Area of Polar Curves

The polar co-ordinates in the mathematics are the two dimensional co-ordinate systems in which we determines the distance of any Point on the plane and we also finds the angle in respect of any of the fixed direction. The fixed point is called as the pole of the plane and the Ray which goes from pole to any fixed direction is the polar axis of the plane. The distance of the point from the pole is the radius, and the angle is called as the polar angle or the angular co-ordinate of the polar system. In the polar co-ordinate systems we uses two trigonometrical function Rcos @ and Rsin @ to convert any of the two polar co-ordinates into Cartesian coordinates x and y. As

X = Rcos @ is for x axis on the co-ordinate system,

Y = Rsin @ is for y axis on the co-ordinate system.

Now, put some light on the how to find the area of the polar curves which are bounded by them. In the term of area of polar curves, we have two different words, area enclosed by the curve and the area under the curve so in polar co-ordinate system we always calculate the area enclosed by the curve. Let we have a positive and continuous function r = f (θ), which is defined over the interval [a, b] and the bounded area by curve will give the area for the curve of the function.

The shaded area is bounded or enclosed by the curve and we can calculate the area bounded by the curve with the help of a formula:

Area of Polar Curves =             ∫ ½(r2) d θ            for the limit range of [b, a]

Now, take some of the example of how to find the area of the polar curves.

Example 1: Determine the area of the inner loop of the function r = 2 + 4 cos θ?

Solution: Here the value of the r is given as r = 2 + 4 cos θ. On Graphing the curve of function we have to first find the limits of the curve for which interval we are going to calculate the area of the curve. So, putting values of r:

0 = 2 + 4 cos θ,

cos θ = -1/2 means that the values of the θ will in interval of [2π/3, 4π/3].

The shaded curve of the sketch will give the desired area of the curve, so applying the Integration formula for the area of the curve, we will have:

A             =             ∫ ½ (2 + 4 cos θ)2 dθ,

=             ∫ 1/2(4 + 16 cos θ + 16 cos 2θ) dθ,

=             ∫ 2 + 8 cos θ + 4(1 + cos 2θ) dθ,

Now, after the integration of the function applying the limits [2π/3, 4π/3] on the area of the curve we have:

Area      =             | (6θ + 8sin θ + 2sin 2θ)| for the limits [2π/3, 4π/3],

=             4π - 6√3,

=             2.174.

Example 2: Find the area of a polar curve that is inside the curves r = 3 + 2 sin θ and r = 2?

Solution: First we sketch the graph of the function and the graph of Functions is as:

We cannot find the area of polar curves by the similar pattern as we have done in the above curve because no any bounding limits of the area of the function, for which we calculate the area, are given. To find the area we have to enclose some of the part of the curves for any particular range of the angle θ.

So, in this case we take another approach as we find the angle where the both graphs intersect each other. From graph we can say that they are intersecting at 7π/6 and 11π/6, so The Range of the limits are [7π/6, 11π/6]. Now, we can calculate area by applying the formula for the area as:

Area      =             ∫ ½((2)2 - (3 + 2sin θ) 2) dθ,

A             =             ∫ 1/2(-5 -12 sin θ - 4 sin 2θ) dθ,

A             =             ∫ 1/2(-7 - 12 sin θ + 2 cos 2θ) dθ,

On integration we get:

A             =             1/2| (-7θ + 12 cos θ + sin (2θ))|,

Applying the limits of the angle we have:

A             =             11√3 - 7π/3,

Area of the curve             =             2.196.

Example: Find the area bounded by both of the curves x = 2 sin t, and y = 3 sin2 t?

Solution: Again the limits are not present in the example given, so we have to calculate them.

As the graphs of both the Functions are symmetrical in nature so we can take limits as [0, π/2].

Now, simplifying the area of the curve and find the differentiation of the x with respect to t, we have: dx/dt = 2 cos t.

So, area can be given as:

Area      =             2 ∫ 3 sin2 t (2 cos t) dt,

=             12 ∫ (sin2 t cos t) dt,

Now integrating it and applying the limits we have solution and then we get:

Area      =             |4sin3 t| for the limit range of [0, π/2].

The area of the curve will be = 4,

Example: Find the area of the surface generated by the polar function such that r = √(cos 2θ). The limits for the curve are

0 <= θ <= π/4?

Solution: The solution about the y axis can be given as:

Area = ∫ 2π (√cos 2θ cos θ √ ((√cos 2θ) + (- sin 2θ (√cos 2θ)2)))) dθ

Area = 4.269

Similarly, we can calculate the area of a polar curve with the help of the given formula and the limits ranges of the function variables.