Chapter 10.5, Problem 50E

### Calculus (MindTap Course List)

11th Edition
Ron Larson + 1 other
ISBN: 9781337275347

Chapter
Section

### Calculus (MindTap Course List)

11th Edition
Ron Larson + 1 other
ISBN: 9781337275347
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=2asinÎ¸,r=2acosÎ¸ and r=a.

Formula used:

Trigonometric identities:

cos2Î¸=2cos2Î¸âˆ’1cos2Î¸=1+cos2Î¸2cos2Î¸=cos2Î¸âˆ’sin2Î¸

Calculation:

The area of seven regions divided by three interlocking circles r=2acosÎ¸, r=2asinÎ¸ and

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

2acosÎ¸=2asinÎ¸tanÎ¸=1Î¸=Ï€4

And,

2acosÎ¸=acosÎ¸=12Î¸=Ï€3,âˆ’Ï€3

Also,

2asinÎ¸=asinÎ¸=12Î¸=Ï€6,5Ï€6

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

A=12âˆ«ab(r)2dÎ¸.

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=2acosÎ¸ and r=a under limits âˆ’Ï€3 and Ï€6 in summation with the region between r=2acosÎ¸ and r=2asinÎ¸ under limits Ï€6 and Ï€4.

Hence,

A1=12âˆ«âˆ’Ï€3Ï€6[(2acosÎ¸)2âˆ’(a)2]dÎ¸+12âˆ«Ï€6Ï€4[(2acosÎ¸)2âˆ’(2asinÎ¸)2]dÎ¸=a22âˆ«âˆ’Ï€3Ï€6(4cos2Î¸âˆ’1)dÎ¸+a22âˆ«Ï€6Ï€4[4cos2Î¸âˆ’4sin2Î¸]dÎ¸

Now,

cos2Î¸=cos2Î¸âˆ’sin2Î¸.

And,

cos2Î¸=2cos2Î¸âˆ’1cos2Î¸=1+cos2Î¸2

Put these values in above integration and get,

A1=a22âˆ«âˆ’Ï€3Ï€6(4(1+cos2Î¸2)âˆ’1)dÎ¸+a22âˆ«Ï€6Ï€4[4cos2Î¸]dÎ¸=a22âˆ«âˆ’Ï€3Ï€6(1+2cos2Î¸)dÎ¸+2a2âˆ«Ï€6Ï€4[cos2Î¸]dÎ¸

Integrate each term separately,

A1=a22[Î¸+sin2Î¸]âˆ’Ï€3Ï€6+a2[sin2Î¸]Ï€6Ï€4=a2[12(Ï€6+Ï€3)+12(32+32)+(1âˆ’32)]=a2[Ï€4+1]

Hence, A1=A2=a2[Ï€4+1].

From the figure shown above, the area A3 is similar to the area of the sector of circle r=a between lines Î¸=Ï€3 and Î¸=5Ï€6.

Put these values in formula for area of sector.

That is A=(Î¸2Ï€)Ï€r2.

Hence,

A3=(5Ï€6âˆ’Ï€3)2Ï€(Ï€r2)=(Ï€2)2Ï€[Ï€a2]=Ï€a24

Hence, A3=A4=Ï€a24.

The area A5 can be found by evaluating the region under the curve r=2asinÎ¸ under limits 5Ï€6 and Ï€ in summation with the area of the sector of circle r=a between lines Î¸=âˆ’Ï€3 and Î¸=Ï€.

Therefore,

A5=(Ï€2+Ï€3)2Ï€(Ï€a2)+12âˆ«5Ï€6Ï€[(2asinÎ¸)2]dÎ¸=(5Ï€6)2Ï€(Ï€a2)+2a2âˆ«5Ï€6Ï€sin2Î¸dÎ¸

Further solve and get,

A5=512(Ï€a2)+2a2âˆ«5Ï€6Ï€(1âˆ’cos2Î¸2)dÎ¸=5Ï€a212+a2âˆ«5Ï€6Ï€(1âˆ’cos2Î¸)dÎ¸

Integrate each term separately and get,

A5=512(Ï€a2)+2a2[Î¸âˆ’sin2Î¸2]5Ï€6Ï€=5Ï€a212+2a2(Ï€âˆ’5Ï€6âˆ’34)=5Ï€a212âˆ’a2(Ï€3âˆ’32)=a2(Ï€12+â€‹32)

Hence, A5=a2(Ï€12+â€‹32)

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