Question

1-20. Computing surface areas Find the area of the surface gener- ated when the given curve is revolved about the given axis

Just do 11,12,and 13

(3x)?/3, for 0 < x < 2; about the y-axis sxs 4; about the y-axis B) y = V1 – xě, for sxs ; about the x-axis

Answer 1

21 Y = (3x)'3 for os as of about y. Axi 43.320 1 + fcy) and and sley)ya at x = 0, y = 0. Area of revolution A = 21 few) / 1 + $4))? dy 2 vagas de - 20 / 2 y 3 Vity & dy let Ityt - t 4 y 3 dy = dt y3dy - dt limite - Y=0 t = 1 y = 2, t - 17 117 = 21 I st dt 3 J,

A 1"+" dt 2 TO (1791] M = 24.110 square unit 012: < y = x for 24 msu about y-Axis g? = 4y - Jay. fly) vty, f'(4) - 7 dny vy at x=2 y = 1 x= 4, y = 4 Area of revolution about y-Axis- V A : 21 [ & fly) √ 1 + [f'412 dy 20 S ty 14 Chap dy

A = 25 [+ 255 sitty sy 416* ny xv y dy - +7 का let Ity ut e dy = at limity :- at y = 1, tzn. | 1 4 , t 5 A 41 vt dt 411 = 40X31's l_22] - am (s. 212_2013] TA = 69.9600 square unit. I de-Axi Q13 - fcy)- y = 51-22 1<x< 1 about fley) = 2 - - Area of revolution about a AXIS A = 211 fcx) dit [f'(x)}? dac V - 212

22 -12 21 1 + 1-22 22 V 1-12 an | Vox.vit dve = 21 f or r 1 - 2 + 12 de 21 [ VT de - 2 IT x² +22 J-12 dy - 2 Salz 12 = 29 (x 3.112 - 21 [ 1 + 2 - 211 = 6. 2031 square unit.
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