Question Number 159958 by mr W last updated on 22/Nov/21
$${find}\:{the}\:{area}\:{and}\:{perimeter}\:{of} \\ $$$$\left(\frac{\boldsymbol{{x}}}{\boldsymbol{{a}}}\right)^{\frac{\mathrm{2}}{\mathrm{3}}} +\left(\frac{\boldsymbol{{y}}}{\boldsymbol{{b}}}\right)^{\frac{\mathrm{2}}{\mathrm{3}}} =\mathrm{1} \\ $$
Commented by MJS_new last updated on 23/Nov/21
$${y}=\pm\frac{{b}}{{a}}\left({a}^{\mathrm{2}/\mathrm{3}} −{x}^{\mathrm{2}/\mathrm{3}} \right)^{\mathrm{3}/\mathrm{2}} \\ $$$$\mathrm{we}\:\mathrm{can}\:\mathrm{solve}\:\mathrm{the}\:\mathrm{integral}\:\mathrm{for}\:\mathrm{the}\:\mathrm{area} \\ $$$$\mathrm{4}\frac{{b}}{{q}}\underset{\mathrm{0}} {\overset{{a}} {\int}}\left({a}^{\mathrm{2}/\mathrm{3}} −{x}^{\mathrm{2}/\mathrm{3}} \right)^{\mathrm{3}/\mathrm{2}} {dx}=\frac{\mathrm{3}\pi}{\mathrm{8}}{ab} \\ $$$${y}'=\pm\frac{{b}\sqrt{{a}^{\mathrm{2}/\mathrm{3}} −{x}^{\mathrm{2}/\mathrm{3}} }}{{ax}^{\mathrm{1}/\mathrm{3}} } \\ $$$$\mathrm{we}\:\mathrm{can}\:\mathrm{also}\:\mathrm{solve}\:\mathrm{the}\:\mathrm{integral}\:\mathrm{for}\:\mathrm{the}\:\mathrm{perimeter} \\ $$$$\mathrm{4}\underset{\mathrm{0}} {\overset{{a}} {\int}}\sqrt{\mathrm{1}+\left({y}'\right)^{\mathrm{2}} }{dx}=\mathrm{4}\frac{{a}^{\mathrm{2}} +{ab}+{b}^{\mathrm{2}} }{{a}+{b}} \\ $$
Commented by mr W last updated on 23/Nov/21
$${thanks}\:{alot}\:{sir}! \\ $$$${i}\:{got}\:{the}\:{same}\:{result}\:{using}\:{parametric} \\ $$$${way}. \\ $$
Commented by MJS_new last updated on 23/Nov/21
$$\mathrm{I}\:\mathrm{corrected}\:\mathrm{it} \\ $$
Answered by mr W last updated on 23/Nov/21
Commented by mr W last updated on 23/Nov/21
![let A=area, P=perimeter due to symmetry we only need to treat the quarter in the first quadrant. the curve can be described as x=a cos^3 θ y=b sin^3 θ with parameter θ∈[0,(π/2)] (A/4)=∫_0 ^a ydx=∫_0 ^(π/2) b sin^3 θ 3a cos^2 θ sin θ dθ (A/4)=3ab∫_0 ^(π/2) sin^4 θ cos^2 θdθ (A/4)=((3ab)/8)∫_0 ^(π/2) (1+cos 2θ) sin^2 2θ dθ (A/4)=((3ab)/8)∫_0 ^(π/2) ((1/2)+(1/2)cos 4θ+cos 2θ sin^2 2θ) dθ (A/4)=((3ab)/8)[(θ/2)+(1/8)sin 4θ+((sin^3 2θ)/6)]_0 ^(π/2) (A/4)=((3ab)/8)×(π/4) ⇒A=((3πab)/8) (P/4)=∫_0 ^a (√(1+(y′)^2 ))dx=∫_0 ^(π/2) (√(1+(−((3b sin^2 θ cos θ)/(3a cos^2 θ sin θ)))^2 )) 3a cos^2 θ sin θ dθ (P/4)=3∫_0 ^(π/2) (√(a^2 cos^2 θ+b^2 sin^2 θ)) cos θ sin θ dθ (P/4)=−(3/4)∫_0 ^(π/2) (√((a^2 −b^2 )cos^2 θ+b^2 )) d(cos 2θ) (P/4)=−(3/4)∫_0 ^(π/2) (√(((a^2 −b^2 )/2)cos 2θ+((a^2 +b^2 )/2))) d(cos 2θ) (P/4)=−(1/(a^2 −b^2 ))[(((a^2 −b^2 )/2)cos 2θ+((a^2 +b^2 )/2))^(3/2) ]_0 ^(π/2) (P/4)=(1/(a^2 −b^2 ))[(((a^2 −b^2 )/2)+((a^2 +b^2 )/2))^(3/2) −(((a^2 +b^2 )/2)−((a^2 −b^2 )/2))^(3/2) ] (P/4)=((a^3 −b^3 )/(a^2 −b^2 ))=((a^2 +ab+b^2 )/(a+b)) ⇒P=((4(a^2 +ab+b^2 ))/(a+b))](https://www.tinkutara.com/question/Q160000.png)
$${let}\:{A}={area},\:{P}={perimeter} \\ $$$${due}\:{to}\:{symmetry}\:{we}\:{only}\:{need}\:{to} \\ $$$${treat}\:{the}\:{quarter}\:{in}\:{the}\:{first}\:{quadrant}. \\ $$$${the}\:{curve}\:{can}\:{be}\:{described}\:{as} \\ $$$${x}={a}\:\mathrm{cos}^{\mathrm{3}} \:\theta \\ $$$${y}={b}\:\mathrm{sin}^{\mathrm{3}} \:\theta \\ $$$${with}\:{parameter}\:\theta\in\left[\mathrm{0},\frac{\pi}{\mathrm{2}}\right] \\ $$$$\frac{{A}}{\mathrm{4}}=\int_{\mathrm{0}} ^{{a}} {ydx}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {b}\:\mathrm{sin}^{\mathrm{3}} \:\theta\:\mathrm{3}{a}\:\mathrm{cos}^{\mathrm{2}} \:\theta\:\mathrm{sin}\:\theta\:{d}\theta \\ $$$$\frac{{A}}{\mathrm{4}}=\mathrm{3}{ab}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{sin}^{\mathrm{4}} \:\theta\:\mathrm{cos}^{\mathrm{2}} \:\theta{d}\theta \\ $$$$\frac{{A}}{\mathrm{4}}=\frac{\mathrm{3}{ab}}{\mathrm{8}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left(\mathrm{1}+\mathrm{cos}\:\mathrm{2}\theta\right)\:\mathrm{sin}^{\mathrm{2}} \:\mathrm{2}\theta\:{d}\theta \\ $$$$\frac{{A}}{\mathrm{4}}=\frac{\mathrm{3}{ab}}{\mathrm{8}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\mathrm{4}\theta+\mathrm{cos}\:\mathrm{2}\theta\:\mathrm{sin}^{\mathrm{2}} \:\mathrm{2}\theta\right)\:{d}\theta \\ $$$$\frac{{A}}{\mathrm{4}}=\frac{\mathrm{3}{ab}}{\mathrm{8}}\left[\frac{\theta}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{8}}\mathrm{sin}\:\mathrm{4}\theta+\frac{\mathrm{sin}^{\mathrm{3}} \:\mathrm{2}\theta}{\mathrm{6}}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \\ $$$$\frac{{A}}{\mathrm{4}}=\frac{\mathrm{3}{ab}}{\mathrm{8}}×\frac{\pi}{\mathrm{4}} \\ $$$$\Rightarrow{A}=\frac{\mathrm{3}\pi{ab}}{\mathrm{8}} \\ $$$$\frac{{P}}{\mathrm{4}}=\int_{\mathrm{0}} ^{{a}} \sqrt{\mathrm{1}+\left({y}'\right)^{\mathrm{2}} }{dx}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \sqrt{\mathrm{1}+\left(−\frac{\mathrm{3}{b}\:\mathrm{sin}^{\mathrm{2}} \:\theta\:\mathrm{cos}\:\theta}{\mathrm{3}{a}\:\mathrm{cos}^{\mathrm{2}} \:\theta\:\mathrm{sin}\:\theta}\right)^{\mathrm{2}} }\:\mathrm{3}{a}\:\mathrm{cos}^{\mathrm{2}} \:\theta\:\mathrm{sin}\:\theta\:{d}\theta \\ $$$$\frac{{P}}{\mathrm{4}}=\mathrm{3}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \sqrt{{a}^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \:\theta+{b}^{\mathrm{2}} \:\mathrm{sin}^{\mathrm{2}} \:\theta}\:\:\mathrm{cos}\:\theta\:\mathrm{sin}\:\theta\:{d}\theta \\ $$$$\frac{{P}}{\mathrm{4}}=−\frac{\mathrm{3}}{\mathrm{4}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \sqrt{\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)\mathrm{cos}^{\mathrm{2}} \:\theta+{b}^{\mathrm{2}} }\:{d}\left(\mathrm{cos}\:\mathrm{2}\theta\right) \\ $$$$\frac{{P}}{\mathrm{4}}=−\frac{\mathrm{3}}{\mathrm{4}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \sqrt{\frac{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }{\mathrm{2}}\mathrm{cos}\:\mathrm{2}\theta+\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{\mathrm{2}}}\:{d}\left(\mathrm{cos}\:\mathrm{2}\theta\right) \\ $$$$\frac{{P}}{\mathrm{4}}=−\frac{\mathrm{1}}{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }\left[\left(\frac{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }{\mathrm{2}}\mathrm{cos}\:\mathrm{2}\theta+\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{\mathrm{2}}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \\ $$$$\frac{{P}}{\mathrm{4}}=\frac{\mathrm{1}}{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }\left[\left(\frac{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }{\mathrm{2}}+\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{\mathrm{2}}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} −\left(\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{\mathrm{2}}−\frac{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }{\mathrm{2}}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \right] \\ $$$$\frac{{P}}{\mathrm{4}}=\frac{{a}^{\mathrm{3}} −{b}^{\mathrm{3}} }{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }=\frac{{a}^{\mathrm{2}} +{ab}+{b}^{\mathrm{2}} }{{a}+{b}} \\ $$$$\Rightarrow{P}=\frac{\mathrm{4}\left({a}^{\mathrm{2}} +{ab}+{b}^{\mathrm{2}} \right)}{{a}+{b}} \\ $$