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Calculus

10th Edition
Ron Larson + 1 other
ISBN: 9781285057095

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BuyFindarrow_forward

Calculus

10th Edition
Ron Larson + 1 other
ISBN: 9781285057095
Textbook Problem

Area The area inside one or more of the three interlocking circles r = 2 a cos θ , r = 2 a sin θ , and r = a is divided into seven regions. Find the area of each region.

To determine

To calculate: The value of the area of each seven regions divided by three interlocking circles r=2acosθ, r=2asinθ and r=a.

Explanation

Given:

The provided curves: r=2asinθ,r=2acosθ and r=a.

Formula used:

Trigonometric identities:

cos2θ=2cos2θ1cos2θ=1+cos2θ2cos2θ=cos2θsin2θ

Calculation:

The area of seven regions divided by three interlocking circles r=2acosθ, r=2asinθ and

r=a are shaded below:

To find the intersection of these three curves, equate these curves in pair and get:

2acosθ=2asinθtanθ=1θ=π4

And,

2acosθ=acosθ=12θ=π3,π3

Also,

2asinθ=asinθ=12θ=π6,5π6

Since, the area of the region under a polar curve is given by:

A=12ab(r)2dθ.

It is clear from the above figure that by symmetry A1=A2 and A3=A4.

The area A1 can be found by evaluating the region between r=2acosθ and r=a under limits π3 and π6 in summation with the region between r=2acosθ and r=2asinθ under limits π6 and π4.

Hence,

A1=12π3π6[(2acosθ)2(a)2]dθ+12π6π4[(2acosθ)2(2asinθ)2]dθ=a22π3π6(4cos2θ1)dθ+a22π6π4[4cos2θ4sin2θ]dθ

Now,

cos2θ=cos2θsin2θ.

And,

cos2θ=2cos2θ1cos2θ=1+cos2θ2

Put these values in above integration and get,

A1=a22π3π6(4(1+cos2θ2)1)dθ+a22π6π4[4cos2θ]dθ=a22π3π6(1+2cos2θ)dθ+2a2π6π4[cos2θ]dθ

Integrate each term separately,

A1=a22[θ+sin2θ]π3π6+a2[sin2θ]π6π4=a2[12(π6+π3)+12(32+32)+(132)]=a2[π4+1]

Hence, A1=A2=a2[π4+1].

From the figure shown above, the area A3 is similar to the area of the sector of circle r=a between lines θ=π3 and θ=5π6.

Put these values in formula for area of sector.

That is A=(θ2π)πr2.

Hence,

A3=(5π6π3)2π(πr2)=(π2)2π[πa2]=πa24

Hence, A3=A4=πa24.

The area A5 can be found by evaluating the region under the curve r=2asinθ under limits 5π6 and π in summation with the area of the sector of circle r=a between lines θ=π3 and θ=π.

Therefore,

A5=(π2+π3)2π(πa2)+125π6π[(2asinθ)2]dθ=(5π6)2π(πa2)+2a25π6πsin2θdθ

Further solve and get,

A5=512(πa2)+2a25π6π(1cos2θ2)dθ=5πa212+a25π6π(1cos2θ)dθ

Integrate each term separately and get,

A5=512(πa2)+2a2[θsin2θ2]5π6π=5πa212+2a2(π5π634)=5πa212a2(π332)=a2(π12+32)

Hence, A5=a2(π12+32)

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