Question

xy=1 and y 2x V -X -Region S is bounded by the lines xy 2. Draw the region and indicate all the vertices. and the hyperbolas 2 and B) Transfer region S from x-y to u-v plane and indicate all the vertices on the new plane acx. y au,v) =1 C) Show that the area corrections are related by (u,v) x, y) D) Find the centroid of region S

Answer 1

R is the region bounded by y=x y 2x.(2) xy 1 (3) xy 2..(4) =1-x 2 Substituting (1) in (3), we get Taking square root on both sides we get x t\2 Substituting get x =t2 in (1) we get get y =+ - ==+ Therefore, the points of intersect of (1) and (3) are t2, t. =2>x= 4 x =t2 Substituting (1) in (4), we get x Substituting x =+2 in (1) we get y t 1 Therefore, (1) and (4) will intersect at t 2, ± 1) Substituting (2) in (3) we get x(2x) 12x 1> x =t5 Substituting x =±in (2), we get y= v2) Therefore, (2) and (3) will intersect at t Substituting (2) in (4) we get x(2x) = 2 >2x" = 2=> x =t 1 Substituting x =+1 in (2), we get y 2(1)t 2 xy 2 Therefore, (2) and (4) will intersect at (t 1,2) Y-2x Yx/2

Using the transformation x and y= u-v u 2u - 2v u + v u = 3v...(5) will be transferred to u-v: y y 2x will be transferredto u-v= 2(u + v) -> u-v= 2u+ 2v-> u = -3v...(6) will be transferredto (u- v)(u + v) = 1 -> u2- v = 1-> . ..( 7) Vx u-v=2> ...(8) yx 2 will be transferredto (u - v)(u + v) = 2 (5) and (6) are straight lines and (7) and (8) are hyperbolae 1 Substituting (5) in (7),we get (3v) - v= 1> 8v = 1=> v =t Corresponding u =t Therefore, the points ±are the points of intersect of (5) and (7) Substituting (5) in (8),we get (3v)^ - v= 2=>8v = 2=> v=t; Corresponding u =±> 3 Therefore, the points , t,), are the points of intersect of (5) and (8) 1 Substituting (6) in (7),we get -3v)^ - v= 1=> 8v = 1=>v =t Correspondingu-T Therefore, the points ( intersect of (6) and (7) ,t are the points Substituting (6) in (8),we get ( - 3v)7-v= 2=> 8v=2=> v =t Correspondingu=T; Therefore, the points (7, t5), are the points of intersect of (6) and (8) |дх дх а (ху) auv)

(u) (a +r (a-n)e (a-n)" = ( 1) -)-(1) (1) =-2 xy x-y From xu vand y= u-vwe get u and v= (a'n)e d(xy) dy = -( 0lxv au,v) Multiplyingfs and we getv äízv) =- 2 (xv) (u) Therefore,,v a(xy) = 1