Question

Suppose that the Type I region defined by R = {(8,4)2 5 5 0,9(2) 545 f()} has (737) as its centroid. Let k >o be an arbitrary positive real number. Use the formulas for finding the centroid to show that if f(x) and g(*) are multiplied by k, then the resulting region, R' = {(2,4)| 5 5 , kg(z) S4 5 #f(x)}, will have a centroid that is given by (7,ky).

Answer 1

has (5 g) as its centrod Let eritten dessidy function Sa, y) = x where yo is a contact DE f=Jw b b - da pelo dy da x da xt [42] Prs Criven R = {1xylasneb, gw£ G 240s } M= Mass = fi 80, y) LA = f S x dg da igo si dy do g=ye M = A (2-g18) de Now Evaluating first Moments My = $$ 4513,4) dA = $ 8W dy dx = xj das y dy ob Mau [{pap Law 82 da a=a 20 I a b 020 b x=a y = g(x) ob yg (x) X=a A JOB laza

My = ff a d (2, 4) LA = SS a (x) dy de R R b fix) DESIG = 13x de dy q=a b y = JU) Es a dx x [y] &lt) ТЕ СА b a da x Cru- g(x) di p=a b My = x ja (pir-gw) dr lca : Centriod = că, (ă, ý My Ma M M al sw-y) do lza XS (pa-gwy doc X=a b u 2 2 Tza de soutienos X 307842) -80lly 1:47, ( 81-90) dz es un bè sau gi' ) f (Pw-gtx de 2=t b 2 by Mixe-gloggda &=a Coniaci US - 011-41350505. 89668-89668

M'= Mass= R ob KPiR ib к Фt ) Tea yckgr) Ik gu) My = SS. G Ser, y) 2A = J J y IV Gyken Region R' = Sex glazRab, KIUL JA KPU where ko Sane dersity sanction Sr, y) = d. where so is a constad SJ, S (a JJLA = SS s dy dre dy da = Af de f dy dia yg 10 =xgo da x (KP(x) – Kg(x) yea M'=ikl (ptr)-4143) doo ara How Evaluating first Moments P10 - As de y dy a=a y = kgW b Kra) aza b b R! Foto xif [ye) koup y=k gw

Centrota of Region R (ar, y My M? M'x 11 b b x 1 x (Hw-ga) 2x at go cros-glos) de -Toa aza Xx 2 ( Covery 2 (gures & Jade X* $6(+10-403) da k s (1 * 0j? (allojado b X=a b Tea g (P00-900)da po prose-goo) de I a s pb (PU)-2009) de ya = (ł, ký) 1.: (87 , y) = (2, kg) /Proked'