Example of Area using Polar Coordinates

 

Problem:   Sketch the curve and find the area enclosed by  r = 3 cos θ.

 

 

Sketching a polar curve can be tricky.  Some guidelines are as follows:

 

 

 If polar equation is same when replacing θ  with  ˗ θ, then the curve is

symmetric about its polar axis, the x˗axis.

 

 

If polar equation is same when r is replaced by  ˗ r  or when  θ  is replaced

by  θ + π, then curve is symmetric about its pole, the origin.

 

 

If polar equation is same when  θ is replaced by π ˗ θ, then curve is symmetric

about the vertical line  θ = π/2.

 

 

 

 

The figure below shows the polar curve   r = 3 cos θ .  Note the curve is symmetric about

the  x-axis. (polar axis)

                                              

                                                                                 θ = b

So the total area under the curve,  r = r(θ) is   A  =  ∫ ½ [ r(θ)2 ] dθ

                                                                                θ = a

 

Be careful to get the correct limits of integration on θ.  This can be tricky as well.

In this example,  θ  goes from  θ = 0  to  θ = π (to sweep out entire area enclosed by curve)

 

                θ = π                                  θ = π                       θ = π

         A  =  ∫ ½ [ 3 cos(θ) ] 2 dθ  =  9/2 ∫ cos2 θ dθ   =  9/2 ∫ [( 1 + cos 2θ)/2] dθ

               θ = 0                                  θ = 0                        θ = 0

 

 

     Result:            A  =  (9/4) π    (Simply the area of the circle of diameter 3)

 

Click here for an example involving arc length.

                                                               




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