Question

3. (a) Given I = S, V10(2x + y) ds where c is the straight line segment y = 3x from (0,0) to (2,6) as shown below. 2 (1 mark) 0) With x = t, express y in terms of the parametert for the straight line. () With ds = dt, express ds in terms of parameter t and its derivative. (4 marks) C) Use the above (i) and (ii) results to find the value of I. (5 marks) (b) Given V = S," S*(8xy - 4y) dy dx. Find the value of v. (6 marks) () Sketch the region of integration. (2 marks) C) Reverse the order of integration, but do not re-evaluate. (2 marks)

Answer 1

5)) Given I = Sio (2x+y) do с. ca is the line segment y = 3x from loo) to (2, with a=t, the line segment ỹ = 32 = 3t = 9=36 do= It &y=3t olt) then da=dt 2 x=t& dy = 3cht - 1 & dy=3 alt It do = 12 dt Sit Colt doa dt 12 dt = - (1) +(3)2 1 +9 ſłodt do=fo de I= Sito (2x+y)cho use parsmelizalion x=t,y=32 do=5łodt s To (2++ 38 3+)(Tode) С?. Oct=2 t=02 softdt = solt= 0(4-0) sio(st)dt - 2584 JI = 100 Given D) Y = S'S '18ry -43)dly de o

EN ។។ edx v=sS skydy dx de – ugššay de a fac Sšajad -45 (547) de 3. *Ax] -4 ' [ Sexq? 08-'15'iz'de] - (8x4 -8xo)-(4x4-9) = 1-43 = s y² 8 x 2 2 다. 4 - = 8X22 Luxx² 2x410 1 2X3 -o TX8 ys 8 » [V=Y ☺ R= {(29) 0<<1, oxysx} Region of integration . yoto Oo's - (iiy (*) 03491-4954)=1444)/=231} of yeast R={(1,1) 741,721 ey="9) sydx = SS(8x9=49) dx I dedy 7 VE 0 o
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