Midterm Exam 1 Equation Sheet
• Euclidian coordinates: dA = dxdy and dV = dxdydz.
• Polar coordinates: ~r(r, θ) = hr cos(θ), r sin(θ)i, r2 = x2 + y 2 , dA = rdrdθ.
• Cylindrical coordinates: ~r(r, θ, z) = hr cos(θ), r sin(θ), zi, dV = rdrdθdz.
• Spherical coordinates: ~r(r, θ, φ) = hρ sin(φ) cos(θ), ρ sin(φ) sin(θ), ρ cos(φ)i, ρ2 = x2 + y 2 + z 2 , dV =
ρ2 sin(φ)dρdθdφ.
• Surface area of S: Let D be a set in the u-v plane and S = {(x, y, z)| < x, y, z >= ~r(u, v), (u, v) ∈ D}.
ZZ
ZZ
Area(S) =
dS =
||~ru × ~rv ||dudv.
S
D
• Work: If C is a path in 2D parameterized by ~r(t) =< x(t), y(t) > on t ∈ [a, b] and F~ (x, y) is a vector
R
Rb
field, then the work done by the vector field is W = C F~ · d~r = a F~ (x(t), y(t)) · ~r0 (t)dt.
• Conservative vector field. If F~ = hP, Qi is conservative in the simply connected set D then
R
– C F~ · d~r = 0 for all closed paths C.
– Py = Qx .
– Independence of path: A potential function f (x, y) exists such that
paths starting at ~a and ending at ~b contained in D.
R
C
F~ · d~r = f (~b) − f (~a) for all
• Green’s theorem: C is simple closed counterclockwise contour. D is simply connected interior set to
C. If Qx − Py is continuous on D then
ZZ
Z
~
F · d~r =
[Qx − Py ]dA.
W =
D
C
• Surface element: Let ~r(u, v) = hx(u, v), y(u, v), z(u, v)i be the vector function parameterization of the
surface S, where the set D in the u-v plane draws S in x-y-z space. The surface element is
~ndS = ±(~ru × ~rv )dudv
where a ± indicates you must select your desired orientation.
• Curl and divergence: Let ∇ = h∂x , ∂y , ∂z i, and F~ = hP, Q, Ri then Div(F~ ) = ∇ · F~ , and Curl(F~ ) =
∇ × F~ .
• Stokes theorem: C is a simple smooth closed path in 3D. F~ is a smooth vector field. S is a smooth
surface who’s boundary is C
Z
ZZ
W =
F~ · d~r =
(∇ × F~ ) · ~ndS,
C
S
where ~n is the outward normal taken with orientation consistent with the right hand rule with respect
to C.
• Divergence theorem: S is a closed surface enclosing volume V with outward normals ~n. F~ is a smooth
vector field.
ZZ
ZZZ
~
f lux =
F · ~ndS =
(∇ · F~ )dV.
S
V