Find the area inside the inner loop of the following limacon: r = 5 - 10 \\sin \

ID: 3193155 • Letter: F

Question

Find the area inside the inner loop of the following limacon: r = 5 - 10 sin heta

Explanation / Answer

The inner loop occurs when r = 0: 10 - 20 sin ? = 0 ==> sin ? = 1/2 ==> ? = p/6, 5p/6 (in one revolution, where the limacon is entirely traced). Hence, the area equals ?(? = p/6 to 5p/6) (1/2) (10 - 20 sin ?)^2 d? = ?(? = p/6 to 5p/6) 50 [1 - 4 sin ? + 4 sin^2(?)] d? = ?(? = p/6 to 5p/6) 50 [1 - 4 sin ? + 2 (1 - cos(2?))] d? = ?(? = p/6 to 5p/6) 50 [3 - 4 sin ? - 2 cos(2?)] d? = 50 [3? + 4 cos ? - sin(2?)] {for ? = p/6 to 5p/6} = 50 {[5p/2 + 4 * -v3/2 + v3/2] - [p/2 + 4 * v3/2 - v3/2]} = 50(2p - 3v3)

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Find the area inside the inner loop of the following limacon: r = 5 - 10 \\sin \

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Explanation / Answer

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