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Question Number 80404 by peter frank last updated on 02/Feb/20

Answered by MJS last updated on 02/Feb/20

$$f\left(x\right):y=\frac{x^2}{5}\Rightarrow f^{-1}\left(y\right):x=\sqrt{5y}$$$$x^{\prime }=\frac{\sqrt{5}}{2\sqrt{y}}$$$$\mathrm{surface}=2\pi \underset{a}{\stackrel{b}{\int }}{f}^{-1}\left(y\right)\sqrt{1+{\left(f^{-1}\right)}^{\prime 2}}dy$$$$\pi \sqrt{5}\underset{5}{\stackrel{10}{\int }}\sqrt{4y+5}dy=\pi \sqrt{5}{\left[\frac{{(4y+5)}^{\frac{3}{2}}}{6}\right]}_5^{10}=\frac{25\pi w}{6}(27-5\sqrt{5})$$$$+\mathrm{the}\mathrm{bottom}\mathrm{circle}\mathrm{which}\mathrm{is}25\pi$$$$\Rightarrow$$$$\mathrm{surface}\mathrm{is}$$$$\frac{25}{6}\pi (33-5\sqrt{5})=285.618...$$$$\mathrm{cost}\mathrm{is}\mathrm{surface}\times .02\approx 5.71$$

Commented by peter frank last updated on 03/Feb/20

$$thankyousir$$

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