Polar Coordinates
Definition, Conversions, and
Integration
Where is it?
Coordinate systems are used to locate the position of a point.
(3,1)
In rectangular coordinates:
•We break up the plane into a grid of
horizontal and vertical line lines.
•We locate a point by identifying it
as the intersection of a vertical and a
horizontal line.
(1,/6)
In polar coordinates:
•We break up the plane with circles
centered at the origin and with rays
emanating from the origin.
•We locate a point as the
intersection of a circle and a ray.
Locating points in Polar Coordinates
Suppose we see the point
(r, )= (2,/6)
and we know it is in polar
coordinates. Where is it in the
plane?
The first coordinate,
r =2, indicates the
distance of the point
from the origin.
The second coordinate,  = /6,
indicates the distance counterclockwise around from the
positive x-axis.
r =2
(2,/6)
 = /6
Locating points in Polar Coordinates
Note, however, that every point in
the plane as infinitely many polar
representations.

(r , )  2, 
6

(2,/6)
 = /6
Locating points in Polar Coordinates
Note, however, that every point in
the plane as infinitely many polar
representations.
 6
  2,   2 
6
(r , )  2, 
 2,13 6 
2  
6
Locating points in Polar Coordinates
Note, however, that every point in
the plane as infinitely many polar
representations.
 6
  2,   2 
6
  2,   2 
6
(r , )  2, 
And we can go clockwise or
counterclockwise around the circle as
many times as we wish!
 2, 11 6 
  2
6
Converting Between Polar and
Rectangular Coordinates
It is fairly easy to see that if
(x,y) and (r, ) represent the
same point in the plane:
 r, 
x  r cos()
y  r sin()
r x y
x
tan() 
y
2
2
2
These relationships allow us to
convert back and forth between
rectangular and polar coordinates
Integration in Polar
Coordinates
Non-rectangular Integration
Elements
Small Changes in r and 
Suppose we consider a small
change from r
. . . to r + dr
r + dr
r
Small Changes in r and 
Suppose we consider a small
change from r
. . . to r + dr
This gives us a thin “ring”
around the origin.
r + dr
r
Small Changes in r and 
Suppose we consider a small
change from 
. . . to  + d
 + d

Small Changes in r and 
Suppose we consider a small
change from 
d
. . . to  + d
This gives us a “pie-shaped
wedge” that is subtended by
the angle d.
 + d

Small Changes in r and 
Intersecting the “thin ring”
. . . and the “pie-shaped
wedge”,
. . . we get . . .
Small Changes in r and 
Intersecting the “thin ring”
. . . and the “pie-shaped
wedge”,
. . . we get . . .
Small Changes in r and 
In order to integrate a function
given in polar coordinates
(without first converting to
rectangular coordinates!), we
need to know the area of this
little piece.
Why?
Integration in Polar Coordinates
In order to integrate a function given in polar coordinates, we will first
“chop up” our region into a bunch of concentric circles and rays
emanating from the origin.
(r*,*, f(r*,*))
(r*,*)
Now do this for
each little
“wedge” and add
up the volumes
of the “towers”.
Problem: the volume of the “tower” is the area of the base times the
height. But the base is not a rectangle, so its area is not dr d!
Area of the “Small Bit”
A= area of sector of a circle

r 2

A
 2  A
2
2 r
Area of a “Small Bit”
In order to integrate a function
given in polar coordinates
(without first converting to
rectangular coordinates!), we
need to know the area of this
little piece.
(r  dr )2 d  r 2 d 
dA 

2
2
Area of a “Small Bit”
(r  dr ) 2 d  r 2 d 
dA 

2
2
2
2
r

2
rdr

(
dr
)
d  r 2d 




2
2 So . . .
dr 2 d 

r dr d  
V   f (r , ) r dr d 
2
R
dA  r dr d 
dV  f (r , ) dA 
f (r , ) r dr d 