Find the area inside the loop by the fallowing limacon r=10-20sin(theta) show wo

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Explanation / Answer

Note that the two points where the limacon intersects at the origin is: 10 - 20sin? = 0 ==> ? = p/6 and 5p/6. Thus, the area inside the loop is: A(loop) = 1/2 ? (10 - 20sin?)^2 d? (from ?=p/6 to 5p/6) = 1/2 ? (40sin^2? - 400sin? + 100) d? (from ?=p/6 to 5p/6). By the double-angle formula for cosine, we can re-write this as: A(loop) = 1/2 ? (40sin^2? - 400sin? + 100) d? (from ?=p/6 to 5p/6) = 1/2 ? {20[1 - cos(2?)] - 400sin? + 100} d? (from ?=p/6 to 5p/6) = 1/2 ? [-2cos(2?) - 400sin? + 120] d? (from ?=p/6 to 5p/6) = ? [-cos(2?) - 200sin? + 60] d? (from ?=p/6 to 5p/6) = [(1/2)sin(2?) + 200sin? + 60?] (evaluated form ?=p/6 to 5p/6) = (-v3/4 + 100 + 50p) - (v3/4 + 100 + 10p) = (80p - v3)/2.

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