Question Number 76883 by john santu last updated on 31/Dec/19
$${what}\:{is}\:{the}\:{perimeter}\:{of}\:{the}\:{loop}? \\ $$$$\mathrm{3}{ay}^{\mathrm{2}} \:=\:{x}\left({x}−\mathrm{3}{a}\right)^{\mathrm{2}\:} ? \\ $$
Commented by MJS last updated on 31/Dec/19
![length of a curve y=f(x) within [p, q] ∫_p ^q (√(1+(y′)^2 ))dx in this case y=±(x−3a)(√(x/(3a))) due to symmetry we can take the upper curve ×2 p=0; q=3a y′=((3(x−a))/(2(√(3ax)))) this leads to (√(3/a))∫_0 ^(3a) (√((x^2 −((2a)/3)x+a^2 )/x))dx and I found no way to exactly solve this but it′s c×a with c=8.14677558193...](https://www.tinkutara.com/question/Q76891.png)
$$\mathrm{length}\:\mathrm{of}\:\mathrm{a}\:\mathrm{curve}\:{y}={f}\left({x}\right)\:\mathrm{within}\:\left[{p},\:{q}\right] \\ $$$$\underset{{p}} {\overset{{q}} {\int}}\sqrt{\mathrm{1}+\left({y}'\right)^{\mathrm{2}} }{dx} \\ $$$$\mathrm{in}\:\mathrm{this}\:\mathrm{case} \\ $$$${y}=\pm\left({x}−\mathrm{3}{a}\right)\sqrt{\frac{{x}}{\mathrm{3}{a}}} \\ $$$$\mathrm{due}\:\mathrm{to}\:\mathrm{symmetry}\:\mathrm{we}\:\mathrm{can}\:\mathrm{take}\:\mathrm{the}\:\mathrm{upper}\:\mathrm{curve}\:×\mathrm{2} \\ $$$${p}=\mathrm{0};\:{q}=\mathrm{3}{a} \\ $$$${y}'=\frac{\mathrm{3}\left({x}−{a}\right)}{\mathrm{2}\sqrt{\mathrm{3}{ax}}} \\ $$$$\mathrm{this}\:\mathrm{leads}\:\mathrm{to} \\ $$$$\sqrt{\frac{\mathrm{3}}{{a}}}\underset{\mathrm{0}} {\overset{\mathrm{3}{a}} {\int}}\sqrt{\frac{{x}^{\mathrm{2}} −\frac{\mathrm{2}{a}}{\mathrm{3}}{x}+{a}^{\mathrm{2}} }{{x}}}{dx} \\ $$$$\mathrm{and}\:\mathrm{I}\:\mathrm{found}\:\mathrm{no}\:\mathrm{way}\:\mathrm{to}\:\mathrm{exactly}\:\mathrm{solve}\:\mathrm{this} \\ $$$$\mathrm{but}\:\mathrm{it}'\mathrm{s}\:{c}×{a}\:\mathrm{with}\:{c}=\mathrm{8}.\mathrm{14677558193}… \\ $$