Question

Use a double integral to find the area of a petal of the rose r =? 2 (cos 3?(theta))

Answer 1

f Given:- 7= 2LOS 30 we know that, Curve of a rose curve 1/6 is As we have onl only one petal, 2 I 16 The limits will be, cos300 o = 4/6 => - sos % g have variable the If we See We angle 1 bounds fon r from to 2cos 3o the region of petal. moving along < < 2 Cos (30) -7% <o< 1% and oso < 2 COS(39) 2 cas(30) T1/6 , , Area do do – 1/6 T1/6 sc 2008 (30) do ४२ 148 -T/ 11/6 4 cas(30) do . - 2 -T116

피 2 cos(39) d9 피 COS A = + COS (24) 끼 1+ Cos (Go) do 22. 2. 핍 T/ 지 6 1 l 9 + COS (60) d9 ) . -T6 피 피 6 (9] 나 sin 60 6 팅 패 피 6 1 Sin(n) - SIP-T) ( 'sin(77)=o => Sin(IT) = 0 이디 sy ㅠ 6 + sin(-1) =0 AN Area 3 of the rose petal of , Area 1 0 is F] - 2 COS (39) 3

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