Question Number 23283 by ajfour last updated on 28/Oct/17
$${Compute}\:{the}\:{area}\:{of}\:{a}\:{loop}\:{of} \\ $$$${the}\:{curve}\:\boldsymbol{\rho}=\boldsymbol{{a}}\mathrm{sin}\:\mathrm{2}\boldsymbol{\theta}\:;\:{and}\:{even} \\ $$$${sketch}\:{the}\:{curve},\:{please}. \\ $$
Answered by mrW1 last updated on 28/Oct/17
![a small loop for θ from 0 to (π/2): A_1 =(1/2)∫_0 ^( (π/2)) ρ^2 dθ =(a^2 /2)∫_0 ^( (π/2)) sin^2 2θdθ =(a^2 /4)∫_0 ^( (π/2)) (1−cos 4θ)dθ =(a^2 /4)[θ−((sin 4θ)/4)]_0 ^(π/2) =((πa^2 )/8) for a total loop from 0 to 2π, A=4×((πa^2 )/8)=((πa^2 )/2)](https://www.tinkutara.com/question/Q23286.png)
$$\mathrm{a}\:\mathrm{small}\:\mathrm{loop}\:\mathrm{for}\:\theta\:\mathrm{from}\:\mathrm{0}\:\mathrm{to}\:\frac{\pi}{\mathrm{2}}: \\ $$$$\mathrm{A}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \rho^{\mathrm{2}} \mathrm{d}\theta \\ $$$$=\frac{\mathrm{a}^{\mathrm{2}} }{\mathrm{2}}\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \mathrm{sin}^{\mathrm{2}} \:\mathrm{2}\theta\mathrm{d}\theta \\ $$$$=\frac{\mathrm{a}^{\mathrm{2}} }{\mathrm{4}}\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \left(\mathrm{1}−\mathrm{cos}\:\mathrm{4}\theta\right)\mathrm{d}\theta \\ $$$$=\frac{\mathrm{a}^{\mathrm{2}} }{\mathrm{4}}\left[\theta−\frac{\mathrm{sin}\:\mathrm{4}\theta}{\mathrm{4}}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \\ $$$$=\frac{\pi\mathrm{a}^{\mathrm{2}} }{\mathrm{8}} \\ $$$$\mathrm{for}\:\mathrm{a}\:\mathrm{total}\:\mathrm{loop}\:\mathrm{from}\:\mathrm{0}\:\mathrm{to}\:\mathrm{2}\pi, \\ $$$$\mathrm{A}=\mathrm{4}×\frac{\pi\mathrm{a}^{\mathrm{2}} }{\mathrm{8}}=\frac{\pi\mathrm{a}^{\mathrm{2}} }{\mathrm{2}} \\ $$
Commented by mrW1 last updated on 28/Oct/17
Commented by ajfour last updated on 28/Oct/17
$${thank}\:{you}\:{very}\:{much}\:{Sir},\:{right} \\ $$$${answer},\:{nice}\:{question}\:;\:{i}\:{thought} \\ $$$${of}\:{sharing}\:{it}\:{with}\:{you}. \\ $$$${i}'{ve}\:{created}\:{a}\:{caromboard}\:{question}, \\ $$$${please}\:{view}\:{it},\:{sir}. \\ $$