13.2.3 Area By Polar Curves
| 13.2.3 |
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| 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 Light-Blue Color. |
Differential Area
| Fig. 1.4 Differential Area In Light-Blue Color |
Using Symmetry
The colored region in Fig. 1.5 is symmetric about the x-axis. The area A of the region is twice that of the upper-half region. For
simplification we can find the area A/2 of the upper-half 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 top-half region. The area of the top-half 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 light-blue 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
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Solution
By symmetry the required area A is 8 times that of the upper-half of the leaf along the polar axis (positive x-axis). The area of
this half-leaf is:
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Solution
The desired area A is:
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Solution
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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 = a2 square units.
Note
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Solution
By symmetry the desired area A is twice that of the top-half region. Let’s find the angle of the point of intersection of the 2
curves for the top-half region:
The area of the top-half region is:
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