Math Question and Answers


Question and Answers Forum

All Questions Topic List
Others Questions
Previous in All Question Next in All Question
Previous in Others Next in Others
Question Number 172234 by Sotoberry last updated on 24/Jun/22

Commented by cortano1 last updated on 24/Jun/22
$S = ∫_0 ^( 2π) (√(((dx/dθ))^2 +((dy/dθ))^2 )) dθ { (((dx/dθ) = −3acos^2 θ sin θ )),(((dy/dθ)= 2asin θ cos θ)) :} S=∫_0 ^(2π) (√(9a^2 sin^2 θ cos^4 θ+4a^2 sin^2 θ cos^2 θ)) dθ = ∫_0 ^(2π) ∣a∣sin θ cos θ (√(9 cos^2 θ+4)) dθ = −((4∣a∣)/(18)) ∫_0 ^(π/2) (√(9cos^2 θ+4)) d(9cos^2 θ+4) =−((2∣a∣)/9).(2/3) [ 9cos^2 θ+4 ]_0 ^(π/2) =−((4∣a∣)/(27)) [ 4−13 ] = ((4∣a∣)/3)$
$S = \int_{0}^{2 π} \sqrt{{(\frac{d x}{d θ})}^{2} + {(\frac{d y}{d θ})}^{2}} d θ$ ${\begin{cases} \frac{d x}{d θ} = - 3 a \cos^{2} θ \sin θ \\ \frac{d y}{d θ} = 2 a \sin θ \cos θ \end{cases}$ $S = \underset{0}{\int^{2 π}} \sqrt{9 a^{2} \sin^{2} θ \cos^{4} θ + 4 a^{2} \sin^{2} θ \cos^{2} θ} d θ$ $= \underset{0}{\int^{2 π}} ∣ a ∣ \sin θ \cos θ \sqrt{9 \cos^{2} θ + 4} d θ$ $= - \frac{4 ∣ a ∣}{18} \underset{0}{\int^{π / 2}} \sqrt{9 \cos^{2} θ + 4} d (9 \cos^{2} θ + 4)$ $= - \frac{2 ∣ a ∣}{9} . \frac{2}{3} {[9 \cos^{2} θ + 4]}_{0}^{\frac{π}{2}}$ $= - \frac{4 ∣ a ∣}{27} [4 - 13] = \frac{4 ∣ a ∣}{3}$
	Answered by mr W last updated on 24/Jun/22

	$\frac{d x}{d θ} = - 3 a \cos^{2} θ \sin θ$ $\frac{d y}{d θ} = 2 a \sin θ \cos θ$ $d s = \sqrt{{(d x)}^{2} + {(d y)}^{2}}$ $= \sqrt{a^{2} \sin^{2} θ \cos^{2} θ (9 \cos^{2} θ + 4)} d θ$ $= \frac{a}{2 \sqrt{2}} ∣ \sin 2 θ ∣ \sqrt{9 \cos 2 θ + 17} d θ$ $t h e c u r v e f r o m θ = π t o 2 π i s t h e$ $s a m e a s f r o m θ = 0 t o π, t h e r e f o r e$ $i j u s t c a l c u l a t e t h e l e n g t h f r o m$ $θ = 0 t o π .$ $s = \int_{0}^{π} d s$ $= 2 \int_{0}^{\frac{π}{2}} d s$ $= \frac{2 a}{2 \sqrt{2}} \int_{0}^{\frac{π}{2}} \sin 2 θ \sqrt{9 \cos 2 θ + 17} d θ$ $= \frac{a}{18 \sqrt{2}} \int_{\frac{π}{2}}^{0} \sqrt{9 \cos 2 θ + 17} d (9 \cos 2 θ + 17)$ $= \frac{a}{18 \sqrt{2}} \times \frac{2}{3} {[{(9 \cos 2 θ + 17)}^{\frac{3}{2}}]}_{\frac{π}{2}}^{0}$ $= \frac{a}{27 \sqrt{2}} [{(9 + 17)}^{\frac{3}{2}} - {(- 9 + 17)}^{\frac{3}{2}}]$ $= \frac{a}{27 \sqrt{2}} \times (26^{\frac{3}{2}} - 8^{\frac{3}{2}})$ $= \frac{a}{27 \sqrt{2}} \times (26 \sqrt{26} - 16 \sqrt{2})$ $= \frac{2 (13 \sqrt{13} - 8) a}{27}$ $\approx 2.879 a$
	Commented by mr W last updated on 24/Jun/22

	Commented by mr W last updated on 25/Jun/22

	$l e n g t h s h o u l d b e > 2 \sqrt{2} a = 2.83 a$ $s = 2.879 a s e e m s t o b e c o r r e c t .$
	Commented by Tawa11 last updated on 25/Jun/22

	$Great sir$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum