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Unformatted text preview: 1. (15 marks) Determine the area interior to both curves whose equations in polar coordinates are given by C 1 : r = sin ( 2 ) [0 , 2 ] C 2 : r = sin ( 2 ) [2 , 4 ] (A sketch is enough to argue about the intersection of the two curves.) Solution. A sketch of the two curves is as follows (the first curve is plotted in blue):1 11 1 x y They are both symmetric with respect to the xaxis (because r ( ) = r (2  ) for the first one, r (2 + ) = r (4  ) for the second). Moreover, the second curve is obtained from the first by reflecting along the origin (because r 2 (2 + ) = r 1 ( )). 1 From this we deduce that to compute the area of the intersection we can just take the area inside the sector delimited by the first curve in the interval [0 ,/ 2] and multiply by 4. Thus A = 4 1 / 2 Z 2 sin 2 ( / 2) d Recall that cos ( ) = cos 2 / 2 sin 2 ( / 2) = 1 2 sin 2 ( / 2). Then A = Z 2 (1 cos ( )) d = 2 1 1 If you used the correct symmetries without proving them rigorously, you will not lose any points. 1 2. (15 marks) Show that the lines L 1 and L 2 given by L 1 : x = y 1 2 = z 2 3 L 2 : x 3 4 = y 2 3 = z 1 2 intersect, find the point of intersection and the angle at which they intersect. Solution. A point of intersection is a point which satisfies the equa tions of both lines, that is: x = y 1 2 = z 2 3 x 3 4 = y 2 3 = z 1 2 x = y 1 2 y 1 2 3 4 = y 2 3 y 1...
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 Spring '11
 Apropolos
 Calculus, Equations, Polar Coordinates

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