lesson 20: integration in polar coordinates
DESCRIPTION
sometimes a region (or a function) is more concisely described in polar coordinates. This can make integrating it a much easier task.TRANSCRIPT
. . . . . .
Section 12.4Integration in Polar Coordinates
Math 21a
March 31, 2008
Announcements
◮ Office hours Tuesday, Wednesday 2–4pm SC 323◮ Problem Sessions: Mon, 8:30; Thur, 7:30; SC 103b
..Image: Flickr user duncan
. . . . . .
Announcements
◮ Office hours Tuesday, Wednesday 2–4pm SC 323◮ Problem Sessions: Mon, 8:30; Thur, 7:30; SC 103b
. . . . . .
Outline
Last Time
Polar geometrySectorsPolar Rectangles
Integration over polar regionsFact of the day
Worksheet
Next Time
. . . . . .
FactIf D is a region of Type I:
D = { (x, y) | a ≤ x ≤ b, g1(x) ≤ y ≤ g2(x) }
Then for any “mostly” continuous function f∫∫D
f(x, y) dA =
∫ b
a
∫ g2(x)
g1(x)f(x, y) dy dx
. . . . . .
FactIf D is a region of Type II:
D = { (x, y) | c ≤ y ≤ d, h1(y) ≤ x ≤ h2(y) }
Then for any “mostly” continuous function f∫∫D
f(x, y) dA =
∫ d
c
∫ h2(y)
h1(y)f(x, y) dx dy
. . . . . .
This time
Can polar coordinates help in computing integrals?
. . . . . .
Outline
Last Time
Polar geometrySectorsPolar Rectangles
Integration over polar regionsFact of the day
Worksheet
Next Time
. . . . . .
Arc length and sector area
s = rθ
A =12
r2θ
(if θ is in radians).
.
.r
.θ
.s
. . . . . .
Area of a polar rectangle
∆A =12
r22 (θ2 − θ1) −12
r21 (θ2 − θ1)
=12
(r22 − r21
)(θ2 − θ1)
=12
(r2 + r1) (r2 − r1) (θ2 − θ1)
= r̄ ∆r ∆θ.
.
..r1
..r2. .θ1
..θ2
.̄r
.∆θ
. . . . . .
Outline
Last Time
Polar geometrySectorsPolar Rectangles
Integration over polar regionsFact of the day
Worksheet
Next Time
. . . . . .
Polar slicing
.
Here the boundaries of x and yare complicated
.
Here the boundary r is afunction of θ
. . . . . .
Fact of the day
FactIf f is continuous on a polar region of the form
D = { (r, θ) | α ≤ θ ≤ β, h1(θ) ≤ r ≤ h2(θ) }
then ∫∫D
f(x, y) dA =
∫ β
α
∫ h2(θ)
h1(θ)f(r cos θ, r sin θ) r dr dθ.
◮ Notice the “area element” is r dr dθ, not dr dθ!
. . . . . .
Fact of the day
FactIf f is continuous on a polar region of the form
D = { (r, θ) | α ≤ θ ≤ β, h1(θ) ≤ r ≤ h2(θ) }
then ∫∫D
f(x, y) dA =
∫ β
α
∫ h2(θ)
h1(θ)f(r cos θ, r sin θ) r dr dθ.
◮ Notice the “area element” is r dr dθ, not dr dθ!
. . . . . .
Example: area of a cardioid
Find the area enclosed by the curve r = 1 + sin θ.
SolutionWe can use the symmetry, findhalf the area, then double:
A = 2∫ π/2
−π/2
∫ 1+sin θ
0r dr dθ
=
∫ π/2
−π/2r2
∣∣∣r=1+sin θ
r=0dθ
=
∫ π/2
−π/2(1 + sin θ)2 dθ
=
∫ π/2
−π/2(1 + 2 sin θ + sin2 θ) dθ
.
.
. . . . . .
Example: area of a cardioid
Find the area enclosed by the curve r = 1 + sin θ.
SolutionWe can use the symmetry, findhalf the area, then double:
A = 2∫ π/2
−π/2
∫ 1+sin θ
0r dr dθ
=
∫ π/2
−π/2r2
∣∣∣r=1+sin θ
r=0dθ
=
∫ π/2
−π/2(1 + sin θ)2 dθ
=
∫ π/2
−π/2(1 + 2 sin θ + sin2 θ) dθ
.
.
. . . . . .
Example: area of a cardioid
Find the area enclosed by the curve r = 1 + sin θ.
SolutionWe can use the symmetry, findhalf the area, then double:
A = 2∫ π/2
−π/2
∫ 1+sin θ
0r dr dθ
=
∫ π/2
−π/2r2
∣∣∣r=1+sin θ
r=0dθ
=
∫ π/2
−π/2(1 + sin θ)2 dθ
=
∫ π/2
−π/2(1 + 2 sin θ + sin2 θ) dθ
.
.
. . . . . .
The computation
∫ π/2
−π/21 dθ = π∫ π/2
−π/22 sin θ dθ = −2 cos θ|θ=π/2
θ=−π/2 = 0∫ π/2
−π/2sin2 θ dθ =
12
∫ π/2
−π/2(1 − cos 2θ) dθ =
12
[θ − 1
2sin 2θ
]θ=π/2
θ=−π/2
=π
2
So
A =3π
2
. . . . . .
ExampleFind the average value of the distance from the origin over the unitdisk.
. . . . . .
Cartesian Solution I
Recall that the average value of a function over a region D is given by
f̄ =1
A(D)
∫∫D
f(x, y) dA
So we are looking for the value of
1π
∫∫D
√x2 + y2 dA,
. . . . . .
Cartesian Solution II
where D is the unit disk x2 + y2 ≤ 1. We can do this Cartesian-like:
1π
∫∫D
√x2 + y2 dA =
1π
∫ 1
−1
∫ √1+x2
−√
1−x2
√x2 + y2 dy dx.
To do the inner integral, we use the trigonometric substitution
y = x tan θ. Then√
x2 + y2 = x sec θ, and dy = x sec2 θ dθ.
. . . . . .
Cartesian Solution III
Moreover, if tan θ =
√1 − x2
x, we have x = cos θ. So
1π
∫ 1
−1
∫ √1+x2
−√
1−x2
√x2 + y2 dy dx =
∫ 1
−1
∫ arccos(x)
− arccos(x)x2 sec3 θ dθ dx.
OK,∫
sec3 θ dθ is one of those nasty trigonometric integrals.
There’s gotta be a better way!
. . . . . .
Polar Solution
Switch to polar coordinates and
1π
∫∫D
r dA =1π
∫ 2π
0
∫ 1
0r r dr dθ
=1π
∫ 2π
0
[r3
3
]r=1
r=0dθ
=1π
∫ 2π
0
13
dθ
=1π
[θ
3
]θ=2π
θ=0=
23.
. . . . . .
Outline
Last Time
Polar geometrySectorsPolar Rectangles
Integration over polar regionsFact of the day
Worksheet
Next Time
. . . . . .
Worksheet #1
ProblemEvaluate
∫∫R
cos(x2 + y2) dA, where R is the region that lies above the
x-axis and within the circle x2 + y2 = 9.
Solution
∫∫R
cos(x2 + y2) dA =
∫ π
0
∫ 3
0cos(r2)r dr dθ
=
∫ π
0
12
sin(r2)
∣∣∣∣r=3
r=0dθ
=
∫ π
0
12
sin(9) dθ =π
2sin(9)
. . . . . .
Worksheet #1
ProblemEvaluate
∫∫R
cos(x2 + y2) dA, where R is the region that lies above the
x-axis and within the circle x2 + y2 = 9.
Solution
∫∫R
cos(x2 + y2) dA =
∫ π
0
∫ 3
0cos(r2)r dr dθ
=
∫ π
0
12
sin(r2)
∣∣∣∣r=3
r=0dθ
=
∫ π
0
12
sin(9) dθ =π
2sin(9)
. . . . . .
Worksheet #2
ProblemEvaluate
∫∫R
arctan(y/x) dA, where
R ={
(x, y)∣∣∣ 1 ≤ x2 + y2 ≤ 4, 0 ≤ y ≤ x
}
Solution
I =
∫ π/4
0
∫ 2
1θr dr dθ
=
(∫ π/4
0θ dθ
)(∫ 2
1r dr
)
=
[θ2
2
]π/4
0
[r2
2
]2
1=
(π/4)2
2· 3
2=
3π2
64
.
.
.1 .2
. . . . . .
Worksheet #2
ProblemEvaluate
∫∫R
arctan(y/x) dA, where
R ={
(x, y)∣∣∣ 1 ≤ x2 + y2 ≤ 4, 0 ≤ y ≤ x
}
Solution
I =
∫ π/4
0
∫ 2
1θr dr dθ
=
(∫ π/4
0θ dθ
)(∫ 2
1r dr
)
=
[θ2
2
]π/4
0
[r2
2
]2
1=
(π/4)2
2· 3
2=
3π2
64
.
.
.1 .2
. . . . . .
Worksheet #2
ProblemEvaluate
∫∫R
arctan(y/x) dA, where
R ={
(x, y)∣∣∣ 1 ≤ x2 + y2 ≤ 4, 0 ≤ y ≤ x
}Solution
I =
∫ π/4
0
∫ 2
1θr dr dθ
=
(∫ π/4
0θ dθ
)(∫ 2
1r dr
)
=
[θ2
2
]π/4
0
[r2
2
]2
1=
(π/4)2
2· 3
2=
3π2
64
.
.
.1 .2
. . . . . .
Worksheet #3
ProblemIn each of the regions R below,decide whether to use polarcoordinates or rectangularcoordinates and write∫∫
R
f(x, y) dA as an iterated
integral, where f is an arbitrarycontinuous function on R.
SolutionRegions 1, 4, 5, and 6 are allgood regions for polarcoordinates.
Exercises
856 M CHAPIIR 12 MUI.TIPLT IilTEGRAI-S
l-6 m A region R is shown. Decide whether to use polar coor_dinates or rectangular coordinates and write lJ* f @, y) dAas an iterated integral, where / is an arbitrary continuous func_
21.
22.
5.
^ ln/2 f4cos|6 . | | rdrd?JO JO
t2'II*'Rwhere
:I!. lJ^ atcrwhere
14. 
. . . . . .
Worksheet #3
ProblemIn each of the regions R below,decide whether to use polarcoordinates or rectangularcoordinates and write∫∫
R
f(x, y) dA as an iterated
integral, where f is an arbitrarycontinuous function on R.
SolutionRegions 1, 4, 5, and 6 are allgood regions for polarcoordinates.
Exercises
856 M CHAPIIR 12 MUI.TIPLT IilTEGRAI-S
l-6 m A region R is shown. Decide whether to use polar coor_dinates or rectangular coordinates and write lJ* f @, y) dAas an iterated integral, where / is an arbitrary continuous func_
21.
22.
5.
^ ln/2 f4cos|6 . | | rdrd?JO JO
t2'II*'Rwhere
:I!. lJ^ atcrwhere
14. 
. . . . . .
Worksheet #4
ProblemSketch the region whose area is given by the integral∫ 2π
π
∫ 7
4r dr dθ
Evaluate this integral.
Solution
A =
∫ 2π
π
∫ 7
4r dr dθ
=
[r2
2
]r=7
r=4[θ]θ=2π
θ=0
=49 − 16
2· π =
33π
2
.
.
.4 .7
. . . . . .
Worksheet #4
ProblemSketch the region whose area is given by the integral∫ 2π
π
∫ 7
4r dr dθ
Evaluate this integral.
Solution
A =
∫ 2π
π
∫ 7
4r dr dθ
=
[r2
2
]r=7
r=4[θ]θ=2π
θ=0
=49 − 16
2· π =
33π
2
.
.
.4 .7
. . . . . .
Worksheet #4
ProblemSketch the region whose area is given by the integral∫ 2π
π
∫ 7
4r dr dθ
Evaluate this integral.
Solution
A =
∫ 2π
π
∫ 7
4r dr dθ
=
[r2
2
]r=7
r=4[θ]θ=2π
θ=0
=49 − 16
2· π =
33π
2
.
.
.4 .7
. . . . . .
Worksheet #5
ProblemUse polar coordinates to find the volume above the cone z =
√x2 + y2
and below the sphere x2 + y2 + z2 = 1.
SolutionConverting to polar coordinates,we get
V =
∫ 2π
0
∫ √2/2
0
(√1 − r2 − r
)r dr dθ
= 2π
∫ √2/2
0
(r√
1 − r2 − r2)
dr
=π
3
(2 −
√2) -0.5
0.0
0.5
-0.50.0
0.5
0.0
0.5
1.0
. . . . . .
Worksheet #5
ProblemUse polar coordinates to find the volume above the cone z =
√x2 + y2
and below the sphere x2 + y2 + z2 = 1.
SolutionConverting to polar coordinates,we get
V =
∫ 2π
0
∫ √2/2
0
(√1 − r2 − r
)r dr dθ
= 2π
∫ √2/2
0
(r√
1 − r2 − r2)
dr
=π
3
(2 −
√2)
-0.5
0.0
0.5
-0.50.0
0.5
0.0
0.5
1.0
. . . . . .
Worksheet #5
ProblemUse polar coordinates to find the volume above the cone z =
√x2 + y2
and below the sphere x2 + y2 + z2 = 1.
SolutionConverting to polar coordinates,we get
V =
∫ 2π
0
∫ √2/2
0
(√1 − r2 − r
)r dr dθ
= 2π
∫ √2/2
0
(r√
1 − r2 − r2)
dr
=π
3
(2 −
√2) -0.5
0.0
0.5
-0.50.0
0.5
0.0
0.5
1.0
. . . . . .
Worksheet #6
ProblemEvaluate ∫ 3
−3
∫ √9−x2
0sin(x2 + y2) dy dx.
SolutionRewrite the integral in polarcoordinates:
I =
∫ π
0
∫ 3
0sin(r2)r dr dθ
=π
2(1 − cos(9)) .
.
.−3 .3
. . . . . .
Worksheet #6
ProblemEvaluate ∫ 3
−3
∫ √9−x2
0sin(x2 + y2) dy dx.
SolutionRewrite the integral in polarcoordinates:
I =
∫ π
0
∫ 3
0sin(r2)r dr dθ
=π
2(1 − cos(9))
.
.
.−3 .3
. . . . . .
Worksheet #6
ProblemEvaluate ∫ 3
−3
∫ √9−x2
0sin(x2 + y2) dy dx.
SolutionRewrite the integral in polarcoordinates:
I =
∫ π
0
∫ 3
0sin(r2)r dr dθ
=π
2(1 − cos(9)) .
.
.−3 .3
. . . . . .
Outline
Last Time
Polar geometrySectorsPolar Rectangles
Integration over polar regionsFact of the day
Worksheet
Next Time
. . . . . .
Next time:Surface area