sign in sign up
<<
Home , Leibniz integral rule

VECTOR CALC STUFF - , GRAD, DIV,

1. Theory Note 1. The hardest part (for me at least) about this vector stu is the notation. There are a few dierent ways of writing any one thing. You should nd a system of notation that YOU like and stick to it. You also have to be able to translate from other notations into your own so that you can understand the problem when it is given to you. I also did everything here in 3 dimensions for illustration purposes, but its not too hard to generalize to more dimensions if you are so inclined. Once you understand what is going on with these things you should do a lot of problems from old exams. There is a category called  lled with problems on the written exam wiki. Many are very routine once the machinery below is set up; the problems themselves usually are pretty simple. Note 2. WARNING! On functions I talk about here, I assume the functions are well behaved, usually meaning that they are dierentiable. You should watch out because the denition of a mutli-dimensional being dierentiable can be a bit slippery. One thing that will give us what we want is if f(x, y, z) has continuous partial , ∂f , ∂f and ∂f . ∂x ∂y ∂z Fact 3. ( and ) Let x : R3 → R,y : R3 → R, and z : R3 → R be three dierentiable functions. We will use u, v, w here as our coordinates so that we write x(u, v, w), y(u, v, w), z(u, v, w). Dene the matrix:  ∂x ∂x ∂x  ∂u ∂v ∂w ∂(x, y, z) ∂y ∂y ∂y :=  ∂u ∂v ∂w  ∂(u, v, w) ∂z ∂z ∂z ∂u ∂v ∂w

Suppose f is a function f : R3 → R. Dene: h ∂f ∂f ∂f i ∇(x,y,z)f = ∂x ∂y ∂z (Think of as ∂f )Then we have the following chain rule: ∇(x,y,z)f ∂(x,y,z) ∂(x, y, z) ∇ f = ∇ f (u,v,w) (x,y,z) ∂(u, v, w)  ∂x ∂x ∂x  ∂u ∂v ∂w  ∂f ∂f ∂f  h ∂f ∂f ∂f i ∂y ∂y ∂y i.e. ∂u ∂v ∂w = ∂x ∂y ∂z  ∂u ∂v ∂w  ∂z ∂z ∂z ∂u ∂v ∂w Reading this o entry by entry is the classical version of the chain rule, e.g: ∂f ∂f ∂x ∂f ∂y ∂f ∂z = + + ∂u ∂x ∂u ∂y ∂u ∂z ∂u Exercise 4. Prove the Leibniz Rule: ∂ h(x) h(x) ∂ ∂h ∂g . ∂x g(x) f(x, t)dt = g(x) ∂x f(x, t)dt + f(x, h(x)) ∂x − f(x, g(x)) ∂x What conditions do we need on f for this? Note:´ Sometimes knowing´ this formula is worth a point on the written exam! Fact 5. The ∇f is also useful for the following fact: If we look at the level surface S = {(x, y, z): f(x, y, z) = c} = f −1 {c} the ∇f is perpendicular to the the surface S. The direction ∇f is the direction of most increase of the function f. You can also easily check that it obey the ∇ (fg) = (∇f) g + f (∇g) Fact 6. ( for integration) Let x : R3 → R,y : R3 → R, and z : R3 → R be three dierentiable functions. We will use u, v, w here as our coordinates so that we write x(u, v, w), y(u, v, w), z(u, v, w). Dene ∂x ∂x ∂x ∂u ∂v ∂w ∂(x, y, z) ∂y ∂y ∂y := det ∂u ∂v ∂w ∂(u, v, w) ∂z ∂z ∂z ∂u ∂v ∂w

∂(x, y, z) f(x, y, z)dxdydz = f[x(u, v, w), y(u, v, w), z(u, v, w)] dudvdw ˚R ˚R? ∂(u, v, w) 1 2 VECTOR CALC STUFF - INTEGRALS, GRAD, DIV, CURL

Exercise 7. Check that for polar coordinates, with (x, y, z) = (r sin θ cos φ, r sin θ sin φ, r cos θ) where r ∈ (0, ∞),

θ ∈ (0, π) and φ ∈ (0, 2π) that ∂(x,y,z) = r2 sin θ. ∂(r,θ,φ)

2 √ Exercise 8. Show that ∞ −x . How can changing variables help? −∞ e dx = π ´ Denition 9. (Volumes and Volume Integrals). Sometimes, a problem will say V = {some volume region} and then the problem will talk about f dV . How is this dened? A parametrization of V is a way to write V as in terms of a function ~r(x, y, z) = (r1˝(x, y, z), r2(x, y, z), r3(x, y, z)) so that V = {~r(x, y, z):(x, y, z) ∈ R} where R is some region in R3 (usually we pick the parametrization so that it is a box!). Take any parametrization of the region V then we have: (Q: Why does this not depend on the particular parametrization of the region?)

∂(r1, r2, r3) f dV := f[~r(x, y, z)] dxdydz ˚ ˚R ∂(x, y, z) Denition 10. (Surfaces and Surface Integrals) A surface S ⊂ R3 is another common thing that shows up. A parametrization of S is a way to write S as in terms of a function ~r(s, t) = (r1(s, t), r2(s, t), r3(s, t)) so that S = {~r(s, t):(s, t) ∈ R} where R ⊂ R2 is a region in the plane. Compare this to the denition of a volume above. For surfaces the most important thing to remember is that the normal direction to the surface can be found by a cross product: ∂~r ∂~r ∂r ∂r ∂r  ∂r ∂r ∂r  ~n ∼ × = 1 , 2 , 3 × 1 , 2 , 3 ∂s ∂t ∂s ∂s ∂s ∂t ∂t ∂t

We are often interested in the normalized version of this, which is ∂~r × ∂~r ∂~r ∂~r . There are two avors of nˆ = ∂s ∂t/| ∂s × ∂t | surface integrals that can come up. They are both dened by taking a parametrization ~r(s, t) and using the formula for the area element d ∂~r ∂~r d d . (How do you remember this?!?) The two avours of integral are: (Here A = ∂s × ∂t s t 3 is a scalar functions, while ~ 3 3 is a vector function ~ ) f : R → R G : R → R G = (G1,G2,G3)

∂~r ∂~r [Scalar Integral] fdA := f (~r(s, t)) × dsdt ¨ ¨R ∂s ∂t   [Vector Integral (Flux)] G~ · dA~ := G~ · nˆ · dA~ ¨ ¨ ∂~r ∂~r ! ∂s × ∂t ∂~r ∂~r = G~ (~r(s, t)) · × dsdt ¨ ∂~r ∂~r ∂s ∂t R ∂s × ∂t ∂~r ∂~r  = G~ (~r(s, t)) · × dsdt ¨R ∂s ∂t Example 11. A graph is a special type of surface which is parametrized by S = {x, y, f(x, y)} for a positive scalar function f > 0. Show that the area element of such a surface is dA = p1 + |∇f|2dxdy Denition 12. (Curves and Line Integrals) A parametrization of a curvesC is a way to write C as in terms of a function ~r(t) = (r1(t), r2(t), r3(t)) so that C = {~r(t): t ∈ [a, b]}. Like surface integrals there are two types of line integrals we can do. (Here 3 is a scalar functions, while ~ 3 3 is a vector function ~ ) f : R → R G : R → R G = (G1,G2,G3)

b d Scalar Integral] d ~r [ f l = f (~r(t)) d dt ˆC ˆa t ~ [Vector Integral] G~ · dl = G1dx + G2dy + G3dz ˆC ˆ b d  ~ ~r d = G (~r(s, t)) · d t ˆa t Theorem 13. (Fundamental Theorem of Calculus) Let γ : [0, 1] → R be a curve. Then:

(∇f) · d~l = f (γ(1)) − f (γ(0)) ˆ γ VECTOR CALC STUFF - INTEGRALS, GRAD, DIV, CURL 3

Denition 14. ( and Curl) Suppose ~ 3 3 is a vector function ~ . We dene the G : R → R G = (G1,G2,G3) divergence and curl of the vector eld as:

∂G ∂G ∂G [Divergence (scalar)]∇ · G~ = 1 + 2 + 3 ∂x ∂y ∂z ∂G ∂G ∂G ∂G ∂G ∂G  [Curl (vector)]∇ × G~ = 3 − 2 , 1 − 3 , 2 − 1 ∂y ∂z ∂z ∂x ∂x ∂y Fact 15. Here is a collection of a facts about Divergence and Curl:   (1) Let F~ = ∇ × G~ . Then ∇ · F~ ≡ 0. Short form: The curl of G~ is divergence-free : ∇ · ∇ × G~ ≡ 0. Under some conditions, the converse is also true: If certain mild conditions hold, and ∇ · F~ ≡ 0, then there exists G~ so that F~ = ∇ × G~ . Look up Helmholtz Decomposition to learn more about the mild conditions if you are interested. (2) Let φ be a scalar function, and let G~ = ∇φ be its gradient eld. Then ∇ × G~ ≡ 0. Short form: Gradient elds are curl-free : ∇×(∇φ) ≡ 0. The converse holds if we are on a simply connected domain: If ∇×G~ ≡ 0 on a simply connected domain, then there exists φ so G~ = ∇φ. (3) Let ~ be a vector eld. Let be a curve in 3 going from to . Consider the ~ d~ . If F γ R v1 v2 γ F · l ~ then ~ d~ depends only on the endpoints and . The converse holds too:´ If for every ∇ × F ≡ 0 γ F · l v1 v2 , , the integral´ ~ d~ depends only on the endpoints and , then ~ . v1 v2 γ F · l v1 v2 ∇ × F ≡ 0 ´ Theorem 16. () Suppose we have some volume V whose boundary is a surface S. Then, for a vector eld F~ we have that:     F~ · d~A = F~ · nˆ dA = ∇ · F~ dV ¨ ¨ ˚ S S V Corollary 17. (Green's First Identity) By plugging in F~ = ψ∇φ into the divergence theorem, and using some manipulations we get: ψ (∇φ · nˆ) dA = ψ∇2φ + ∇φ · ∇ψ dV ¨ ˚ S V Theorem 18. (Stokes Theorem) Suppose that S is a surface whose boundary is a curve C. Then, for a vector eld F~ we have that:     F~ · d~l = ∇ × F~ · nˆdA = ∇ × F~ · d~A ˆ ¨ ¨ C S S Corollary 19. (Green's Theorem) If we take F~ = (L, M, 0), and S is a surface that is lying in the plane (so that the normal vector is in the zˆ direction) then Stoke's theorem says that: ∂M ∂L Ldx + Mdy = − dA ˆ ¨ ∂x ∂y C S 2. Problems Note 20. I take the questions word for word here, even though sometimes the wording is confusing, because that is part of the diculty of the questions! Problem 21. (Jan 11 #1) Compute 2 2 . Here 2 is the unit sphere in 3, and is S2 x1x2dS(x) S R x = (x1, x2, x3) dS the surface area element. ´ Problem 22. (Jan 09 #2) Compute where is the curve given by the intersection L(y−z)dx+(z−x)dy+(x−y)dz L of the two surfaces: ´ x2 + y2 + z2 = a2 x + y + z = 0 with counterclockwise orientation viewed from the positive x-axis. Problem 23. (Sept 07 #5 Part I) Evaluate the integral 2 2 , where is the sphere of radius centered S(x + y )dσ S 1 at (0, 0, 0) and dσv is surface area. ´ 4 VECTOR CALC STUFF - INTEGRALS, GRAD, DIV, CURL

Problem 24. (Sept 10 #5) Part I: For every positive integer n, nd m(n) such that the following integral is nite for m > m(n): dx1 . . . dxn Pn m ˆ n 1 + |x | R i=1 i Part II: Evaluate the integral dx1dx2dx3 . 3 2 R (1+P3 x2) ´ i=1 i Problem 25. In R3 let C be the circle in the xy plane with radius 2 and the origin as center, i.e. C = {x2 + y2 = 4, z = 0}. Let Ω consist of all the points (x, y, z) ∈ R3 whose distance to C is at most 1. Compute: |x|dxdydz ˆΩ Problem 26. Let f :[a, b] → R be a continuously dierentiable function and S be the surface in R3 obtained by revolving the curve y = f(x) around the x axis. Determine the surface area of S.