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Math 241, Exam 3 Information.
11/28/12, LC 310, 11:15 - 12:05.
Exam 3 will be based on:
• Sections 15.2 - 15.4, 15.6 - 15.8.
• The corresponding assigned homework problems
(see http://www.math.sc.edu/∼boylan/SCCourses/241Fa12/241.html)
At minimum, you need to understand how to do the homework problems.
Topic List (not necessarily comprehensive):
You will need to know: theorems, results, and definitions from class.
§15.3: Double integrals over general regions.
Suppose that f (x, y) is continuous on D ⊆ R2 .
• Type I: A region D is of type I if and only if
D = {(x, y) : a ≤ x ≤ b, g1 (x) ≤ y ≤ g2 (x)}.
We have
Z bZ
ZZ
g2 (x)
f (x, y) dy dx.
f (x, y) dA =
a
D
g1 (x)
• Type II: A region D is of type II if and only if
D = {(x, y) : h1 (y) ≤ x ≤ h2 (y), c ≤ y ≤ d}.
We have
ZZ
Z
d
Z
h2 (y)
f (x, y) dx dy.
f (x, y) dA =
c
D
h1 (y)
Facts: Let D ⊆ R2 be a type I or type II region.
• Volume: Let f (x, y) be continuous on D. Then the volume of the region above D and
below z = f (x, y) ≥ 0 is
ZZ
f (x, y) dA.
D
• Area: The area of D is
ZZ
dA.
D
§15.4: Double integrals in polar coordinates.
Polar coordinates: P = (r, θ).
• r ≥ 0: the distance from the origin (pole), O, to the point P .
• 0 ≤ θ ≤ 2π: the angle, measured counter-clockwise, that the ray OP makes with the
positive x-axis.
Transformation formulas:
• (r, θ) 7→ (x, y): x = r cos θ; y = r sin θ.
p
• (x, y) 7→ (r, θ): r = x2 + y 2 , y/x = tan θ (the value of 0 ≤ θ ≤ 2π depends on
the quadrant in which (x, y) lies).
Fact: Suppose that
D = {(x, y) : a ≤ x ≤ b, g1 (x) ≤ y ≤ g2 (x)} = {(r, θ) : α ≤ θ ≤ β, r1 (θ) ≤ r ≤ r2 (θ)},
and that f (x, y) is continuous on D. Then we have
Z bZ
g2 (x)
g1 (x)
β
Z
r2 (θ)
f (r cos θ, r sin θ)r dr dθ.
f (x, y) dA =
f (x, y) dy dx =
a
Z
ZZ
α
D
r1 (θ)
Note: We have dy dx = dA = r dr dθ. The extra factor r arises as the absolute value of the
Jacobian of the change of coordinates
x(r, θ) = r cos θ,
y(r, θ) = r sin θ.
The Jacobian is
∂(x, y) ∂x/∂r ∂x/∂θ cos θ −r sin θ
=
=
= r(cos2 θ + sin2 θ) = r.
∂y/∂r ∂y/∂θ sin θ r cos θ ∂(r, θ)
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§15.6: Triple integrals.
Definition: Let E be a simple solid in R3 :
E = {(x, y, z) : a ≤ x ≤ b, u1 (x) ≤ y ≤ u2 (x), v1 (x, y) ≤ z ≤ v2 (x, y)},
and suppose that f (x, y, z) is continuous on E. Then we have
Z bZ
ZZZ
u2 (x)
Z
v2 (x,y)
f (x, y, z) dz dy dx.
f (x, y, z) dV =
u1 (x)
a
E
v1 (x,y)
Note: In rectangular coordinates (x, y, z), there are six possible integration orders. Care
must be taken when switching the order unless all limits of integration are constant.
Possible strategy: Once we pick an order, say dV = dz dy dx, then we may consider
working in two steps.
1. We first try to identify E; at least we try to identify equations for the “top” of E:
z = v2 (x, y), and the “bottom” of E: z = v1 (x, y). We find that
#
ZZ
ZZZ
Z Z "Z
v2 (x,y)
E
F (x, y) dA,
f (x, y, z) dz dA =
f (x, y, z) dV =
D
v1 (x,y)
D
where D is the projection (or “shadow”) of E on the xy-plane.
2. Second, we evaluate the double integral (as in Sections §15.2 - 15.4)
ZZ
F (x, y) dA,
D
which may require a quick sketch of D in the xy-plane.
Fact: Suppose that E ⊆ R3 is a solid which can be contained in a finite box. The volume
of E as a triple integral is
ZZZ
Vol(E) =
dV.
E
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§15.7: Triple integrals in cylindrical coordinates.
Cylindrical coordinates: P = (r, θ, z). (Let Q = (x, y, 0) be the projection of P in
rectangular coordinates on the xy-plane.
• r ≥ 0: the distance from the origin (pole), O, to the projection Q.
• 0 ≤ θ ≤ 2π: the angle, measured counter-clockwise, that the ray OQ makes with the
positive x-axis.
• z ∈ R: the height of the point P above the xy-plane.
Transformation formulas:
• (r, θ, z) 7→ (x, y, z): x = r cos θ; y = r sin θ; z = z.
p
• (x, y, z) 7→ (r, θ, z): r = x2 + y 2 ; y/x = tan θ (the value of 0 ≤ θ ≤ 2π
depends on the quadrant in which (x, y) lies); z = z.
Fact: Suppose that
E = {(r, θ, z) : α ≤ θ ≤ β, r1 (θ) ≤ r ≤ r2 (θ), v1 (r, θ) ≤ z ≤ v2 (r, θ)},
and that f (x, y, z) is continuous on E. Then we have
ZZZ
Z β Z r2 (θ) Z v2 (r,θ)
f (x, y, z) dV =
f (r cos θ, r sin θ, z)r dz dr dθ.
α
E
r1 (θ)
v1 (r,θ)
Note: We have dV = r dz dr dθ. The extra factor r arises as the absolute value of the
Jacobian, just as it does for polar coordinates.
∂(x, y, z) ∂(r, θ, z) = r
Possible strategy: As for triple integrals in rectangular ((x, y, z)) coordinates, we work in
two steps.
1. We try to identify equations for the “top” of E: z = v2 (r, θ), and the “bottom” of E:
z = v1 (r, θ). We find that
#
ZZZ
Z Z "Z v2 (r,θ)
ZZ
f (x, y, z) dV =
f (r cos θ, r sin θ, z) dz dA =
F (r, θ) dA.
E
D
v1 (r,θ)
D
2. We view the projection (or “shadow”) of E in the xy-plane as a polar region:
D = {(r, θ) : α ≤ θ ≤ β, r1 (θ) ≤ r ≤ r2 (θ)},
and we compute the double integral in polar coordinates:
ZZ
Z β Z r2 (θ)
F (r, θ) dA =
F (r, θ)r dr dθ.
α
D
4
r1 (θ)
§15.8: Triple integrals in spherical coordinates.
Spherical coordinates: P = (ρ, θ, φ). (Let Q = (x, y, 0) be the projection of P in
rectangular coordinates on the xy-plane.)
• ρ ≥ 0: the distance from the origin, O, to the point P .
• 0 ≤ θ ≤ 2π: the angle, measured counter-clockwise, that the ray OQ makes with the
positive x-axis.
• 0 ≤ φ ≤ π: the angle that the ray OP makes with the positive z-axis.
Transformation formulas:
• (ρ, θ, φ) 7→ (x, y, z): x = ρ sin φ cos θ; y = ρ sin φ sin θ; z = ρ cos φ.
p
• (x, y, z) 7→ (ρ, θ, φ): ρ = x2 + y 2 + z 2 ; θ : tan θ = y/x (the value of θ depends
on the quadrant in which (x, y) lies); φ : tan φ = (x2 + y 2 )1/2 /z (the value of φ
depends the sign of z).
Fact: Suppose that
E = {(ρ, θ, φ) : α ≤ θ ≤ β, φ1 (θ) ≤ φ ≤ φ2 (θ), ρ1 (θ, φ) ≤ ρ ≤ ρ2 (θ, φ)},
and that f (x, y, z) is continuous on E. Then we have
ZZZ
Z
β
Z
φ2 (θ)
Z
r2 (θ,φ)
f (x, y, z) dV =
α
E
φ1 (θ)
f (ρ sin φ cos θ, ρ sin φ sin θ, ρ cos φ)ρ2 sin φ dρ dφ dθ.
r1 (θ,φ)
Note: We have dV = ρ2 sin φ dρ dφ dθ. The extra factor ρ2 sin φ arises as the absolute value
of the Jacobian of the change of coordinates
x(ρ, θ, φ) = ρ sin φ cos θ, y(ρ, θ, φ) = ρ sin φ sin θ, z(ρ, θ, φ) = ρ cos φ,
given by
∂x/∂ρ ∂x/∂θ ∂x/∂φ
∂(x, y, z) = ∂y/∂ρ ∂y/∂θ ∂y/∂φ = −ρ2 sin φ.
∂(ρ, θ, φ) ∂z/∂ρ ∂z/∂θ ∂z/∂φ
Possible strategy: We sketch the solid E, just as when we work in other coordinate systems.
We use this sketch to determine all limits. The notion of a projection (or “shadow”) of E
does not make sense in this system.
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§15.9 Changes of variable for multiple integrals.
Definition: Suppose that T : R2 → R2 is a transformation from the uv-plane to the xy-plane
given by T (u, v) = (x(u, v), y(u, v)). The Jacobian of T is
∂x ∂x ∂(x, y) ∂u ∂v ∂x ∂y ∂x ∂y
J(T ) =
=
=
−
.
∂(u, v) ∂y ∂y ∂u ∂v ∂v ∂u
∂u ∂v Theorem: Assume the following hypotheses:
1. T : R2 → R2 given by T (u, v) = (x(u, v), y(u, v)) is C 1 (x(u, v) and y(u, v) have
continuous partials).
2. S is an elementary region in the uv-plane, and R is an elementary region in the xyplane with T (S) = R (the image of S under T is R).
3. T is one-to-one on S except possibly on the boundary of S.
4. J(T ) 6= 0 on S.
We have
ZZ
ZZ
f (x, y) dA =
R
∂(x, y) du dv.
f (x(u, v), y(u, v)) ∂(u, v) S
Note: The theorem suitably generalizes to n ≥ 3 variables.
Examples:
1. Polar. T (r, θ) = (r cos θ, r sin θ) =⇒ J(T ) = r.
2. Cylinderical. T (r, θ, z) = (r cos θ, r sin θ, z) =⇒ J(T ) = r.
3. Spherical. T (ρ, θ, φ) = (ρ sin φ cos θ, ρ sin φ sin θ, ρ cos φ) =⇒ J(T ) = −ρ2 sin φ.
Notes: Suppose that T (u, v) = (x(u, v), y(u, v).
1. If x(u, v), y(u, v) are linear, then T maps lines to lines, parallelograms to parallelograms, vertices to vertices...
2. If T : S → R is one-to-one, then T −1 : R → S exists, and
J(T −1 ) =
∂(u, v)
1
1
=
=
.
∂(x, y)
∂(x, y)/∂(u, v)
J(T )
Alternatively, we have J(T ) = 1/J(T −1 ). This is useful for computations when one is
given T −1 (x, y) = (u(x, y), v(x, y)) rather than T (u, v) = (x(u, v), y(u, v)).
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