More generally any two particles of masses m and M attracts each other and the force of attraction depends how apart they are kept: GmM ∥F(r)∥ = ∥r∥2 Since the direction is as that of the position vector, GmM r −GmM F(r) = − 2 = r ∥r∥ ∥r∥ ∥r∥3 L INE INTEGRAL AND S URFACE INTEGRALS G RADIENT VECTOR FIELD We know that a multivariable function gives rise to the Gradient vector Represented as ∇f for f = f (x, y , z) ∇f = fx i + fy j + fz k It is defined at every point of the domain of the function. Gradient field is defined at every point of the domain Draw the gradient field of f (x, y ) = xi + 2y j L INE INTEGRAL AND S URFACE INTEGRALS C ONSERVATIVE FIELDS AND S CALAR POTENTIAL There are some special properties for the Gradient fields. So a given vector field which is actually the gradient field of some multivariable function is known with a special name. If a vector field can be written in the form f = ∇f , then it is called a conservative vector field Conservative fields are generated from scalar function f of ϕ Ex: 2xi + 2y j is a conservative field where ϕ = x 2 + y 2 We will see that conservative fields are path independent! How to know a given vvf is conservative or not? How to find the potential function of a conservative field? L INE INTEGRAL AND S URFACE INTEGRALS D IVERGENCE AND C URL OF A VECTOR FIELD The are quantities which helps in identifying the behaviour and properties of vector fields. Eg., the curl tells if a given vector field is conservative or not. L INE INTEGRAL AND S URFACE INTEGRALS D IVERGENCE AND C URL OF A VECTOR FIELD Divergence is a scalar and curl is a vector. Example: Find divergence and curl of 2xi + 2y j L INE INTEGRAL AND S URFACE INTEGRALS D EL OPERATOR , L APLACE OPERATOR AND H ARMONIC FUNCTIONS ∂ ∂ ∂ From the definition of div and curl, the operator ( ∂x , ∂y , ∂z ) has special importance. The operator ∇ = ∂ ∂ ∂ ∂x i + ∂y j + ∂z k is called the Del operator. div (F) = ∇ · F curl(F) = ∇ × F The square ∇2 = ∇ · ∇ = ∂2 ∂2 ∂2 ∂x 2 + ∂y 2 + ∂z 2 is called Laplacian operator. It acts on scalar valued functions ∇2 ϕ = 0 is called Laplace equation A function with continuous derivatives satisfying the Laplace equation is called Harmonic function Example: f (x, y ) = x 2 + y 2 L INE INTEGRAL AND S URFACE INTEGRALS I DENTITIES INVOLVING D IV AND C URL 1 div (k F ) = k div (F ) 2 div (F + G) = div (F ) + div (G) 3 div (ϕF ) = ϕ div (F ) + ∇ϕ · F 1 curl(k F ) = k curl(F ) 2 curl(F + G) = curl(F ) + curl(G) 3 curl(ϕF ) = ϕ curl(F ) + ∇ϕ × F 1 div (curlF ) = 0 = curl(∇ϕ) L INE INTEGRAL AND S URFACE INTEGRALS C ONSERVATIVE VECTOR FIELDS AND C URL 1 If F (x, y , z) is conservative, then curl(F ) = curl(∇ϕ) = 0 2 Conversely, if F is over R3 and curl(F ) = 0 3 We can replace R3 with any simply connected domain. 4 For F (x, y ) also, it is true! 5 A proof (for the converse part) can be given using Stoke’s theorem. F (x, y , z) is conservative if and only if curl(F ) = 0 L INE INTEGRAL AND S URFACE INTEGRALS L INE INTEGRALS - M ASS OF WIRE Suppose we want to find the total mass of a thin wire, whose density is given at each point. How do we find it? The wire may be consisting of various type of materials Density vary along the curve. We divide the wire into small segments We can divide so small that a segment contains only one material. Find the mass of segments (density x length) and adding all, we get total mass. There may be still errors. We can minimize error by taking segment length infinitesimely small. The last is actually a limiting procedure. L INE INTEGRAL AND S URFACE INTEGRALS L INE INTEGRAL Integrals of functions which are defined over a curve are known as line integrals. Divide the curve into small segments such that function is ’almost’ same on a segment. Evaluate the function value over the segment (taking some value of the function on the segment x length of segment) Add all the segment values. You get approximate value To minimize the error, take limit as segment length → 0. L INE INTEGRAL AND S URFACE INTEGRALS D EFINITION OF LINE INTEGRAL L INE INTEGRAL AND S URFACE INTEGRALS E VALUATION OF LINE INTEGRALS If the curve is given in terms of the distance, i.e., r = r(s), then we can evaluate the line integral easily. Also, the length of the curve by: L INE INTEGRAL AND S URFACE INTEGRALS E VALUATION OF LINE INTEGRALS Suppose that the equation of the curve is given as r = r(t). We need to convert the integral from s to t ds ∆s ∥r′ (t)∥ = dt = lim. ∆t→0 ∆t ∥r′ (t)∥ ≈ ∆sk ∆tk. The ∆sk ≈ ∥r′ (t)∥∆tk a, b are the limits of t, in the above. L INE INTEGRAL AND S URFACE INTEGRALS E VALUATION OF LINE INTEGRALS To calculate line integral if the function is available as a function of t? and Zthe length Z of the curve Z is given by L= ds = 1 ds = ∥r′ (t)∥ dt C C C s Z  2  2  2 dx dy dz = + + dt C dt dt dt ∗ The line integral can be viewed as area below f , above the curve C! L INE INTEGRAL AND S URFACE INTEGRALS E XAMPLE - LINE INTEGRAL EVALUATION L INE INTEGRAL AND S URFACE INTEGRALS I NTEGRALS WITH RESPECT TO SINGLE VARIABLE The following are defined in a similar way: Z Z f (x, y , z) dx = f (x(t), y (t), z(t)) x ′ (t) dt C C Z Z f (x, y , z) dy = f (x(t), y (t), z(t)) y ′ (t) dt C C Z Z f (x, y , z) dz = f (x(t), y (t), z(t)) z ′ (t) dt C C That is only ’partial segment’ of piece of C is taken instead of ∆sk. Z n X f (x, y , z) dx = lim f (xk∗ , yk∗ , zk∗ ) ∆xk C n→∞ k =1 L INE INTEGRAL AND S URFACE INTEGRALS P ROPERTIES OF LINE INTEGRALS It satisfy most properties of the real integrals Z Z Z f + g ds = f ds + g ds C C C Z Z αf ds = α f ds C C Z Z f ds = − f ds −C C Z Z Z f ds = f ds + f ds C1 +C2 C1 C2 ∗ C1 + C2 means the join of the two curves. L INE INTEGRAL AND S URFACE INTEGRALS E XAMPLE L INE INTEGRAL AND S URFACE INTEGRALS L INE INTEGRALS OF VECTOR FIELDS So far we have considered line integral of f (x, y , z) : R3 → R. Let F(x, y , z) = f (x, y , z)i + g(x, y , z)j + h(x, y , z)k We interpret dr = dxi + dy j + dzk and define The line integral is defined as this integral: L INE INTEGRAL AND S URFACE INTEGRALS W HAT IS THE USE OF LINE INTEGRALS OF VVF ? Work done is a quantity defined using Force’s (vector) effect in moving a particle along a curve! Work done by F in moving a particle along a line segment r is W =F·r Suppose particle is moving in a curve C For finding work, go along the curve. Divide curve into segments (line segment, if small) Calculate F · r in each and add them up. X Work done = F · r, approximately Taking the limit as segments gets infinitesimely small, we get exact work done. L INE INTEGRAL AND S URFACE INTEGRALS E VALUATION OF LINE INTEGRALS OF VVF Taking r = x(t)i + y (t)j + z(t)k and F(r)(t) = f (x(t), y (t), z(t))i + g(x(t), y (t), z(t))j + h(x(t), y (t), z(t))k Z Work done Z= F · dr = f (x(t), y (t), z(t)) dx + g(...) dy + h(...) dz C C Using dr ′ = r (t) dt Z Z W = F · dr = F · r′ (t) dt C C Using dr ′ = r (s) = T ds Z Z W = F · dr = F · T ds, in terms of arclength. C C * F · T = ∥F∥cos θ, so component of F along tangential direction. L INE INTEGRAL AND S URFACE INTEGRALS E XAMPLE L INE INTEGRAL AND S URFACE INTEGRALS PATH INDEPENDENT INTEGRALS Z A line integral f ds is said to be path independent, if for any path C C connecting P to Q, the integral has the same value. This means the value depends only on the end points We know that for ordinary functions, if the integrand is the derivative of some function (i.e., F = f ′ ), then Z b f (x) dx = F (b) − F (a) a Can such a thing happen for the line integral of a vvf? Yes, if the vector field is conservative. F is conservative, if F = ∇ϕ, for some svf ϕ L INE INTEGRAL AND S URFACE INTEGRALS PATH INDEPENDENCE OF CONSERVATIVE FIELDS L INE INTEGRAL AND S URFACE INTEGRALS P ROOF OF FUNDAMENTAL THEOREM L INE INTEGRAL AND S URFACE INTEGRALS E XAMPLE L INE INTEGRAL AND S URFACE INTEGRALS C LOSED PATH - T HEOREM If the vvf is conservative in a connected domain, then the line integral over a closed path is 0, as both end points are same. L INE INTEGRAL AND S URFACE INTEGRALS H OW TO CHECK CONSERVATIVE FIELDS ? Using the partial derivatives, we can establish F conservative or not. For the converse part, region should be ‘simply connected’. Simply connected regions does not have ‘holes’ inside! L INE INTEGRAL AND S URFACE INTEGRALS C HECK CONSERVATIVE IN 3D SPACE Suppose F(x, y , z) : R3 → R3 is conservative. F = f i + gj + hk Then ∂f ∂g ∂f ∂h ∂g ∂h ∂y = ∂x , ∂z = ∂x , ∂z = ∂y That is Curl(F) = 0 We need to assume that f , g, h are continuous in some open region D. Then F is conservative if curl(F)=0. Check if F(x, y ) = (x + y )i + (x − y )j is conservative. You may use either the partial derivatives condition or the curl L INE INTEGRAL AND S URFACE INTEGRALS F INDING THE SCALAR POTENTIAL FUNCTION Suppose F(x, y ) : R2 → R2 is conservative. How to find ϕ? Let F = f i + gj We need a ϕ such that ∇ϕ = F That means ∂ϕ (∗) ∂ϕ (∗∗) ∂x =f and ∂y =g From (*), ϕ(x, y ) = f (x, y ) dx + K (y ) (∗∗∗) , integration done R keeping y constant. This gives ϕ, but incomplete. Find ∂ϕ ∂y from the above and equate with g On equating, we can get K ′ (y ). Integrating K ′ (y ), get K (y ). Substitute K (y ) in (***) to get the complete ϕ L INE INTEGRAL AND S URFACE INTEGRALS E XAMPLE L INE INTEGRAL AND S URFACE INTEGRALS E XAMPLE L INE INTEGRAL AND S URFACE INTEGRALS P ROBLEMS L INE INTEGRAL AND S URFACE INTEGRALS P ROBLEMS L INE INTEGRAL AND S URFACE INTEGRALS P ROBLEMS L INE INTEGRAL AND S URFACE INTEGRALS S URFACES Like the equation of a curve, which may be implicit/Explicit/Parametric, a surface is also described in three forms: Implicit F (x, y , z) = 0 Explicit z = f (x, y ) Parametric r (u, v ) = x(u, v ) i + y (u, v ) j + z(u, v ) k , a ≤ u ≤ b, c ≤ v ≤ d Examples: Sphere x 2 + y 2 + z 2 = a2 Paraboloid z = x 2 + y 2 cone: x = r cos θ, y = r sin θ, z = r , giving r (r , θ) = r cos θ i + r sin θ j + r k with 0 ≤ r ≤ 1, 0 ≤ θ ≤ 2π L INE INTEGRAL AND S URFACE INTEGRALS R ECALL LINE INTEGRALS For Evaluating the line integral Z f (x, y ) ds, C we considered the parametric equation of the curve C : r (t), a ≤ t ≤ b and converted arclength expression to t. Z Z b f (x, y ) ds = f (x(t), y (t)) ∥r ′ (t)∥ dt C a To find integrals over surfaces, we convert surface integrals to double integral over a planar region. L INE INTEGRAL AND S URFACE INTEGRALS E VALUATION OF SURFACE A REA To evaluate surface area/integral, we consider the surface as a surface above a region R and try to evaluate surface area as a double integral over R. L INE INTEGRAL AND S URFACE INTEGRALS S MOOTH SURFACES We assume that the surface is "smooth" (defined by continuous function F (x, y , z) = 0 with ∇F ̸= 0) so that integral can be evaluated as double integral. L INE INTEGRAL AND S URFACE INTEGRALS E VALUATION OF SURFACE A REA OF F (x, y , z) = 0 Assume that the surface is above a region R in the plane. Divide R into small rectangular areas. Then correspondingly, the surface also gets ’divide’ in to ’patches’. Now find areas of patches and add to get the total surface area. But how do you find the curved area ∆Sk ? L INE INTEGRAL AND S URFACE INTEGRALS S URFACE AREA SEGMENT The surface can be approximated by the tangent plane, just touching the surface at one point. L INE INTEGRAL AND S URFACE INTEGRALS S URFACE AREA SEGMENT - A CLOSER LOOK L INE INTEGRAL AND S URFACE INTEGRALS S URFACE AREA OF IMPLICIT SURFACES See that ∆Pk | cos γk | = ∆Ak ∆Ak ∆Pk = | cos γk | X X ∆Ak ∆Pk = | cos γk | Z Z 1 Limit gives surface area = dA R |cosγ| γ is the angle between the planes ∆Pk and ∆Ak. Same as the angle between their normals ∇F and p̂ (with |p̂| = 1) |∇F · p̂| = |∇F ||p̂|| cos γ| |∇F | Z Z surface area = dA R |∇F · p̂| L INE INTEGRAL AND S URFACE INTEGRALS S URFACE AREA Suppose the surface is given by F (x, y , z) = 0. Then |∇F | Z Z Surface area = dA R |∇F · p̂| When the surface is projected over to XY plane, p̂ = k̂. When the surface is projected over to XZ plane, p̂ = ĵ. When the surface is projected over to YZ plane, p̂ = î. L INE INTEGRAL AND S URFACE INTEGRALS S URFACE AREA OF EXPLICIT SURFACES Suppose surface is z = f (x, y ). Then obviously it is above XY plane. Consider it as F (x, y , z) = f (x, y ) − z = 0 p̂ = k̂ q |∇F | = |fx î + fy ĵ − 1| = fx 2 + fy 2 + 1 |∇F · p̂| = |∇F · k̂ | = 1 Z Z q surface area = 1 + fx 2 + fy 2 dydx R Similarly, Z Z q If y = f (x, z), then it becomes 1 + fx 2 + fz 2 dzdx R L INE INTEGRAL AND S URFACE INTEGRALS E XAMPLE Find the area of the surface cut from the bottom of the paraboloid x 2 + y 2 − z = 0 by the plane z = 4. L INE INTEGRAL AND S URFACE INTEGRALS E XAMPLE L INE INTEGRAL AND S URFACE INTEGRALS E XAMPLE L INE INTEGRAL AND S URFACE INTEGRALS W HAT IF THE SURFACE IS GIVEN IN PARAMETRIC FORM ? Suppose S : r (u, v ) = x(u, v )î + y (u, v )î + z(u, v )î, where a ≤ u ≤ b, c ≤ v ≤ d. We assume that the surface is smooth (means r u × r v ̸= 0 forming a tangent plane for each point (u, v ). Here also we find the area of parallelogram approximating the patch’s (yellow color on surface) area L INE INTEGRAL AND S URFACE INTEGRALS S URFACE AREA - PARAMETRIC FORM The sides of the parallelogram are in the directions of the derivatives r u and r v Corresponding to the base rectangle of sides ∆u and ∆v , the parallelogram will have sides ∆ur u and ∆v r v Hence area of parallelogram (on the tangent plane) is ∥∆ur u × ∆v r v ∥ = ∥r u × r v ∥∆u∆v L INE INTEGRAL AND S URFACE INTEGRALS S URFACE AREA - PARAMETRIC FORM Hence the total area is X X X ∆Sk ≈ ∆Pk = r u × r v ∆u∆v Taking limit as subdivisions goes to ∞, we have Z Z surface area = ∥r u × r v ∥ du dv v u L INE INTEGRAL AND S URFACE INTEGRALS S TANDARD ( PARAMETRIZED ) SURFACES Standard Sphere - center origin, radius a: x = a sin ϕ cos θ, y = a sin ϕ sin θ, z = a cos ϕ, 0 ≤ ϕ ≤ π; 0 ≤ θ ≤ 2π Its implicit equation is x 2 + y 2 + z 2 = a2 Standard Cone - vertex at origin: x = r cos θ, y = r sin θ, z =p r , 0 ≤ r < ∞; 0 ≤ θ ≤ 2π Its implicit equation is z = x 2 + y 2 Standard infinite cylinder - radius a: x = a cos θ, y = a sin θ, z = z, 0 ≤ z < ∞; 0 ≤ θ ≤ 2π Its implicit equation is x 2 + y 2 = a2 , z can be any. General explicit surface - z = f (x, y ), a ≤ x ≤ b, c ≤ y ≤ d x = u, y = v , z = f (u, v ), a ≤ u ≤ b, c ≤ v ≤ d L INE INTEGRAL AND S URFACE INTEGRALS E XAMPLE L INE INTEGRAL AND S URFACE INTEGRALS S URFACE INTEGRALS We have defined integrals over curves, which are known as line integrals. There are functions which are defined on surfaces Many a time we need to take some "aggregate value" of such functions over the total surface Example: Total Painting cost of a surface (function is ’painting cost’), assuming paints of different colors have different costs! Example: If the density of a thin sheet (consisting of various metals) is known as a function, we can calculate the mass of the object (density = mass/area) L INE INTEGRAL AND S URFACE INTEGRALS S URFACE INTEGRALS Suppose ϕ(x, y , z) is a scalar valued function defined over a surface σ. We can assume ϕ(x, y , z) to be the density of a point (x, y , z). We want to find the mass of the lamina Divide the lamina into small patches, each patch may be very very small so that it will be almost of the same material. So, the density of a small patch is almost the same over the lamina. Choose a point in the lamina as (xk ∗ , yk ∗ , zk ∗ ), which represents the patch. L INE INTEGRAL AND S URFACE INTEGRALS S URFACE INTEGRALS ϕ(xk ∗ , yk ∗ , zk ∗ ) is the density of the point (xk ∗ , yk ∗ , zk ∗ ), which is almost the same throughout the patch (the patch is very small) Now ϕ(xk ∗ , yk ∗ , zk ∗ ) · ∆Sk gives the mass of the patch. Do the same for each patch ∗ X ϕ(xk , yk ∗ , zk ∗ ) · ∆Sk gives the mass of the total lamina approximately. There may be errors of approximation! Taking the lime as the number of patches becomes infinite, errors vanish. ∗ X Mass = lim ϕ(xk , yk ∗ , zk ∗ ) · ∆Sk which is defined as n→∞ Z Z ϕ(x, y , z) dS σ L INE INTEGRAL AND S URFACE INTEGRALS E VALUATION OF SURFACE INTEGRALS The surface may be given implicit/Explicit/parametrically. X Z Z Recall that Surface area = lim 1 · ∆Sk = 1 dS = n→∞ σ |∇F | Z Z dA if surface is F (x, y , z) = 0 |∇F · p̂| Z ZR q 1 + fx 2 + fy 2 dA if the surface is z = f (x, y ) type Z ZR ∥r u × r v ∥ du dv if surface is r (u, v ), parametric. R L INE INTEGRAL AND S URFACE INTEGRALS S URFACE INTEGRALS - H OW TO EVALUATE ? The surface σ may be given implicit/Explicit/parametrically. Z Z n ∗ X ϕ(x, y , z) dS = lim ϕ(xk , yk ∗ , zk ∗ ) · ∆Sk = σ n→∞ 1 |∇F | Z Z ϕ(x, y , z) dA if surface is F (x, y , z) = 0 R |∇F · p̂| Z Z q ϕ(x, y , z) 1 + fx 2 + fy 2 dA if the surface is z = f (x, y ) type R Z Z ϕ(x(u, v ), y (u, v ), z(u, v ))∥r u × r v ∥ du dv if surface is R r (u, v ), parametric. L INE INTEGRAL AND S URFACE INTEGRALS E XAMPLE L INE INTEGRAL AND S URFACE INTEGRALS E XAMPLE L INE INTEGRAL AND S URFACE INTEGRALS E XAMPLE L INE INTEGRAL AND S URFACE INTEGRALS E XAMPLE L INE INTEGRAL AND S URFACE INTEGRALS THANK YOU L INE INTEGRAL AND S URFACE INTEGRALS