Transcript Double Integrals in Polar Coordinates

Double Integrals in Polar
Coordinates
Double Integrals in Polar Coordinates
Recall: Polar Coordinates:
𝑥 = 𝑟 cos 𝜃
𝑟2 = 𝑥2 + 𝑦2
𝑦 = 𝑟 sin 𝜃
tan 𝜃 =
𝑦
𝑥
Sometimes equations and regions are expressed more simply in polar rather
than rectangular coordinates.
For instance, the circle 𝑥 2 + 𝑦 2 = 5 can be expressed in polar
coordinates as 𝑟 = 5.
Double Integrals in Polar Coordinates
The basic region in polar coordinates is the “Polar Rectangle”:
𝑅 = (𝑟, 𝜃) 𝑎 ≤ 𝑟 ≤ 𝑏, 𝛼 ≤ 𝜃 ≤ 𝛽
where 𝛽 − 𝛼 ≤ 2𝜋
Goal: Compute the double integral  f ( x , y ) dA where R is a polar rectangle.
R
Divide the interval [𝑎, 𝑏] into m subinterval [𝑟𝑖 , 𝑟𝑖+1 ] with length ∆𝑟
Divide the interval [𝛼, 𝛽] into n subinterval [𝜃𝑗 , 𝜃𝑗+1 ] with length ∆𝜃
The circles 𝑟 = 𝑟𝑖 and the rays 𝜃 = 𝜃𝑗 divide R into small polar rectangles
of area ∆𝐴𝑖𝑗
Double Integrals in Polar Coordinates
Let  ri , j  be the center of the polar rectangle 𝑅𝑖𝑗.


1
2
Recall that the area of a sector of a circle of radius r and central angle 𝜃 is 𝑟 2 𝜃
We will compute the area ∆𝐴𝑖𝑗 by subtracting the area of the sector with
central angle ∆𝜃 and radius 𝑟𝑖 from the area of the sector with the same
central angle and radius 𝑟𝑖+1 .
2
2
1
1
 Aij   ri 1      ri   
2


2
1 2
ri 1  ri2   

2
ri 1  ri
 ri 1  ri   
2
 ri  r  
The rectangular coordinates of the center of 𝑅𝑖𝑗 are
( ri cos  j , ri sin  j )
Therefore a typical Riemann Sum will be
m
n

i 1 j 1
f ( ri cos  j , ri sin  j )  Ai j 
m
n

i 1 j 1
f ( ri cos  j , ri sin  j ) ri  r  
Double Integrals in Polar Coordinates
Taking the limit as m and n approach infinity, gives the following important
result:
Change to Polar Coordinates in a Double Integral:
If f is continuous on a polar rectangle 𝑅 = (𝑟, 𝜃) 𝑎 ≤ 𝑟 ≤ 𝑏, 𝛼 ≤ 𝜃 ≤ 𝛽
with 𝛽 − 𝛼 ≤ 2𝜋, then

f ( x , y )d A 
R

b
  a
f ( r cos  , r sin  ) r d rd 
The formula above says that, in order to change the integral from x, y to r, θ
we need to substitute 𝑥 = 𝑟 cos 𝜃, 𝑦 = 𝑟 sin 𝜃 and 𝑑𝐴 = 𝑟𝑑𝜃𝑑𝑟, and use the
appropriate limits of integration for r and θ.
Do not forget the additional factor r in the formula.
 1d A
Note 1: A rea of R =
R
rectan gu lar

 rd rd 
R
p olar
Note 2: The limits on r may depend on θ:

R
f ( x , y )d A 

g ( )
  g1 ( )
2
f ( r cos  , r sin  ) r d rd 
Double Integrals in Polar Coordinates- Example 1
Sketch the region of integration and evaluate the integral by changing
to polar coordinates.
2
4 x2
  2 0
1
dydx
x2  y2  2
The region is bounded below by y = 0 (the x-axis) and above by 𝑦 =
4 − 𝑥 2 (the top half of a circle of radius 2)
𝑟=2
𝜃=0
𝜃=𝜋
𝑟=0
2 
0 0
1
rd  d r 
2
r  2 dA
2
0
In polar coordinates, the region is
bounded by
0r2
0 
Since 𝑥 2 + 𝑦 2 = 𝑟 2 , the integral in
polar coordinates is
r  
 dr  
2
  0

r 2
2
0
2
r

2

d r  ln( r  2)
2
 0
2
r 2


2
 ln(6)  ln(2)   0.55
Double Integrals in Polar Coordinates - Example 2
Use polar coordinates to find the volume of the solid below 𝑧 = 9 and above
𝑧 = 𝑥 2 + 𝑦 2 in the first octant.
The two surfaces intersect
along a curve C:
0 x3
x2  y2  9
This quarter of a circle of radius 3
is the projection of C on the xy
plane and it is also the boundary
of the region of integration R.
In polar coordinates the
region is described by:
0  r  3, 0   
3  /2
V 
 (
R
9
upper
surface
 ( x2  y2 )
low er
surface
)
dA 
  9  r 
2
0



𝜃=
2
rd  dr
0
 9 r  r  dr
2 0
3
3

81

8
𝜋
2
𝜃=0
𝑟=3
Double Integrals in Polar Coordinates – Example 3
Find the volume inside the sphere 𝑥 2 + 𝑦 2 + 𝑧 2 = 9 and outside the
cylinder 𝑥 2 + 𝑦 2 = 4.
The region R is bounded by the circles
𝑥 2 + 𝑦 2 = 9 and 𝑥 2 + 𝑦 2 = 4.
In polar coordinates the
circles have equation
𝑟 = 2 and 𝑟 = 3.
By symmetry we can
compute 8 times the
volume in the first
octant.
Our region of
integration is then:
The surface is bounded above by
z 
9  x  y
2
2

V  8  9  ( x 2  y 2 ) d A  8

R
8

3
2 2
9r
2
rdr   2
3
2
0
5
 /2
0
2  3/2  5 2 0
u d u  2  u  
5   1 5


0
3
3
( u  9  r , du   2 rdr )
2
9  r 2 rd  dr
Double Integrals over General Regions Example 4
Calculate the volume bounded by 𝑧 = 𝑥 2 + 𝑦 2 and 𝑧 = 18 − 𝑥 2 − 𝑦 2 .
The two surfaces intersect along a curve C.
18  x 2  y 2  x 2  y 2

x2  y2  9
The circle of radius 3 is the boundary of
the region of integration R.
0  r  3,
V 
 ( ( 1 8  x
R

2
)  ( x 2  y 2 ) ) dA
 y2
upper surface
3
2
0 0

18  2 r
2

0    2
low er surface
rd  d r  2  1 8  2 r 2  rd r  8 1
3
0