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subject: Area Using Polar Coordinates [print this page]


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

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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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