Find-the-perimeter-of-the-figure-which-is-bounded-with-the-curves-y-3-x-2-and-y-2-x-2-

Question Number 159742 by amin96 last updated on 20/Nov/21

Find the perimeter of the figure which is

bounded with the curves y^{3} = x^{2} and y = \sqrt{2 - x^{2}}

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

y = x^{\frac{2}{3}}

y = \sqrt{2 - x^{2}} i s a c i r c l e w i t h r a d i u s \sqrt{2}

\sqrt{2 - x^{2}} = x^{\frac{2}{3}}

\Rightarrow x = \pm 1, y = 1

b o t h c u r v e s i n t e r s e c t a t (- 1, 1), (1, 1) .

l e n g t h o f c u r v e y = x^{\frac{2}{3}} f o r x = 0 t o 1 :

\int_{0}^{1} \sqrt{1 + {(y^{'})}^{2}} d x

= \int_{0}^{1} \sqrt{1 + {(\frac{2}{3} x^{- \frac{1}{3}})}^{2}} d x

= \int_{0}^{1} \sqrt{1 + \frac{4}{9 x^{\frac{2}{3}}}} d x

= {[x {(1 + \frac{4}{9 x^{\frac{2}{3}}})}^{\frac{3}{2}}]}_{0}^{1}

= {(1 + \frac{4}{9})}^{\frac{3}{2}}

= \frac{13 \sqrt{13}}{27}

s o t h e t o t a l p e r i m e t e r i s

2 \times \frac{13 \sqrt{13}}{27} + \frac{2 π \sqrt{2}}{4}

= \frac{26 \sqrt{13}}{27} + \frac{π \sqrt{2}}{2}

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

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