Area in Polar CoordinatesIf we have a circle of radius r, and select a sector of angle θ ,

then the area of that sector can be shown to be 1 2 .2

r θ

Area = (1/2)r2θ

As a check, we see that if θ = 2π, then the area of the entire circle is (1/2)r2(2π) = π r2.

Now suppose that we have a curve in polar coordintes. Then we can derive the formula for area by using the usual mnemonic device of infinitesimals. The figure below indicates that the area under the curve r = f(θ ) between θ 1 and θ2 can be thought of as the integral

2

1

A dAθ

θ= ∫

dA

r = f(θ )

Since dθ is infinitesimally small, the wedge shaped dA shown is essentially a sector of angle dθ. Thus its area is dA = (1/2)r2dθ. This means that the formula for area is

2 21 22

1 1

A dA r dθ θ

θθ θ

= =∫ ∫

Of course this derivation is not mathematically rigorous. To prove it correctly, we should divide the area into wedge shaped regions having angle ∆θi then sum them up and pass to the limit. We will not do this, but will rely on the informal derivation.

Problem. Find the area in the first quadrant that lies within the cardioid r = 1 + cosθ.

This is clearly the area that lies between the rays θ = 0, and θ = π/2.

Thus the area is

/ 2/ 2 11 2 2(1 cos( ))22 00

A r d dππ

θ θ θ= = +∫∫

/ 2 1 cos(2 )1 1 2cos( )22 0

dπ θ

θ θ+ = + + =∫

/ 21 2(1 2cos( ) cos ( ))2 0

dπ

θ θ θ= + +∫

21 1 1sin( ) sin(2 )2 4 8 0

π

θ θ θ θ = + + +

3 18π= +

Problem. Find the area under the rose r = sin(2θ ).

For this problem (and in every possible case) we try to exploit the symmetry of the figure. The entire area is shown below, together with the area between θ = 0 and θ = π/4.

It is easier to find the smaller area and multiply by 8.

Thus the area is

/ 4/ 41 2 28 4 sin(2 )2 00

A r d dππ

θ θ θ = = ∫∫

/ 22 (1 cos(4 ))

0d

πθ θ= −∫

412 sin(4 )2 0

π

θ θ = −

.2π=

Problem. Find the total area that lies within the cardioidr = 2 + 2cosθ.

The desired area is shown below.

Thus the area is

1 2 2(2 2cos( ))2

A r d dππ

θ θ θππ

= = +∫∫ −−

Here we take advantage of symmetry again and let the desired area be written as 2 times the area shown below.

Thus the area is

1 2 2 22 (2 2cos( )) 4 8cos( ) 4cos ( )2 00 0

A r d d dππ π

θ θ θ θ θ θ = = + = + +∫∫ ∫

[4 8cos( ) 2 2cos(2 )] 6 8sin( ) sin(2 ) 6 .00

dππ

θ θ θ θ θ θ π= + + + = + + =∫

Problem. Find the area that is inside the cardioidr = 1 + cos(θ) and outside the circle r = 1.

As in virtually all polar area problems, a sketch of the curves is needed. However, when working with two curves, a sketch of curves and area is absolutely essential. For this problem, the required area is shown below.

It is particularly important not to confuse the area requested with the area inside the circle and outside the cardioid, or the area common to both figures. These two areas are shown below.

Again we need to take advantage of symmetry. This produces easier integral limits and avoids problems that can arise in situations where r can be negative. Below we show that the desired area can be written as 2 times a simpler and smaller area.

It seems clear from the figure that we want the area under the curve r1 = 1 + cos(θ) between θ = 0 and θ = π/2 minus the area under the curve r2 = 1 between θ = 0 and θ = π/2. [We have used subscripts to distinguish between the two curves]. Thus the originally desired area (twice the area shown above) is

/ 2 / 2 / 21 1 12 2 2 22 2 ( )1 2 1 22 2 20 0 0A r d r d r r d

π π πθ θ θ

= − = − ∫ ∫ ∫

/ 2 / 22 2 2((1 cos( )) 1 ) (2cos( ) cos ( ))0 0

d dπ π

θ θ θ θ θ= + − = +∫ ∫

2/ 2 1 1 1(2cos( ) cos(2 )) 2sin( ) sin(2 )2 2 2 4 00

dπ

π θθ θ θ θ θ = + + = + +∫

24π= +

/ 2 / 2 / 21 1 12 2 2 22 2 ( )1 2 1 22 2 20 0 0A r d r d r r d

π π πθ θ θ

= − = − ∫ ∫ ∫

As we saw in the previous problem, the way to find the area between two curves r1 = f(θ ) and r2 = g(θ ), where r1 is the “outside” curve and r2 is the “inside” curve, is to evaluate the integral

Where the angle limits define the area.

21 2 2( )1 221

r r dθ

θθ

−∫

The required area is shown below.

θ = θ1

θ = θ2

This problem is almost exactly the same as the previous one, but here it is not clear what the angles are. Again we can write the desired area as 2 times a smaller area.

Problem. Find the area that is inside the cardioidr = 4 + 4cos(θ) and outside the circle r = 6.

θ = θ2

This smaller integral is between 0 and θ2, so the first job is to identify θ2. We note that at that angle, the two curves meet, so the two radii are equal. We thus need to solve the equation

1 4 4cos( ) 6 or cos( ) .2

θ θ+ = =

This implies that .2 3

πθ =

θ = θ2

Thus the desired area is

/ 3 /31 22 22 ( ) ((4 4cos( )) 36)1 22 0 0A r r d d

π πθ θ θ= ⋅ − = + −∫ ∫

/ 3 2[ 20 32cos( ) 16cos ( )]0

dπ

θ θ θ= − + +∫

312 32sin( ) 4sin(2 ) 18 3 4 .

0

π

θ θ θ π=− + + = −

/ 3[ 20 32cos( ) 8 8cos(2 )]

0d

πθ θ θ= − + + +∫

Problem. Find the area that is shown in the figure below, where the outside curve is the cardioid r = 1 + cos(θ) and the inside curve is the circle r = cos(θ).

This problem is tricky, because it really has to be reduced to two different integrals with two different sets of limits. It is clearly the sum of the following two areas.

r = 1 + cos(θ)r = cos(θ)

Area 1 Area 2

Area 1 is the area inside 1 + cos(θ) and outside cos(θ) between the angles θ = 0 and θ = π/2. Thus its value is

/ 2 / 21 1 2 22 2( ) ((1 cos( )) (cos( ) )1 21 2 20 0A r r d d

π πθ θ θ θ= − = + −∫ ∫

2/ 21 11 2cos( ) ( 2sin( )) 1.2 2 400

dπ

π πθ θ θ θ= + = + = +∫

Area 2

Area 2 has nothing to do with the circle. It is the area inside the cardioid r = 1 + cos(θ) between the angles θ = π/2, and θ = π. Thus its area is

/ 2 / 2

1 12 2(1 cos( ) ) (1 2cos( ) cos ( ))2 2 2

A d dπ π

π πθ θ θ θ θ= + = + +∫ ∫

/ 2

1 11 (1 2cos( ) cos(2 ))2 22

dπ

πθ θ θ= + + +∫

/ 2 2

3 1 3 11 2cos( ) cos(2 ) sin( ) sin(2 )2 2 4 82

dππ

ππ

θθ θ θ θ θ

= + + = + +∫

3 3 31 1.4 8 8π π π= − − = −

Thus the total area is

3 51 1 .4 8 8π π π+ + − =

Method 2.r = 1 + cos(θ)

r = cos(θ)

We can also see that the area can be formed by subtracting the area of the semicircle from the area under the cardioid within the upper half plane. Thus we can find the area under the cardioid exactly as we did above, except for limits, and get

00

3 11 32(1 cos( ) ) sin( ) sin(2 )1 4 82 4

A dππ θ πθ θ θ θ

= + = + + =∫

The area of the semicircle is clearly π/8, so the area we desire is

3 54 8 8π π π− =

Problem. Find the area that is shown in the figure below, where the cardioid is r = 1 + cos(θ) and the circle is r = 3cos(θ). This is the area inside both curves simultaneously.

In this problem, the area must again be broken into parts. In this case no part of the area is between two curves. We begin by reducing the area using symmetry.

The desired area is clearly twice the area shown below.

The angle represented by the black line is the angle where 1 + cos(θ) = 3cos(θ), or cos(θ) = 1/2. Thus it is π/3. The area shown above can thus be broken into two parts. It is the area inside the cardioid from 0 to π/3 plus the area under the circle from π/3 to π/2. These areas are shown below.

Area 1 Area 2

These areas are as follows./ 3 / 3

0 0

1 12 2(1 cos( ) ) (1 2cos( ) cos ( ))1 2 2

A d dπ π

θ θ θ θ θ= + = + +∫ ∫

/3

0

1 11 (1 2cos( ) cos(2 ))2 22

dπ

θ θ θ= + + +∫

/ 3

0

3 1 3 3 9 3sin( ) sin(2 )4 8 4 2 16 4 16

πθ π πθ θ = + + = + + = +

/ 2 / 2

/ 3 / 3

1 92(3cos( ) ) (1 cos(2 ))2 2 4

A d dπ π

π πθ θ θ θ= = +∫ ∫

/ 2

/ 3

1 39 9 3 9 3sin(2 ) .2 2 3 44 4 8 16

π

π

π π πθ θ = + = − − = −

Thus the total area shown in

is A1 + A2 = 9 3 3 9 3 54 16 8 16 8π π π+ + − =

Finally, the area shown below must be

2(A1 + A2) = 54π