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

$$\mathbf{Find}\mathbf{the}\mathbf{perimeter}\mathbf{of}\mathbf{the}\mathbf{figure}\mathbf{which}\mathbf{is}$$$$\mathbf{bounded}\mathbf{with}\mathbf{the}\mathbf{curves}{\mathbf{y}}^3={\mathbf{x}}^2\mathbf{and}\mathbf{y}=\sqrt{2-{\mathbf{x}}^2}$$

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

$$y=x^{\frac{2}{3}}$$$$y=\sqrt{2-x^2}isacirclewithradius\sqrt{2}$$$$\sqrt{2-x^2}=x^{\frac{2}{3}}$$$$\Rightarrow x=\pm 1,y=1$$$$bothcurvesintersectat(-1,1),(1,1).$$$$lengthofcurvey=x^{\frac{2}{3}}forx=0to1:$$$${\int }_0^1\sqrt{1+{\left(y^{\prime }\right)}^2}dx$$$$={\int }_0^1\sqrt{1+{\left(\frac{2}{3}{x}^{-\frac{1}{3}}\right)}^2}dx$$$$={\int }_0^1\sqrt{1+\frac{4}{9x^{\frac{2}{3}}}}dx$$$$={\left[x{(1+\frac{4}{9x^{\frac{2}{3}}})}^{\frac{3}{2}}\right]}_0^1$$$$={(1+\frac{4}{9})}^{\frac{3}{2}}$$$$=\frac{13\sqrt{13}}{27}$$$$sothetotalperimeteris$$$$2\times \frac{13\sqrt{13}}{27}+\frac{2\pi \sqrt{2}}{4}$$$$=\frac{26\sqrt{13}}{27}+\frac{\pi \sqrt{2}}{2}$$

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

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