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In Exercises 30-33 use the fact that in polar coordinates small lengths √r² + (dr/de)² de along a curve can be expressed in the form ds = (see formula 9.14) to evaluate the line integral. * 30. * lc √x² + y² r = 2-sin starting from the point on the x-axis (x² + y²) ds, where C is the cardioid r = 1 + cos 30. With ds = 22 + ds, where C is the first quadrant part of the limaçon dr de 31. (x² + y²) ds = [ 1² **** 2 Love²² - So ds = + - 8 do - (2-sin 0)2+(- cos 0)2 do √5-4 sin de, we obtain स/2 {-(5-4 sin 0) 3/2) 17/2 = 0 fahre r cos 0 7" √5-4 sin 0 do = *(1 + cos 0)² √2 + 2 cos 6 de 0 = √2 2 dr √² + (6) * 4 = ["₁₁. 1 "d6 = "” (1 + cos 0)² √(1 + cos 0)² + (-- sin 0)² do de do = 2 / (1+ cos 0)5/² de T √21+2 sin² (0/2) - 115/² d0 = 8 ² do = 8 [²² sin5 (0/2) de sin (0/2) [1cos² (0/2)]² d0 = 8 : 8 ["²8 -T 5√5-1 6 sin (0/2) [1-2 cos² (0/2) + cos¹ (0/2)] do 256 = - {-2 cos (0/2) +cos³ (0/2) - cos5 (0/2)} 2 5 (2)}" 15