How do you find the area of the region bounded by the polar curves #r=1+cos(theta)# and #r=1-cos(theta)# ?

Answer 1

The region bounded by the polar curves looks like:

Since the region consists of two identical leaves that are symmetric about the #y#-axis, I will try to find a half of one leaf then multiply it by #4#.

#A=4int_0^{pi/2}int_0^{1-cos theta}rdrd theta#

#=4int_0^{pi/2}[r^2/2]_0^{1-cos theta}d theta#

#=2int_0^{pi/2}(1-2cos theta+cos^2theta)d theta#

by #cos^2theta=1/2(1+cos2theta)#,

#=int_0^{pi/2}(3-4cos theta+cos2theta)d theta#

#=[3theta-4sintheta+1/2sin2theta]_0^{pi/2}#

#={3pi}/2-4#


I hope that this was helpful.

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Answer 2

To find the area of the region bounded by the polar curves ( r = 1 + \cos(\theta) ) and ( r = 1 - \cos(\theta) ), integrate the difference of the two curves from their points of intersection. The points of intersection are found by setting the two equations equal to each other and solving for ( \theta ). Then, integrate the difference between the larger curve and the smaller curve with respect to ( \theta ) over the interval of intersection. This gives the area enclosed by the curves.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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