Calculus 3
Lia Vas
Formulas for Exam 3
You can bring the formula sheets for Exam 1 and Exam 2 to the third exam as well.
−
1. Parametric Surface →
r (u, v) = hx(u, v), y(u, v), z(u, v)i where (u, v) is in a region R.
−
−
- The tangent plane has the normal vector →
ru × →
rv = hxu , yu , zu i × hxv , yv , zv i
RR →
R
R
−
- The surface area is R |−
ru × →
rv |dudv = R |hxu , yu , zu i × hxv , yv , zv i| dudv
2. General substitution for double: x = x(u, v), y = y(u, v). Jacobian determinant
J=
∂x
∂u
∂y
∂u
Then
RR
D
f (x, y) dx dy =
∂x
∂v
∂y
∂v
xu xv =
yu yv f (x(u, v), y(u, v)) |J|du dv
RR
D
3. General substitution for triple: x = x(u, v, w) y = y(u, v, w)
determinant J is
∂x ∂x ∂x x
∂u ∂v ∂w u xv xw ∂y ∂y ∂y J = ∂u ∂v ∂w = yu yv yw ∂z ∂z ∂z zu zv zw ∂u
∂v
∂w
RRR
E
f (x, y, z) dx dy dz =
z = z(u, v, w). Jacobian
f (x(u, v, w), y(u, v, w), z(u, v, w)) |J|du dv dw
RRR
E
4. Triple Integrals
-
R R d(x) R h(x,y)
b
E f (x, y, z) dx dy dz = a
c(x)
g(x,y) f (x, y, z) dz dy
x ≤ b, c(x) ≤ y ≤ d(x), g(x, y) ≤ z ≤ h(x, y) }
RRR
- Cylindrical Coordinates:
- Spherical Coordinates:
dx if E = { (x, y, z) | a ≤
x = r cos θ, y = r sin θ, z = z.
Then x2 + y 2 = r2 and the Jacobian is J = r.
x = r cos θ sin φ, y = r sin θ sin φ, z = r cos φ.
Then x2 + y 2 + z 2 = r2 and the Jacobian J is r2 sin φ.
5. Volume, average value, mass and center of mass.
- The volume. The volume of the solid region E is
V (E) =
Z Z Z
dx dy dz.
E
- The average value of function f (x, y, z) over the solid region E is
1 Z Z Z
fave =
f (x, y, z) dx dy dz
V (E)
E
where V (E) is the volume of the solid region E.
- RIf RaR solid object occupies the region E and has density ρ(x, y, z), Rthen
the mass is m =
RR
1
ρ(x,
y,
z)
dx
dy
dz
and
the
center
of
mass
is
given
by
x
=
E R R R
E x ρ(x, y, z) dx dy dz
m
1
1 R R R
y=m
z=m
E y ρ(x, y, z) dx dy dz
E z ρ(x, y, z) dx dy dz
1
6. Line Integrals.
R
- Integral with respect to the arc length over a curve C is C f (x, y, z)ds. If the curve
C is given by x = x(t), y = y(t), and z = z(t), for a ≤ t ≤ b, then
Z
f (x, y, z)ds =
Z b
C
f (x(t), y(t), z(t))
q
(x0 (t))2 + (y 0 (t))2 + (z 0 (t))2 dt.
a
R
The length of C is L(C) =
C ds =
Rbq
(x0 (t))2 + (y 0 (t))2 + (z 0 (t))2 dt.
a
- Integral with respect to the coordinates over a curve C is
→
− →
f · d−
r =
Z
C
Z
P dx + Qdy + Rdz.
C
If
the curve C isR given by x = x(t), y = y(t), and z = z(t), for a ≤ t ≤ b, then
R
b
0
C P (x, y, z)dx = a P (x(t), y(t), z(t)) x (t)dt and similarly for the terms with dy and dz.
→
−
- The work done by the force f = (P, Q, R) in moving the particle along the curve C is
R
R →
− −
r = C P dx + Qdy + Rdz.
the line integral W = C f · d→
- If a wire C in space has the density
ρ(x, y, z), then the mass
m and the center of
R
1 R
y,
z)
are
given
by
m
=
ρ(x,
y,
z)
ds
x
=
y =
x
ρ(x, y, z) ds
mass
(x,
C
m C
1 R
1 R
y ρ(x, y, z) ds z = m C z ρ(x, y, z) ds.
m C
7. Potential. Independence of Path.
f~ = (P, Q, R) is conservative, if Py = Qx , Pz = Rx , and Qz = Ry . If so, a potential function F
can be found as follows.
(1) Fx =RP ⇒ integrate P with respect to x. Denote the integration constant by g(y, z). Thus
F = P dx + g(y, z).
(2) Fy = Q ⇒ differentiate F with respect to y and equate it to Q. Solve for g(y, z). Denote
the integration constant by h(z).
(3) Fz = R ⇒ differentiate F with respect to z and equate your answer to R. Solve for h(z).
If C is a curve starting at (x1 , y1 , z1 ) and ending (x2 , y2 , z2 ), then
Z
P dx + Qdy + Rdz =
C
Z
f~(t)d~r =
C
Z
C
∇F d~r = F (x2 , y2 , z2 ) − F (x1 , y1 , z1 ).
8. Green’s Theorem
I
P dx + Qdy =
C
Z Z
D
∂Q ∂P
−
∂x
∂y
!
dxdy =
Z Z
D
(Qx − Py )dxdy
∂
9. Gradient, Curl, and Divergence. Gradient operator: ∇ = h ∂x
,
→
−
→
−
→
−
The gradient field of F (x, y, z) is ∇F = Fx i + Fy j + Fz k
Divergence: div f~ = ∇ · f~ = ∂P + ∂Q + ∂R
∂x
Curl: curl f~ = ∇ × f~ =
~i
∂
∂x
P
~j
∂
∂y
Q
∂y
~k ∂ ∂z R ∂
, ∂i
∂y ∂z
=
∂~
i
∂x
∂Q
∂x
−
+
∂~
j
∂y
∂z
=
∂R
∂y
2
−
∂Q
∂z
~i +
∂P
∂z
−
∂R
∂x
~j +
∂P
∂y
~k
+
∂ ~
k.
∂z