A movable point P has cartesian coordinates (x,y), where x and y are functions of t. The polar coordinates of P with respect to the origin O are r and θ. Starting with the expression 21∫r2dθ for the area swept out by OP, obtain the equivalent expression 21∫(xdtdy−ydtdx)dt.(*)The ends of a thin straight rod AB lie on a closed convex curve C. The point P on the rod is a fixed distance a from A and a fixed distance b from B. The angle between AB and the positive x direction is t. As A and B move anticlockwise round C, the angle t increases from 0 to 2π and P traces a closed convex curve D inside C, with the origin O lying inside D, as shown in the diagram.
Let (x,y) be the coordinates of P. Write down the coordinates of A and B in terms of a, b, x, y and t.
The areas swept out by OA, OB and OP are denoted by [A], [B] and [P], respectively. Show, using (∗), that [A]=[P]+πa2−af where f=21∫02π((x+dtdy)cost+(y−dtdx)sint)dt. Obtain a corresponding expression for [B] involving b. Hence show that the area between the curves C and D is πab.