find-the-area-and-perimeter-of-x-a-2-3-y-b-2-3-1-

Question Number 159958 by mr W last updated on 22/Nov/21

f i n d t h e a r e a a n d p e r i m e t e r o f

{(\frac{x}{a})}^{\frac{2}{3}} + {(\frac{y}{b})}^{\frac{2}{3}} = 1

Commented by MJS_new last updated on 23/Nov/21

y = \pm \frac{b}{a} {(a^{2 / 3} - x^{2 / 3})}^{3 / 2}

we can solve the integral for the area

4 \frac{b}{q} \underset{0}{\int^{a}} {(a^{2 / 3} - x^{2 / 3})}^{3 / 2} d x = \frac{3 π}{8} a b

y^{'} = \pm \frac{b \sqrt{a^{2 / 3} - x^{2 / 3}}}{{a x}^{1 / 3}}

we can also solve the integral for the perimeter

4 \underset{0}{\int^{a}} \sqrt{1 + {(y^{'})}^{2}} d x = 4 \frac{a^{2} + a b + b^{2}}{a + b}

Commented by mr W last updated on 23/Nov/21

t h a n k s a l o t s i r!

i g o t t h e s a m e r e s u l t u s i n g p a r a m e t r i c

w a y .

Commented by MJS_new last updated on 23/Nov/21

I corrected it

Answered by mr W last updated on 23/Nov/21

Commented by mr W last updated on 23/Nov/21

l e t A = a r e a, P = p e r i m e t e r

d u e t o s y m m e t r y w e o n l y n e e d t o

t r e a t t h e q u a r t e r i n t h e f i r s t q u a d r a n t .

t h e c u r v e c a n b e d e s c r i b e d a s

x = a \cos^{3} θ

y = b \sin^{3} θ

w i t h p a r a m e t e r θ \in [0, \frac{π}{2}]

\frac{A}{4} = \int_{0}^{a} y d x = \int_{0}^{\frac{π}{2}} b \sin^{3} θ 3 a \cos^{2} θ \sin θ d θ

\frac{A}{4} = 3 a b \int_{0}^{\frac{π}{2}} \sin^{4} θ \cos^{2} θ d θ

\frac{A}{4} = \frac{3 a b}{8} \int_{0}^{\frac{π}{2}} (1 + \cos 2 θ) \sin^{2} 2 θ d θ

\frac{A}{4} = \frac{3 a b}{8} \int_{0}^{\frac{π}{2}} (\frac{1}{2} + \frac{1}{2} \cos 4 θ + \cos 2 θ \sin^{2} 2 θ) d θ

\frac{A}{4} = \frac{3 a b}{8} {[\frac{θ}{2} + \frac{1}{8} \sin 4 θ + \frac{\sin^{3} 2 θ}{6}]}_{0}^{\frac{π}{2}}

\frac{A}{4} = \frac{3 a b}{8} \times \frac{π}{4}

\Rightarrow A = \frac{3 π a b}{8}

\frac{P}{4} = \int_{0}^{a} \sqrt{1 + {(y^{'})}^{2}} d x = \int_{0}^{\frac{π}{2}} \sqrt{1 + {(- \frac{3 b \sin^{2} θ \cos θ}{3 a \cos^{2} θ \sin θ})}^{2}} 3 a \cos^{2} θ \sin θ d θ

\frac{P}{4} = 3 \int_{0}^{\frac{π}{2}} \sqrt{a^{2} \cos^{2} θ + b^{2} \sin^{2} θ} \cos θ \sin θ d θ

\frac{P}{4} = - \frac{3}{4} \int_{0}^{\frac{π}{2}} \sqrt{(a^{2} - b^{2}) \cos^{2} θ + b^{2}} d (\cos 2 θ)

\frac{P}{4} = - \frac{3}{4} \int_{0}^{\frac{π}{2}} \sqrt{\frac{a^{2} - b^{2}}{2} \cos 2 θ + \frac{a^{2} + b^{2}}{2}} d (\cos 2 θ)

\frac{P}{4} = - \frac{1}{a^{2} - b^{2}} {[{(\frac{a^{2} - b^{2}}{2} \cos 2 θ + \frac{a^{2} + b^{2}}{2})}^{\frac{3}{2}}]}_{0}^{\frac{π}{2}}

\frac{P}{4} = \frac{1}{a^{2} - b^{2}} [{(\frac{a^{2} - b^{2}}{2} + \frac{a^{2} + b^{2}}{2})}^{\frac{3}{2}} - {(\frac{a^{2} + b^{2}}{2} - \frac{a^{2} - b^{2}}{2})}^{\frac{3}{2}}]

\frac{P}{4} = \frac{a^{3} - b^{3}}{a^{2} - b^{2}} = \frac{a^{2} + a b + b^{2}}{a + b}

\Rightarrow P = \frac{4 (a^{2} + a b + b^{2})}{a + b}

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