Vector CalculusChange of Variables

G(u, v) = (x(u, v), y(u, v))

Jac(G) =

∣∣∣∣∣ ∂x∂u ∂x∂v

∂y∂u

∂y∂v

∣∣∣∣∣ =∂(x, y)

∂(u, v)=∂x

∂u

∂y

∂v−∂x

∂v

∂y

∂u

For F = G−1, Jac(G) = Jac(F )−1.

CoV Formula: If G : D0 → D is injective and both

x and y have continuous partial derivatives then for

continuous f :∫∫Df(x, y)dxdy =

∫∫D0

f(x(u, v), y(u, v)) |Jac(G)| dudv

dxdy = |Jac(G)| dudvIf the domain D is small then for a point P :

Area(G(D)) ≈ |Jac(G)(P )|Area(D)

In three dimensions:

G(u, v, w) = (x(u, v, w), y(u, v, w), z(u, v, w))

Jac(G) =

∣∣∣∣∣∣∣∂x∂u

∂x∂v

∂x∂w

∂y∂u

∂y∂v

∂y∂w

∂z∂u

∂z∂v

∂z∂w

∣∣∣∣∣∣∣ =∂(x, y, z)

∂(u, v, w)

Jac(G) =∂x

∂u

(∂y

∂v

∂z

∂w−∂y

∂w

∂z

∂v

)−∂x

∂v

(∂y

∂u

∂z

∂v−∂y

∂v

∂z

∂u

)+∂x

∂w

(∂y

∂u

∂z

∂v−∂y

∂v

∂z

∂u

)dxdydz = |Jac(G)| dudvdw

Line Integrals

Scalar line integral:∫Cf(x, y, z)ds =

∫ b

af(~c(t))‖~c ′(t)‖dt

Vector line integral:∫C

~F · d~s =

∫C

(~F · ~T

)ds =

∫ b

a

~F (~c(t)) · ~c ′(t)dt

Work exerted on an object: W =

∫C

~F · d~s

Work against a force: W = −∫C

~F · d~s

Vector differential in vector line integrand.

Arc length differential in scalar line integrand.

Conservative Vector Fields

Path-independent vector field:∫~c1

~F · d~s =

∫~c2

~F · d~s

for any two paths ~c1 and ~c2 in D from P to Q.

If ~F = ∇V ,

∫~c

~F · ~s = V (Q)− V (P ).

If ~c is a closed path then

∮~c

~F · d~s = 0.

Test For Conservative Vector Field

In two dimensions:

∂F1

∂y=∂F2

∂x

In three dimensions:

∂F1

∂y=∂F2

∂x,∂F2

∂z=∂F3

∂y,∂F3

∂x=∂F1

∂z

Surface Integrals

G(u, v) = (x(u, v), y(u, v), z(u, v))

~Tu =∂G

∂u=

⟨∂x

∂u,∂y

∂u,∂z

∂u

⟩~Tv =

∂G

∂v=

⟨∂x

∂v,∂y

∂v,∂z

∂v

⟩~n(u, v) = ~Tu × ~Tv

Area(S) ≈ ‖~n(u0, v0)‖AreaD

Area(S) =

∫∫D‖~n(u, v)‖dudv∫∫

Sf(x, y, z)dS =

∫∫Df(G(u, v))‖~n(u, v)‖dudv

Cylinder: G(θ, z) = (R cos(θ), R sin(θ), z)

Sphere: G(θ, φ) = (R cos(θ) sin(φ), R sin(θ) sin(φ), R cos(φ))∫∫S

~F · d~S =

∫∫S

(~F · ~en

)dS ≈

n∑i=1

(~F · ~en)i

n∫∫S

~F · d~S =

∫∫D

~F (G(u, v)) · ~n(u, v)dudv

~F~n =~F (G(u, v)) · ~n(u, v)

‖~n(u, v)‖

~er =⟨xr,y

r,z

r

⟩= 〈cos(θ) sin(φ), sin(θ) sin(φ), cos(φ)〉

Green’s Theorem

curlz(~F ) =∂F2

∂x−∂F1

∂y

If ∂D denotes the boundary of D with its boundary

orientation, then∮∂D

~F · d~s =

∮∂D1

~F · d~s+ . . .+

∮∂Dn

~F · d~s∮∂D

~F · d~s =

∫∫D

(∂F2

∂x−∂F1

∂y

)dA∮

∂DF1dx+ F2dy =

∫∫D

(∂F2

∂x−∂F1

∂y

)dA.

For a region D within a closed curve C

Area(D) =1

2

∮Cxdy − ydx.

For a sufficiently small D with boundary C and P ∈ D∮CF1dx+ F2dy ≈ curlz(~F )(P )Area(D).

∆ =∂2

∂x2+

∂2

∂y2; ~F is harmonic if ∆~F = 0.

Stokes’ Theorem~F = 〈F1, F2, F3〉

curl(~F ) = ∇× ~F =

⟨∂F3

∂y−∂F2

∂z,∂F1

∂z−∂F3

∂x,∂F2

∂x−∂F1

∂y

⟩∮∂S

~F · d~s =

∫∫S

curl(~F ) · d~S

If ~F = curl( ~A) then the flux of ~F through S is given by∫∫S

~F · d~S =

∮∂S

~A · d~s.

For a simple closed curve C around a point P in a plane

through the point P with an enclosed region D and unit

normal vector ~en then∮C

~F · d~s ≈(

curl(~F )(P ) · ~en)

Area(D).

Divergence Theorem

div(~F ) = ∇ · ~F =∂F1

∂x+∂F2

∂y+∂F3

∂z∫∫S

~F · d~S =

∫∫∫W

div(~F )dV

Flow rate across a surface S enclosing a region W is∫∫∫W

div(~F )dV ≈ div(~F )(P )Vol(W ) for small S.

Divergence Theorem

div(~F )(P ) > 0 =⇒ net outflow creation of fluid near P

div(~F )(P ) < 0 =⇒ net inflow destruction of fluid near P

div(~F )(P ) < 0 =⇒ source density of the field

If div(~F ) = 0 everywhere, ~F is incompressible.

Inverse-square vector field ~Fi-sq =~er

r2.∫∫

S

~Fi-sq · d~S =

{4π if S encloses the region0 if S does not enclose the region∫∫

S

~E · d~S =total charge enclosed by S

ε0

Uniformly charged sphere with total charge Q, radius R,

and distance from electric field to origin r :

~E =

{Q

4πε0r2if r > R

0 if r < R

curl(∇(f)) = ~0 and div(curl(~F )) = 0.

Maxwell Equations

div( ~E) = 0

div( ~B) = 0

curl( ~E) = −∂ ~B

∂t

curl( ~B) = µ0ε0∂ ~E

∂t

Wave equation: ∆ϕ =1

c2∂2ϕ

∂t2; ∆ϕ =

∂2ϕ

∂x2+∂2ϕ

∂y2+∂2ϕ

∂z2

∆ ~E = µ0ε0∂2 ~E

∂t2

Identities

For a sphere centered at the origin:

~n = R2 sin(φ) 〈cos(θ) sin(φ), sin(θ) sin(φ), cos(φ)〉 .Directional derivative of unit normal vector:

D~en = d~S.

Quadric Surfaces

Ellipsoid:

(x− ap

)2

+

(y − bq

)2

+

(z − cr

)2

= 1

Hyperboloid 1:

(x− ap

)2

+

(y − bq

)2

−(z − cr

)2

= 1

Hyperboloid 2: −(x− ap

)2

−(y − bq

)2

+

(z − cr

)2

= 1

Elliptic Cone:

(x− ap

)2

+

(y − bq

)2

−(z − cr

)2

= 0

E Paraboloid:

(x− ap

)2

+

(y − bq

)2

−(z − cr

)= 0

H Paraboloid: −(x− ap

)2

+

(y − bq

)2

−(z − cr

)= 0

Constants a, b, c represent the center.

Obtain the intercepts for variables by setting others to zero.

(uv

)′ =v dudx− u dv

dx

v2(fg)′(x) = f ′(x)g(x) + g′(x)f(x)

(f ◦ g)′(x) = f ′(g(x))g′(x)(sin(x))′ = cos(x)(cos(x))′ = − sin(x)(tan(x))′ = sec2(x)(csc(x))′ = csc(x) cot(x)(sec(x))′ = sec(x) tan(x)(cot(x))′ = − csc2(x)

(sin−1(x))′ =1

√1− x2

(cos−1(x))′ = −1

√1− x2

(tan−1(x))′ =1

1 + x2

(sec−1(x))′ =1

|x|√x2 − 1

(csc−1(x))′ = −1

|x|√x2 − 1

(cot−1(x))′ = −1

1 + x2

Trigonometric Identities

sin2(θ) + cos2(θ) = 1 sin(−θ) = − sin(θ)

1 + tan2(θ) = sec2(θ) cos(−θ) = cos(θ)

1 + cot2(θ) = csc2(θ) tan(−θ) = − tan(θ)

sin(x+ y) = sin(x) cos(y) + cos(x) sin(y)

sin(x− y) = sin(x) cos(y)− cos(x) sin(y)

cos(x+ y) = cos(x) cos(y)− sin(x) sin(y)

cos(x− y) = cos(x) cos(y) + sin(x) sin(y)

2 sin(x) cos(y) = sin(x+ y) + sin(x+ y)

2 sin(x) sin(y) = cos(x− y)− cos(x+ y)

2 cos(x) cos(y) = cos(x− y) + cos(x+ y)

2 cos(x) sin(y) = sin(x+ y)− sin(x− y)

sin(2x) = 2 sin(x) cos(x) sin2(x) =1− cos(2x)

2

tan(2x) =2 tan(x)

1− tan2(x)cos2(x) =

1 + cos(2x)

2∫xndx =

xn+1

n+ 1+ c∫

sin(x)dx = − cos(x)∫cos(x)dx = sin(x)∫tan(x)dx = ln |sec(x)|∫sec(x)dx = ln |sec(x) + tan(x)|∫csc(x)dx = ln |csc(x)− cot(x)|∫cot(x)dx = ln |sin(x)|∫

csc2(x)dx = − cot(x)∫sinn(x)dx = −

sinn−1(x) cos(x)

n+n− 1

n

∫sinn−2(x)dx∫

cosn(x)dx =cosn−1(x) sin(x)

n+n− 1

n

∫cosn−2(x)dx∫

tann(x)dx =tanm−1(x)

m− 1−∫

tanm−2(x)dx∫sinm(x) cosn(x)dx =

sinm+1(x) cosn−1(x)

m+ n

+n− 1

m+ n

∫sinm(x) cosn−2(x)dx∫

secn(x)dx =tan(x) secn−2(x)

n− 1+n− 2

n− 1

∫secn−2(x)dx∫

cscn(x)dx =cot(x) cscn−2(x)

n− 1−n− 2

n− 1

∫cscn−2(x)dx∫

cotn(x)dx =− cotn−1(x)

n− 1−∫

cotn−2(x)dx