Find the area of the region that lies inside the first curve and outside the sec

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A = ? (p --> 2p) ? (1 --> 1-sinT) r dr dT Evaluate the inner integral first, ? (1 --> 1-sinT) r dr = ½r² |(1 --> 1-sinT) = ½[(1 - sin T)² - 1²] = ½[1 - 2 sin T + sin² T - 1] = ½sin² T - sin T Evaluate the outer integral A = ? (p --> 2p) [½sin² T - sin T] dT = ½ ? (p --> 2p) [½ - ½ cos 2T] dT - ? (p --> 2p) sin T dT = ½ [½T - ¼sin 2T] |(p --> 2p) - (- cos T) |(p --> 2p) = (¼T - ?sin 2T + cos T) |(p --> 2p) = (p/2 - 0 + 1) - (p/4 - 0 - 1) = 2 + p/4

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