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vector calculus cheat sheet
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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