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Math 241 Chapter 14

Dr
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Wyss-Gallifent

§14
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These can be defined via a Riemann Sum method like in Calculus I but the net result is: We can
define the double integral of f (x, y) over R, denoted R f (x, y) dA to be the signed volume under
the graph of f (x, y) within the region R
...
First
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Defn: An iterated integral is a nested integral
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We evaluate these by working from the inside out, making
sure we integrate with respect to the correct variable each time
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Now then, onto evaluation of

R

f (x, y) dA
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y = top(x) on the interval a ≤ x ≤ b
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In this case
f (x, y) dA = c lef t f (x, y) dx dy
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4
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§14
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Reminder about how polar coordinates work
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Don’t forget x = r cos θ, y = r sin θ and x2 + y 2 = r2
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We describe a region in polar coordinates from the point of view of a person who lives at the
origin
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In this case
f (x, y) dA = α near f (r cos θ, r sin θ) r dr dθ
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It might help to remember it’s the “Jacobian r”
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We can reparametrize (to polar) to do an impossible integral like

√
1−x2
1
−1 0

sin(x2 + y 2 ) dy dx
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4 Triple Integrals
1
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Instead suppose D is a solid object in space and at any point
f (x, y, z) is the density around that point
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The question is how to evaluate which all depends upon how to best describe D
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We have the following:
(a) If D is the solid between the graphs of z = low(x, y) and z = high(x, y) above the region R
b top high
in the xy-plane and if R is VS then
f (x, y, z) dV = a bot low f (x, y, z) dz dy dx
...

D

§14
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Cylindrical coordinates are just polar coordinates plus z
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For example:
(a) r = 2 is a cylinder, as are r = 3 cos θ and r = 2 sin θ
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(c) The cone z =

x2 + y 2 becomes z = r
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2
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D

§14
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Describe how spherical coordinates work and make sure to mention the conversions:
(a) x = ρ sin φ cos θ
(b) y = ρ sin φ sin θ
(c) z = ρ cos φ
(d) x2 + y 2 + z 2 = ρ2
(e) x2 + y 2 = ρ2 sin2 φ
2
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For example:
(a) ρ = 2 is a sphere
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(c) φ =

π
4

is a cone
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3
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In that window we have a near
function ρ = near(φ, θ) and a far function ρ = f ar(φ, θ)
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If D is described this way then
β δ f ar
f (x, y, z) dV = α γ near f (ρ sin φ cos θ, ρ sin φ sin θ, ρ cos φ) ρ2 sin φ dρ dφ dθ
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It’s the “Jacobian” again
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8 Change of Variables in Double Integrals
1
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In calc II when we did a trig sub like x = sin u
we had to make sure that dx got replaced by cos u du and we have to do the same sort of thing
here
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Method: If we substitute x = f (u, v) and y = g(u, v) then three things happen:
(a) The region R in the xy plane changes to a new region S in the uv plane
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(c) dA gets replaced by |Jac| dA where Jac is the Jacobian and is defined by
∂x/∂u ∂x/∂v
Jac =
∂y/∂u ∂y/∂v
3
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We first rewrite the bounds as y − x = 1, y − x = 4, 2x + y = 4 and 2x + y = 10
and then put u = y − x and v = 2x + y
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In order to find Jac though we need to solve for x and y
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3
(b)

x dA for R the elliptical disk
R

x2
9

+

y2
16

≤ 4
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We use x = 3u and y = 4v to find Jac and go from there
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We use u = r cos θ and v = r sin θ
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We first rewrite

(c)
R

the bounds as y/x = 1, y/x = 3, xy = 1 and xy = 5 and then put u = y/x and v = xy
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I like this example because Jac has a variable
in it whereas the other examples have a constant Jac
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9 Parametrized Surfaces
1
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Each value of t gives a
¯
vector which points from the origin to a point on the curve
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We’ll do r(u, v) = x(u, v) ˆ + y(u, v)  + z(u, v) k for some u and v
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Note that the use of u and v is
arbitrary and generic and mostly we’ll see r(x, y), r(r, θ), r(z, θ) and other familiar letters
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The best way to see how this works is to look at a bunch of examples
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(a) The plane z = 3 with 0 ≤ x ≤ 3 and 0 ≤ y ≤ 4
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We have
ˆ
r(x, y) = x ˆ + y  + 3 k with 0 ≤ x ≤ 3 and 0 ≤ y ≤ 4
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This is a little disk
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r cos θ ˆ + r sin θ  + 3 k
ı
ˆ
(c) The piece of x = −2 inside the cylinder y 2 + z 2 = 9
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We use r(r, θ) = −2 ˆ+r cos θ  +r sin θ k for 0 ≤ r ≤ 3 and 0 ≤ θ ≤ 2π
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We use r(θ, z) = 3 cos θ ˆ + 3 sin θ  + z k
¯
ı
ˆ
with 0 ≤ θ ≤ 2π and 0 ≤ z ≤ 7
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We base this off spherical
ˆ
and use r(φ, θ) = 4 sin φ cos θ ˆ + 4 sin φ sin θ  + 4 cos φ k with 0 ≤ θ ≤ 2π and 0 ≤ φ ≤ π/6
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We use r(x, z) = x ˆ+ 12−x−3z  +z k with 0 ≤ x ≤ 4
¯
ı
2
and 0 ≤ z ≤ 2
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For now these are good though

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