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Vector Calc Stuff - Integrals, Grad, Div, Curl VECTOR CALC STUFF - INTEGRALS, GRAD, DIV, CURL 1. Theory Note 1. The hardest part (for me at least) about this vector calculus 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 Multivariable Calculus 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 function 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 derivatives , @f , @f and @f . @x @y @z Fact 3. (Chain Rule and Gradients) 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: 2 @x @x @x 3 @u @v @w @(x; y; z) @y @y @y := 4 @u @v @w 5 @(u; v; w) @z @z @z @u @v @w Suppose f is a function f : R3 ! R. Dene: h @f @f @f i r(x;y;z)f = @x @y @z (Think of as @f )Then we have the following chain rule: r(x;y;z)f @(x;y;z) @(x; y; z) r f = r f (u;v;w) (x;y;z) @(u; v; w) 2 @x @x @x 3 @u @v @w @f @f @f h @f @f @f i @y @y @y i:e: @u @v @w = @x @y @z 4 @u @v @w 5 @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 Integral 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 gradient rf is also useful for the following fact: If we look at the level surface S = f(x; y; z): f(x; y; z) = cg = f −1 fcg the rf is perpendicular to the the surface S. The direction rf is the direction of most increase of the function f. You can also easily check that it obey the product rule r (fg) = (rf) g + f (rg) Fact 6. (Change of variables 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 2 (0; 1), θ 2 (0; π) and φ 2 (0; 2π) that @(x;y;z) = r2 sin θ. @(r,θ,φ) 2 p Exercise 8. Show that 1 −x . How can changing variables help? −∞ e dx = π ´ Denition 9. (Volumes and Volume Integrals). Sometimes, a problem will say V = fsome volume regiong 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 = f~r(x; y; z):(x; y; z) 2 Rg 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 = f~r(s; t):(s; t) 2 Rg 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 n^ = @s @t=j @s × @t j 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~ · n^ · 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 = fx; y; f(x; y)g for a positive scalar function f > 0. Show that the area element of such a surface is dA = p1 + jrfj2dxdy 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 = f~r(t): t 2 [a; b]g. 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: (rf) · d~l = f (γ(1)) − f (γ(0)) ˆ γ VECTOR CALC STUFF - INTEGRALS, GRAD, DIV, CURL 3 Denition 14. (Divergence 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)]r · G~ = 1 + 2 + 3 @x @y @z @G @G @G @G @G @G [Curl (vector)]r × 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~ = r × G~ . Then r · F~ ≡ 0. Short form: The curl of G~ is divergence-free : r · r × G~ ≡ 0. Under some conditions, the converse is also true: If certain mild conditions hold, and r · F~ ≡ 0, then there exists G~ so that F~ = r × 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~ = rφ be its gradient eld. Then r × G~ ≡ 0. Short form: Gradient elds are curl-free : r×(rφ) ≡ 0. The converse holds if we are on a simply connected domain: If r×G~ ≡ 0 on a simply connected domain, then there exists φ so G~ = rφ. (3) Let ~ be a vector eld. Let be a curve in 3 going from to . Consider the line integral ~ 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 r × F ≡ 0 γ F · l v1 v2 , , the integral´ ~ d~ depends only on the endpoints and , then ~ . v1 v2 γ F · l v1 v2 r × F ≡ 0 ´ Theorem 16. (Divergence Theorem) Suppose we have some volume V whose boundary is a surface S.
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