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In Exercises 17-25 find a definite integral expression that represents the area of the given region in the polar plane, and then find the exact value of the expression.

The region bounded by the limaçon r=1+ksinθ, where 0<k<1. Bonus: Explain why the area approaches πas k→0.

Short Answer

Expert verified

Ans:

limk→0A=12limk→02π+k2π=122π+(0)2πlimk→0A=π

Step by step solution

01

Step 1. Given information:

A limaconr=1+ksinθ

02

Step 2. Implying formula: 

Formula to find the area is A=∫αβ12(f(θ))2dθor A=∫αβ12r2dθ.

The limits are from 0 to 2Ï€.

A=12∫02π(1+ksinθ)2dθ

Then,

A=12∫02π1+k2sin2θ+2ksinθdθA=12∫02π1+k2(1-cos2θ)2+2ksinθdθA=12∫02π1+k22-cos2θ2+2ksinθdθ

03

Step 3. Continue:

Thus,

A=12θ+k22θ-sin2θ4+2kcosθ02πA=122π+k222π-sin2·2π4+2kcos2π-2kcos2·0A=122π+k222π-0+2k-2k

04

Step 4. Proving:

Then,

A=122Ï€+k2Ï€

Therefore the area of the limacon is A=122Ï€+k2Ï€.

Here as k→0the area is

limk→0A=12limk→02π+k2π=122π+(0)2πlimk→0A=π

Thus when k→0the area is π.

Hence it is proved.

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