Question

Problem 3. Let C be the curve with implicit equation (x2 + (1 – 2)2 +2 – 1)2 = 2(x² + (1 – 2)2) in the first quadrant of the xz-plane, shown below. Z 1.6 1.4 1.2 1 0.8 0.6 0.4 0.2 х 0.5 1 1.5

Answer 1

(1-3)? + 2- (124 2-1)? 2 (x2+ (1-2)2) *, 230 d first Quadrants where Z 21 7-0 fel? (3x2+ (-632+ 0-1)= 2 (x? +1) (e? +1-1) 2 (1241) 7 d.y= 2x2+2 Lo dY 222 =2 w Y-2E+) = 2+) (x2-1)2 = 3 x? 12 +53 But x2_12-53 is Reje lected as a2=1-53 is Negative which is not possible.

Lo 22!= J3 which gives x²=1+53 X XE 1+53 xoo Finally x-woodlinate is ez J+53 di 1.65289165 1 X ~ 1.6529 x Part 6 SS EXE) onds whale R= yêt zxåt z3 €

Š x ² = 1 را T d Ja е RP d JE g 2x 73 3x2= ($(23) he d (23) da de 29 30 +CaCraw_sa dej Tx= (o-o]ê -[o-o]i+ [2-1) So Text = . K SS (EXF). nds = kinds E M where is created by revolving c alound Z-axis. Best this integral is difficult to find. using stokels's theorem S56xF).nase & F. E

of Where is the boundary curve c. which will be a circle in x-y plane having zo Jading of circle will be equal to X= (1+3=8 fo Equation of Ci x2 + y2 = It J3 where zo

X = J1 +53 Cost y = 1 +53 sint where te [o, art] Lt MA)= JH J3 (Sine) i + 2 J1 +53 Cost ŷ to Lo Sext) nas = g f. aru & R. are still allaide ehele d'(d)e + 53 singl ê + (11+ 53 682) I F.LT= S{1+ 53 sint i +57684]).C-Altsz&inti RI o 9-7X, +11+52 GP AJ THE b-

3(1453) 8" (1-60824) ff & F. do = SAE(+53) sin? 4., 72(1753) Casex) dx where wit= 1-sinet ㄷ 1 E(1+53) Sütf+ 2015) 6 stk) ) dit =12" (201+53) - 3 (1413) Sin24) de = 2(1+53 f It 25 30+53) gsinzt dt 2 is YIT (1753 2

E 40C1t5a) – 3(153) [pin - Storze } " EYTT (1+13) 361453) x 21 = 2 48 (1+53) - 30 (1+55) = H (1433) Frally SS ExF). nads= (1+ S3) {