Question

40-45. Choose a convenient order When converted to an iterated inte- gral, the following double integrals are easier to evaluate in one order than the other. Find the best order and evaluate the integral.

Answer 1

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as x = x=2 2 Now Ans given region R has limits of x. and y to y=0 to y = $h- we will first integrate with respect to a and then integrate with respect to y. so SS y sin (xy) da sy y sin ley) dedy Y=0 R x=o JF - till om dez) de de

9 3 dy (03 (xy') ye Isin (ax) dx = (ar Isinca -Cas (ax) a here a= =y T Cos :(244) + Cos (0.y:)) dy ya y s [-cos (272)+(0) (0)]dy y [- cos (24²) + 1] ay (as Cos(0)=1] 1 g (1 - (03 (24) dy -

کل - dy وله (( و)ه) لا - 9) چکی لاو (وو) 3ه) و y dy ده (33) دی ) - وی را و د ا و ) می -2 الگ كا 4 -) [as Syndy= ynth +) 2 - وہ (مرد) » ) - (ره) (5) ) واد () e9 و - (۰- عه 9 کہ cos

12 - SS y sin(xylda - r у So R y coslay?) dy -3 Now for a cor (272) dy let 14 2y2 = t- - d (242) = dd dy dy - 212y)= dt dy - 4y dy= dt nynnt & y dy dt 15 ។ when y= 2 x - 2y (by) - 2101 t=0

when y = 2 t = 24² 2 - 212 2 ( 2 =) t=2 so SA y cos(24²) dy Some cosley) (ydy) cos (t) (dt) (by 0,0,040) sr t-o

R Y t=0 r Y cos (t) dt => [ain #17 | las Scos ft)dt = sin(49) [sin (2)- sin los because I sin/r)=0 (sin(o)=0 by cos (272) dy = 0 & putting 6 in 6:- SS y sin (x7dA 10-0) Y → Sub 2 r Y