Question Number 225955 by fantastic2 last updated on 17/Nov/25
$${can}\:{we}\:{find}\:{the}\:{perimeter} \\ $$$${of}\:{an}\:{ellipse}? \\ $$
Commented by mr W last updated on 17/Nov/25
$${certainly}\:{we}\:{can}\:{find},\:{but}\:{not}\:{in} \\ $$$${a}\:{simple}\:{formula}\:{as}\:{for}\:{perimeter} \\ $$$${of}\:{a}\:{circle}.\:{we}\:{must}\:{use}\:{elliptic}\: \\ $$$${integral}. \\ $$
Commented by Frix last updated on 17/Nov/25
![There is no exact formula. The length L of a line y=f(x) in the interval [p, q] is given by the integral L=∫_p ^q (√(1+(y′)^2 ))dx Ellipse (upper half): y=(b/a)(√(a^2 −x^2 )) y′=−((bx)/(a(√(a^2 −x^2 )))) Quarter ellipse: L=(1/a)∫_0 ^a (√((a^4 −(a^2 −b^2 )x^2 )/(a^2 −x^2 ))) dx =^([θ=sin^(−1) (x/a)]) =a∫_0 ^(π/2) (√(1−((a^2 −b^2 )/a^2 )sin^2 θ)) dθ= =aE (((√(a^2 −b^2 ))/a))](https://www.tinkutara.com/question/Q225966.png)
$$\mathrm{There}\:\mathrm{is}\:\mathrm{no}\:\mathrm{exact}\:\mathrm{formula}. \\ $$$$\mathrm{The}\:\mathrm{length}\:{L}\:\mathrm{of}\:\mathrm{a}\:\mathrm{line}\:{y}={f}\left({x}\right)\:\mathrm{in}\:\mathrm{the}\:\mathrm{interval} \\ $$$$\left[{p},\:{q}\right]\:\mathrm{is}\:\mathrm{given}\:\mathrm{by}\:\mathrm{the}\:\mathrm{integral} \\ $$$${L}=\underset{{p}} {\overset{{q}} {\int}}\sqrt{\mathrm{1}+\left({y}'\right)^{\mathrm{2}} }{dx} \\ $$$$\mathrm{Ellipse}\:\left(\mathrm{upper}\:\mathrm{half}\right):\:{y}=\frac{{b}}{{a}}\sqrt{{a}^{\mathrm{2}} −{x}^{\mathrm{2}} } \\ $$$${y}'=−\frac{{bx}}{{a}\sqrt{{a}^{\mathrm{2}} −{x}^{\mathrm{2}} }} \\ $$$$\mathrm{Quarter}\:\mathrm{ellipse}: \\ $$$${L}=\frac{\mathrm{1}}{{a}}\underset{\mathrm{0}} {\overset{{a}} {\int}}\sqrt{\frac{{a}^{\mathrm{4}} −\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right){x}^{\mathrm{2}} }{{a}^{\mathrm{2}} −{x}^{\mathrm{2}} }}\:{dx}\:\overset{\left[\theta=\mathrm{sin}^{−\mathrm{1}} \:\frac{{x}}{{a}}\right]} {=} \\ $$$$={a}\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\sqrt{\mathrm{1}−\frac{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\mathrm{sin}^{\mathrm{2}} \:\theta}\:{d}\theta= \\ $$$$={a}\mathrm{E}\:\left(\frac{\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }}{{a}}\right) \\ $$
Commented by fantastic2 last updated on 17/Nov/25
$$\underset{\overset{\frown} {\smile}} {\bullet\:\bullet} \\ $$
Commented by fantastic2 last updated on 17/Nov/25
$${according}\:{to}\:{my}\:{observation} \\ $$$$\:{i}\:{think}\:{u}\:{are}\:{from}\:{USA}\left({Frix}\:{sir}\right) \\ $$
Commented by Frix last updated on 17/Nov/25
$$\mathrm{I}'\mathrm{m}\:\mathrm{from}\:\mathrm{Europe} \\ $$
Commented by fantastic2 last updated on 17/Nov/25
$${in}\:{a}\:{previous}\:{question}\:{from} \\ $$$${me}\:{you}\:{used}\:''{trapezoid}\:':{which} \\ $$$${is}\:{american}\:{English}. \\ $$$${thats}\:{why}\:{i}\:{thought}… \\ $$
Commented by Frix last updated on 17/Nov/25
$$\mathrm{English}\:\mathrm{is}\:\mathrm{not}\:\mathrm{my}\:\mathrm{native}\:\mathrm{language},\:\mathrm{I}\:\mathrm{learned} \\ $$$$\mathrm{it}\:\mathrm{in}\:\mathrm{school}\:\left(\mathrm{a}\:\mathrm{long}\:\mathrm{time}\:\mathrm{ago}\right)\:\mathrm{and}\:\mathrm{by}\:\mathrm{reading}, \\ $$$$\mathrm{listening}\:\mathrm{to}\:\mathrm{music}\:\mathrm{and}\:\mathrm{watching}\:\mathrm{movies}. \\ $$$$\Rightarrow\:\mathrm{My}\:\mathrm{English}\:\mathrm{might}\:\mathrm{be}\:\mathrm{weird}. \\ $$
Commented by fantastic2 last updated on 18/Nov/25
$${math}\:{skill}\:>{English}\:{skill}\left({in}\:{this}\:{forum}\right) \\ $$$${so}\:{you}\:{will}\:{have}\:{no}\:{problem} \\ $$$$\underset{\smile} {\bullet\:\bullet} \\ $$
Commented by TonyCWX last updated on 18/Nov/25
$${To}\:{be}\:{honest},\:{you}\:{can}'{t}\:{just}\:{assume}\:{a}\: \\ $$$${person}'{s}\:{nationality}\:{based}\:{on}\:{what}\:{type}\:{of}\:{English}\:{they}\:{used}. \\ $$$$ \\ $$$${A}\:{person}\:{using}\:{US}\:{English}\:{doesn}'{t}\:{show}\:{that}\:{he}/{she}\:{is}\:{from}\:{USA}. \\ $$$${It}\:{only}\:{shows}\:{that}\:{he}/{she}\:{is}\:{educated}\:{using}\:{US}\:{English}. \\ $$
Commented by fantastic2 last updated on 18/Nov/25
$${true} \\ $$
Answered by TonyCWX last updated on 18/Nov/25
$${There}'{s}\:{an}\:{integral}\:{for}\:{it}. \\ $$$${But}\:{you}\:{can}\:{only}\:{integrate}\:{it}\:{numerically}. \\ $$