Calculus Of One Real Variable – By Pheng Kim
Ving


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1. Area By Polar Curves 

Fig. 1.1 Area Bounded By A Polar Curve And 2 Rays. 

Fig. 1.2 Regular Partition Of Order n = 5 Of An Angle Interval. 

Fig. 1.3 Area Of Wedge In LightBlue Color. 

Differential
Area

Fig. 1.4 Differential Area In LightBlue Color 
Using Symmetry
The colored region in Fig. 1.5 is symmetric about the xaxis. The area A of the region is twice that of the upperhalf
region. For
simplification we can find the area A/2
of the upperhalf region, and then multiply it by 2 to get the area A of the region. In
general, when a region is symmetric about an axis or the origin, we should use
it, for it usually simplifies calculations.
Example 1.1

Fig. 1.5 Area Bounded By A Cardioid. 
Solution 1
The required area A
is:
EOS
Solution 2
By symmetry the required area A is twice that of the upper half region. The area of the upper half region is:
EOS
When the region is symmetric about an axis, we of course can
employ symmetry to simplify calculations. In the above
example, simplification by symmetry is almost negligible. But in some cases
it's useful as it simplifies calculations a lot, as will
be seen in later examples and in the problems & solutions. Whenever there's
symmetry, we should use it.
When Angles Of
Points Of Intersection Are Obvious
Example 1.2

Fig. 1.6 Area Inside A Cardioid And Outside A Circle. 
Solution
By symmetry the desired area A is twice the area of the tophalf region. The area of the tophalf region is:
The desired area is:
EOS
^{p&s_3_or_4}} Problem & Solution 3 Or 4.
When Angles Of
Points Of Intersection Aren't Obvious
Example 1.3

Fig. 1.7 Area Inside A Lemniscate And Outside A Circle. 
Solution

Fig. 1.8 Using Differential Area To Find Required Area. 
The area of the lightblue partial wedge between the circle and the lemniscate is approximately:
EOS
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2. Intervals Of Integration 
Example 2.1

Fig. 2.1 Area Bounded By A Circle. 
Solution
1)
The required area A
is:
EOS
Example 2.2

Fig. 2.2 Area Bounded By A Circle. 
Incorrect Solution
The desired area A
is:
EOS
Correct Solution
The desired area A
is:
EOS
Problems & Solutions 
Solution
By symmetry the required area A is 8 times that of the upperhalf of the leaf
along the polar axis (positive xaxis). The
area of
this halfleaf is:
_{ }
Solution
The desired area A is:
Solution
Solution
By symmetry the required area A is 4 times the area of the part of the region
in the 1st quadrant. The area of the part of the
region in the 1st quadrant is:
So the required area is A = a^{2} square units.
Note
Solution
By symmetry the desired area A is twice that of the tophalf region. Let's
find the angle of the point of intersection of the 2
curves for the tophalf region:
The area of the tophalf region is:
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