Area Using Polar Coordinates
Polar coordinate system is a system fo defining the position of points in a plane like the Cartesian co-ordinate system
. Polar co-ordinates systems measures the radial distance of a point from a fixed point or origin and the angle subtended with the horizontal by the radial line to define a point. Thus the co-ordinates of a point are defined by (r,'theta' ), where r is the radial distance and 'theta' is the angle subtended with the horizontal.
We can convert the Cartesian co-ordinates of the point (x,y) to polar co-ordinates as below:
If (x, y) are the Cartesian coordinates of any point in a plane we have the relations
x = rcos, y = rsin
(r, ) polar coordinates
Hence r =sqrt (x^2 + y^2)
Then the relationship between polar coordinates r and is given by the polar equation
f(r, ) = 0 or r = f()
Computing the Area Using Polar Coordinates:
To compute the area of the region, the region is described by
0 r f (),
The area of the region of polar slice is given as
A f ()^2 ( / 2)
= f ()^2(1/2) ()
Differentiate the above function, we get
dA/d = (1/2) f ()2
dA = (1/2) f ()2 d
A = 'int_(alpha)^(beta)'dA = (1/2) 'int_(alpha)^(beta)' f ()2 d
Ex:
Find the area of the region inside the cardioid r = 1 + cos.
Sol:
A = (1/2) 'int_0^(2(pi))' (1 + cos)2 d
By symmetry,
= 'int_0^(pi)' (1 + cos)2 d
= 'int_0^(pi)'(1 + 2cos + cos2) d
= 'int_0^(pi)'(1 + 2cos +(1/2)(1 + cos 2) d
= 'int_0^(pi)'(3/2) + 2cos + ((cos 2)/2) d
Integrate the above function with respect to '',
= [ (3/2) + 2sin + (sin 2)/4)0(pi)
= 3 / 2
Area Using Polar Coordinates between Polar Curves:
To compute the area of a region between the polar curves, the region is described by
0 g () r f (),
The area of a region is given by
dA = (1/2) ( f ()^2 - g ()^2)d
Therefore,
dA = (1/2)'int_(alpha)^(theta)' ( f ()^2 - g ()2)d
Understand
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Example for area between polar curves:
Find the area of the region that is inside the circle r = 2 cos and outside the circle r = 1.
Sol:
A = (1/2)'int_(-(pi/3))^((pi)/3)' (2^2 cos^2 - 12 ) d
By symmetry,
= 'int_0^((pi)/3)' (4 cos^2 - 1 ) d
= 'int_0^((pi)/3)' (2(1 + cos 2) - 1)d
= 'int_0^((pi)/3)'(1 + 2cos 2) d
Integrate the above function with respect to '',
= ( + sin 2)0(pi/3)
= / 3 + sin(2/3)
= ( / 3) + (3 / 2)
by: johnharmer
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