Vector Integrals. Scott N. Walck. October 13, 2016

Transcription

1 Vector Integrals cott N. Walck October 13, 16 Contents 1 A Table of Vector Integrals Applications of the Integrals.1 calar Line Integral Finding Total Charge of a Line Charge Finding Electric otential of a Line Charge Vector Line Integral Finding Electric Field of a Line Charge Dotted Line Integral Finding Electric otential from Electric Field calar urface Integral Finding Total Charge of a urface Charge Finding Electric otential of a urface Charge Vector urface Integral Finding Electric Field of a urface Charge Flux Integral Finding Electric Flux from Electric Field calar Volume Integral Finding Total Charge of a Volume Charge Finding Electric otential of a Volume Charge Vector Volume Integral Finding Electric Field of a Volume Charge

2 3 Fundamental Theorems of Calculus Gradient Theorem tokes Theorem Divergence Theorem Calculation Line Integrals Vector Line Integrals Dotted Line Integrals Flux Integrals A Table of Vector Integrals scalar line integral vector line integral dotted line integral scalar surface integral vector surface integral flux integral scalar volume integral vector volume integral f dl F dl F dl f da F da F da f dv F dv 1D 1D 1D D D D 3D 3D Applications of the Integrals.1 calar Line Integral The scalar line integral requires a scalar field f and a path. f dl.1.1 Finding Total Charge of a Line Charge The path is along the line charge. Q = λ(r ) dl

3 .1. Finding Electric otential of a Line Charge The path is along the line charge. φ(r) = 1 4πɛ. Vector Line Integral λ(r ) r r dl The vector line integral requires a vector field F and a path. F dl..1 Finding Electric Field of a Line Charge The path is along the line charge. E(r) = 1 4πɛ.3 Dotted Line Integral r r r r 3 λ(r ) dl The dotted line integral requires a vector field F and a path. F dl.3.1 Finding Electric otential from Electric Field The path begins at a point where the electric potential is zero and ends at the field point r. φ(r) = E(r ) dl In electrostatics, this integral is path independent, so we could write the integral using only the endpoints of the path. Let a be a point we choose for the electric potential to be zero. φ(r) = r a E(r ) dl A nice property of this latter form is that we see how the field point r on the left is related to the field point r on the right. We are finding the electric potential at the end point r of the path. 3

4 .4 calar urface Integral The scalar surface integral requires a scalar field f and a surface. The surface does not need an orientation. f da.4.1 Finding Total Charge of a urface Charge The surface is over the surface charge. Q = σ(r ) da.4. Finding Electric otential of a urface Charge The surface is over the surface charge. φ(r) = 1 σ(r ) 4πɛ r r da.5 Vector urface Integral The vector surface integral requires a vector field F and a surface. The surface does not need an orientation. F da.5.1 Finding Electric Field of a urface Charge The surface is over the surface charge. E(r) = 1 r r 4πɛ r r 3 σ(r ) da.6 Flux Integral The flux integral requires a vector field F and an oriented surface. F da 4

5 .6.1 Finding Electric Flux from Electric Field The surface is the surface through which to find the electric flux. The orientation points perpendicular to the surface, so that electric field in that direction would count as positive electric flux. Φ E = E(r ) da.7 calar Volume Integral The scalar volume integral requires a scalar field f and a volume V. f dv.7.1 Finding Total Charge of a Volume Charge The volume V is over the volume charge. Q = ρ(r ) dv.7. Finding Electric otential of a Volume Charge The volume V is over the volume charge. φ(r) = 1 ρ(r ) 4πɛ r r dv.8 Vector Volume Integral The vector volume integral requires a vector field F and a volume V. F dv.8.1 Finding Electric Field of a Volume Charge The volume V is over the volume charge. E(r) = 1 r r 4πɛ r r 3 ρ(r ) dv V V V V 5 V

6 3 Fundamental Theorems of Calculus 3.1 Gradient Theorem The gradient theorem requires a scalar field f and a path. uppose the path starts at a and ends at b. f dl = f(b) f(a) The integral on the left is a dotted line integral. 3. tokes Theorem tokes theorem requires a vector field F and an oriented surface. Let be a (closed) path that forms the boundary of the surface. ( F) da = F dl The integral on the left is a flux integral, and the integral on the right is a dotted line integral. 3.3 Divergence Theorem The divergence theorem requires a vector field F and a volume V. Let V be the (closed) oriented surface that forms the boundary of the volume V. The orientation is outward. ( F) dv = F da V The integral on the left is a scalar volume integral, and the integral on the right is a flux integral. 4 Calculation 4.1 Line Integrals The key to evaluating line integrals, whether they be scalar line integrals, vector line integrals, or dotted line integrals, is to find a single variable or 6 V

7 parameter in which to carry out the integration. In many cases, this single parameter can be one of the coordinates in a Cartesion, cylindrical, or spherical coordinate system Vector Line Integrals Example 1. Find the vector line integral F dl for the vector field over the closed path shown below. F(s, φ, z) = s cos φŝ + s sin φ ˆφ y R olution: The path is made up of three portions. Let 1 be the portion that runs along the x axis. Let be the portion that curves, and let 3 be the portion that runs along the y axis. Denote by the entire closed path. The vector line integral over the entire path is the sum of the vector line integrals over each portion. F dl = F dl + F dl + F dl 1 3 We will carry the integrals out in cylindrical coodinates. However, we want to avoid having the unit vectors ŝ and ˆφ under the integral sign, because these unit vectors change direction from place to place. Therefore, we will rewrite the vector field F using Cartesian unit vectors ˆx, ŷ, and ẑ, but keeping the cylindrical coordinates s, φ, and z. x 7

8 We can write the cylindrical coordinate unit vectors ŝ, ˆφ, and ẑ in terms of the Cartesian coordinate unit vectors ˆx, ŷ, and ẑ as follows. ŝ = cos φˆx + sin φŷ ˆφ = sin φˆx + cos φŷ ẑ = ẑ The vector field F is the following. F = s cos φŝ + s sin φ ˆφ = s cos φ(cos φˆx + sin φŷ) + s sin φ( sin φˆx + cos φŷ) = s cos φˆx + s sin φŷ Here, we have used the following trigonometric identities. cos θ = cos θ sin θ sin θ = sin θ cos θ For the first portion 1, we have dl = ds. R F dl = (s cos φˆx + s sin φŷ) ds 1 We want to get all variables in terms of s, since that is the one variable we are integrating over. Along 1, we have φ =. R F dl = ˆx s ds = R3 1 3 ˆx For the second portion, we have dl = s dφ. π/ F dl = (s cos φˆx + s sin φŷ)s dφ Along, we have s = R. F dl = R 3 = R 3ˆx π/ π/ (cos φˆx + sin φŷ) dφ cos φ dφ + R 3 ŷ 8 π/ sin φ dφ

9 π/ π/ [ sin φ cos φ dφ = [ ] π/ [ cos φ cos φ sin φ dφ = = F dl = R 3 ŷ For the third portion 3, we have dl = ds. 3 F dl = R ] π/ ] = π/ (s cos φˆx + s sin φŷ) ds = 1 ( 1) = 1 This looks very much like the expression for the integral over the first portion above, but it s different in two ways. First, the limits on the integral are different. For 3, we re starting at s = R and going to s =. econd, the value of φ is different. For 3, we have φ = π/. 3 F dl = ˆx R s ds = R3 3 ˆx In total we have the following result. F dl = 3 R3ˆx + R 3 ŷ Note that the result of a vector line integral is a vector Dotted Line Integrals Example. Find the dotted line integral F dl for the vector field over the closed path shown below. F(s, φ, z) = s cos φŝ + s sin φ ˆφ 9

10 y x R olution: The path is made up of three portions. Let 1 be the portion that runs along the x axis. Let be the portion that curves, and let 3 be the portion that runs along the y axis. Denote by the entire closed path. The dotted line integral over the entire path is the sum of the dotted line integrals over each portion. F dl = F dl + F dl + F dl 1 3 We will carry the integrals out in cylindrical coodinates. For dl, we can use the standard expression in cylindrical coordinates. dl = dsŝ + s dφ ˆφ + dzẑ Now we can find an expression for F dl. F dl = (s cos φŝ + s sin φ ˆφ) (dsŝ + s dφ ˆφ + dzẑ) = s cos φ ds + s 3 sin φ dφ Recall that for line integrals of all kinds, we want to express the integral in terms of a single variable that we are integrating over. The expression above is not yet in that form. We have three portions of the path that we want to integrate over, and it will turn out that the variable we want for our single variable of integration is different for the different portions. The expression for F dl will simplify in different ways for each portion of the path. 1

11 For the first portion 1, we have φ =. This means that dφ = and that part of F dl will go away. For 1, we have F dl = s ds. R F dl = s ds = R3 1 3 For the second portion, we have s = R, ds =, and F dl = F dl = R 3 sin φ dφ. π/ R 3 sin φ dφ = R 3 For the third portion 3, we have φ = π/, dφ =, and F dl =. F dl = 3 In total we have the following result. F dl = 4 3 R3 Note that the result of a dotted line integral is a scalar. 4. Flux Integrals Example 3. Find the flux integral for the vector field F(r, θ, φ) = rˆr + r sin θ ˆθ + r sin θ cos φ ˆφ through the region enclosed by a triangle in the xy plane with vertices at (x, y, z) = (,, ), (x, y, z) = (,, ), and (x, y, z) = (,, ). olution: The flux integral is F da so the first order of business is to get an expression for da in terms of the two variables we want to integrate over. The triangular region given suggests 11

12 that we want to use Cartesian coordinates and integrate over x and y. The surface element da is a vector, so we need an orientation for the surface. The orientation is always perpendicular to the surface, so it could be either ẑ or ẑ. An orientation was not given in the problem, so let s choose ẑ for our orientation. da = dx dyẑ Next, we want an expression for F da in terms of the two variables we want to integrate over. We notice that the vector field is given in spherical coordinates and the surface is described with Cartesian coordinates. When we take the dot product, we will have terms such as ˆr ẑ showing up. Let s figure these out in advance. tarting with expressions for the spherical unit vectors in terms of the Cartesian unit vectors, ˆr = xˆx + yŷ + zẑ = sin θ cos φˆx + sin θ sin φŷ + cos θẑ x + y + z ˆθ = cos θ cos φˆx + cos θ sin φŷ sin θẑ ˆφ = sin φˆx + cos φŷ we take the dot product of each of these equations with ẑ to find ˆr ẑ = cos θ ˆθ ẑ = sin θ ˆφ ẑ =. Now we write an expression for F da. F da = (rˆr ẑ + r sin θ ˆθ ẑ + r sin θ cos φ ˆφ ẑ) dx dy = (r cos θ r sin θ) dx dy ( ) x + y = z dx dy x + y + z = x + y dx dy In the last step, we used the fact that z = for the surface we are using. 1

13 Next, we need limits for our integral. F da = x x + y dy dx The inner integral goes from the line y = to the line y = x. The outer integral goes from the point x = to the point x =. All that remains is to evaluate the double integral. The inner integral is x x + y dy This is an integral over y with x held constant. From Gradshteyn and Ryzhik, 5th edition, integral.71.3 (page 15) I find this integral. a + x dx = 1 x a + x + 1 a ln(x + a + x ) For a, we ll substitute x, and for x we ll substitute y. x [ 1 x + y dy = y x + y + 1 x ln(y + x + y ) [ 1 = x x + x + 1 x ln(x + ] x + x ) = x + 1 x ln(x + x) 1 x ln x = x + 1 x ln(1 + ) + ln(1 + ) = x ] y=x y= [ 1 x ln( x ) We used the fact that x is everywhere nonnegative on our surface so that ] 13

14 x = x. F da = = x x + y dy dx + ln(1 + ) x dx + ln(1 + ) = 8 3 = 4 3 [ + ln(1 + )] 3.6 The negative result makes sense because the vector field points downward (in the z direction) in the xy plane. 14