sign in sign up
<<
Home , Gradient theorem

Math 2374: Multivariable and Vector Analysis

Part 26

Fall 2012 The of

• line of scalar-valued function • of vector fields • of scalar-valued function • surface integral of vector fields • double integral • triple integral Line integral of scalar-valued function

Definition (Path integral) Let f : U ⊂ R3 → R be a . Let c : I = [a, b] → U of class C1. We suppose that the composition function t → f (x(t), y(t), z(t)) is continuous on I. We define the path integral as:

Z Z b f ds = f (x(t), y(t), z(t))kc0(t)kdt c a

Remark When f = 1, we recover the definition of the of c. Line integral of vector fields

Definition (Line integral) Let F : U ⊂ R3 → R3 be a continuous vector fields. Let c : I = [a, b] → U of class C1. We suppose that the composition function t → F(x(t), y(t), z(t)) is continuous on I. We define the line integral as:

Z Z b F · ds = F(x(t), y(t), z(t)) · c0(t)dt c a How to compute line integrals?

1 Directly by applying the formula given in the definition. 2 If F is conservative (F = ∇f ) then we have: Z F · ds = f (c(b)) − f (c(a)). c

2 3 If F and c are in R and c is a simple close curve: Z ZZ ∂F ∂F  F · ds = 2 − 1 dxdy (Green’s theorem) c D ∂x ∂y ( c is the boundary of D and is oriented counterclockwise). 3 4 If F and c are in R and c is a simple close curve: Z ZZ F · ds = curlF · dS (Stokes’ theorem) c S (c must be a positively oriented boundary of S) Parametrized Surfaces

Definition (Parametrized Surfaces) A parametrized surface is a function Φ: D ⊂ R2 → R3, where D is a domain in R2. The surface S corresponding to the function Φ is its image: S = Φ(D). We can write:

Φ(u, v) = (x(u, v), y(u, v), z(u, v)).

When Φ is of class C1 we define the two vectors:

∂Φ ∂x ∂y ∂z T = = i + j + k u ∂u ∂u ∂u ∂u ∂Φ ∂x ∂y ∂z T = = i + j + k v ∂v ∂v ∂v ∂v

S is regular at Φ(u0, v0) provided that Tu × Tv 6= 0 at (u0, v0). Surface integral of scalar-valued function

Definition If f (x, y, z) is a real-valued continuous function defined on the parametrized surface S, we define the integral of f over S to be ZZ ZZ f dS = f (Φ(u, v))kTu × Tv kdudv S D

Remark If f = 1 then we recover the definition of the area of the surface S: ZZ A(s) = kTu × Tv kdudv D Surface integral of vector fields

Definition Let F be a vector field defined on the parametrized surface S, we define the surface integral of F over S to be ZZ ZZ F · dS = F(Φ(u, v)) · (Tu × Tv ) dudv S D

Remark 1 If F = G and c is the positively oriented boundary of S ZZ Z F · dS = G · ds (Stokes’ theorem) S c

2 If S is a closed surface that is the boundary of a solid W ZZ ZZZ F · dS = div FdV (Gauss’ theorem) S W in double integral

Definition Let D and D∗ be two elementary regions in the plane and let T : D∗ → D be of class C1; suppose that T is one-to-one on D∗ ∗ and T (D ) = D. Then for any integrable function f : D → R, we have ZZ ZZ ∂(x, y) f (x, y)dxdy = f (x(u, v), y(u, v)) dudv D D∗ ∂(u, v)

Polar coordinates ZZ ZZ f (x, y)dxdy = f (r cos θ, r sin θ)rdrdθ D D∗ Change of variables in triple integral

Definition Let W and W ∗ be two elementary regions in space and let T : W ∗ → W be of class C1; suppose that T is one-to-one on w ∗ and T (W ∗) = W . Then for any integrable function f : W → R, we have ZZZ ∂(x, y, z) f (x(u, v), y(u, v), z(u, v)) dudvdw W ∗ ∂(u, v, w)

Cylindrical coordinates ZZZ f (r cos θ, r sin θ, z)rdrdθdz W ∗

Spherical coordinates

ZZZ f (ρ sin φ cos θ, ρ sin φ sin θ, ρ cos φ)ρ2 sin φdρdθdφ W ∗ The fundamental theorems of

• the theorem for line integrals • Green’s theorem • Stokes’ theorem • Gauss’ theorem ( theorem) for line integrals

Theorem Let f : U ⊂ R3 → R be a function of class C1. Let c : I = [a, b] → U of class C1. We suppose that the composition function t → ∇f (c(t)) is continuous on I. We have Z ∇f · ds = f (c(b)) − f (c(a)). c Green’s theorem

Theorem Let D be a simple region and let c be its boundary. Suppose that P : D → R and Q : D → R are of class C1. Then Z ZZ ∂Q ∂P  Pdx + Qdy = − dxdy c+ D ∂x ∂y

Remark • Can be generalized to any ”decent” region in R2. • If c is a simple closed curve that bounds a region to which Green’s theorem applies, then the area of the region D bounded by c = ∂D is

1 Z A = xdy − ydx. 2 c Stokes’ theorem

Theorem Let S be an oriented surface defined by a one-to-one parametrization Φ: D ⊂ R2 → S, where D is a region to which Green’s theorem applies. Let ∂S denote the oriented boundary of S and let F be a C1 vector field on S. Then ZZ Z (∇ × F) · dS = F · ds. S ∂S

Remark If S has no boundary then the integral on the left is zero. Gauss’ theorem

Theorem Let W be a solid and S its boundary: S = ∂W . Let F be a C1 vector field defined on W then ZZZ ZZ (∇ · F)dV = F · dS. W S