AREA IN POLAR

COORDINATES

Adapted from a presentation by Mercy Snow

AREA ENCLOSED BY POLAR CURVESSimilar to Cartesian equations, we can find the exact area of the polar region using an integral. The only exception is that we are using sectors to approximate the area, not rectangles.

The area of a sector is:

22A r

The area of the highlighted sector is: 22A f

212 f

If we integrate this area over the entire interval, it represents the total area bounded by the curve.

The area of the polar region of from and is given by

212

A r d

AREA ENCLOSED BY POLAR CURVES

EXAMPLE 1Find the area of the region in the plane enclosed by .

2 212 0

r d

2 21

2 02 1 cos d

6

If you created a graph, made a table, or analyzed the

equation; you will see the radius sweeps out the region exactly once as runs from

to .The area is

therefore:

(4,0)

(4,2𝜋 )

EXAMPLE 2Find the area inside the smaller loop of

.

4 /3 212 2 /3

r d

2 212 0

2cos 1 d

0.544

If you created a graph, made a table, or analyzed the

equation; you will see the smaller loop is traced by the

radius as runs from to .The area

is therefor

e:

(0 , 2𝜋3 )

(0 , 4 𝜋3 )

EXAMPLE 3Find the area inside all of the loops of .

If you created a graph, made a table, or analyzed the equation; you will see the size of each loop

are identical.Instead of finding the whole area, you could triple the area of one

loop.Find where the curve goes through the pole:sin 3 0 3 2 , ,0, , 2 ,...

2 23 3 3 3, , 0, , ,...

Use consecutive values for the interval and triple the

integral:

(0 , 𝜋3 )

(0,0)

/3 21

2 03 sin3 d

4

AREA ENCLOSED BY POLAR CURVESThe area of the polar region between and from and is given by

2 2

2 2

12

1 12 2

o i

o i

A r r d

or

A r d r d

It is still outside curve minus the inside curve.

EXAMPLEFind the area of the region that lies inside and outside .

2 212 o ir r d

/2 2 2

/2

1 1 1 cos2

d

1.215

If you created a graph, made a table, or analyzed the equations; you will see they intersect twice. Find that

intersection:

On the interval, is the outside curve and is the inside curve:

1 1 cos cos 0

3 52 2 2 2, , , ,...

The graph can confirm to is a desired interval. Other

intervals like to would give a different area.

(1 , 𝜋2 )

(−1 ,− 𝜋2 )

WARNING!The formulas below only work if the intersection occurs at the same for each relation AND the interval only includes one cycle for each relation.

2 2

2 2

12

1 12 2

o i

o i

A r r d

or

A r d r d

NEVER ASSUME ANYTHING ABOUT A POLAR CURVE

EXAMPLEFind the area of the region that lies outside and inside .

If you created a graph, made a table, or analyzed the equations; you will see they intersect once. Find that

intersection:

or

3 3coscos 1

2 ,0,2 , 4 , 6 ,... Both curves do contain the point or but…

Both also contains the point . And this point is at the same location as .

(−3 ,𝜋 )

This is because completes a cycle on to but completes a cycle on to . The graph

at right is only on to .

EXAMPLE (CONTINUED)Find the area of the region that lies outside

and inside .Therefore, you can not find the area with one integral

that has the same intervals. It must be broken

up into two separate integrals (Remember is on

the outside):

(−3 ,𝜋 )

2 2

0

1 32

d

20

1 3cos2

d

274

RULE OF THUMB FOR POLAR CURVES

Never ASSUME anything about the curve. Always check the graph.

Ask yourself…• Do the intersections have the same angle?• How many cycles are being graphed?• Am I seeing the complete picture?• Does my change in theta need to be decreased?• Etc.