Question

-22-y2 (3pts each) (i) Show that =v*dżdy = (4 **-**d:) (Hint: The intermediate step should be showing Loose-r2-y*dxdy = (soccer)(S%ce=?)) (ii) Given ( <R<o, compute: - 22-yé da where DR is the disk of radius R centered at the origin. iii) Argue that lim *dA= S--ýdrdy iv) Compute: dc

Answer 1

If you have any doubts in the solution please ask me in comments......

Here i use the general definition of interigation ....

2 0 TO Show, OD solve :- - dy s rex? ey²dx dy i f e-x-y* dx dy = (ſezoz) Now Š s ezt-age dx dy = . 1&16*s*) sez de is only function of a constant fory. - 11eky Cheton) () {i ffryddy • Strande -Šezda) (fex*dx) ſ Š e-ze-y*dx dy (je**dx)? b Hence proove. - (2) given 0R<00 to compute SS e-ze-y2 dA. DR Dr is disk of radius R - centered at Origion Solve! (o, R) ЈГ e-z“-у? dA = Ss e-ré-yé dx dye DR DR DBR Νοω assume x = roso y = rsino & drdo then dx oly OSTER and 0 € (0,217]. 20 R S: S .7 or do .72 e sse-x-y² aA O o DR 2 TT ş e 22 2008).do. if ²=t do ardo=dt fjetron Idl fetar) 370 → t=0 r=R » t = R²

re-22-44 dx dy. 2 TV R2 sse-22-4²dA De ***** Ź do = (-e-t).R? [te-t). T(4-e-R?). TT (1-e-R) dA Answer SS e-x²-ya DR -2²-4² clin to argue that dA = Lim R->00 DR - Solve da By previous part sse -x²-y² DR TT (1-e-R?). TT Lim SS e-z-ya da Lim Tri-bike) Rao R0 DR Lim SS e-2²-4² da = R300 DR Now By previous part ci) we have. IŠ e-z7-ya dzdy (Sex*dz). je I= ୧-2 s Let dz Now is an even function then: e-22 dz 2 18** ose ce I V dz - 2x dx 2 dx = 22 - then Z let לוס 2-007 2=0. 2-0, & 2=0 00 e-2.2-² az e-z. dz I= co ſ ez. zied dz z 2-1 dz. o

-2 n-1 z Νοω know that dz as we > I = gno e- ZZ dz = 11 / (gamma function) I= E TT → 22 88 es dx TE By equation Oo 2 Ś (Se-z* da) (T)* - T. Š Š e-zé-yé du dy = » į je za dady = T. CO By es Lim R=0 SS e-x²-y? dA s se re-y dedy DR Hence proove * (41) = I; Set de ſexdx sco Solve: S 2² : (202) I = X Let 2x dx 24 = dz = 2dx = z NILN dz. >> je zase je? zisa I = 20 z²nd ſez. have d2= By gamma function. So (-IN cue V AN ST; → I = senza dxa ST -00 0 senzz da , Answer