Question

Q3(a) Let W be the region above the sphere x2 + y2 + z2 = 6 and below the paraboloid z = 4 - x2 - y2 as shown in Figure Q5(a) below: Z=4-x-y? x2 + y + z = 6 Figure Q3(a) (i) Find the equation of the projection of Won the xy-plane. (ii) Compute the volume of W using polar coordinates. [16 marks] (b) Using double integral in polar coordinates, compute the following: $$*** (2x+3y) dedy [7 marks] (c) Evaluate the triple integral below: SSJ4x’zdv Where E is bounded by the parabolic cylinder z = 1 - y2, the xy-plane, yz-plane and the plane x = 2. [8 marks]

Answer 1

(a) We will find the equation of the projection of win poday wortunate In my plane 20. So, x+y=6 anty=4 out at So, the region after forojection will be sent {(x,y) : 45 x+yu 56.5. (6) Volume of W= on. x²-yr l a dz Z-V6-Xyr 45xity [6 dy dx NS6. 16 44x²tyr Lb. Ir 16-pr dr. 722 2-|zozdz let YO torraz. -2rdr=22dz ³rdre-zdz - SS [4-(*240) -J6 $=x2xul de demain . So set (4-'- vo -) nido de - (60-012- vdo-pa)ar. - 20. Sa ryb. - 1 1 2 22 } - 23. (2.2 - 7.(3.6-16) - 2$). - 20 (-1-24) = -2 ( 3 +262). Hence volume is 27 (3 +252)

fo is 12-yv f (24+34) dady. Non V yo nay. the xay aut The slope of nay line . so, the double integral becomes in ut uar loro, yarsino. osr<2, a 702 g. So, the double integral becomes x²+4² 2. Yrty 2. 21 22 y21. Y271.,x=1 2 0+3 Sino) parardo. o/a poo 1 S S (2 6910+ 3 fino) pa v drdo + 35me) do far (2.8 - 3600 ; (1-ta) - 3 (0-6)] 16-0. $ 2 - +*). - $ (2 + ).

US$*WSS z ar days 20 y=-1220 Z2-1 x20 ye-1 A wwwov = 2 (1-9) dy an. 120 2- Y2-1 yo-1 25 $ (1-73-. as ou *2*** S (1-2y's st) dy. 3.1 232 [y *£y+$45 ) - 3:8. (2 -3.2 + ]). - (2- + 3). "(3020.76.) on -25