Question Number 80404 by peter frank last updated on 02/Feb/20
Answered by MJS last updated on 02/Feb/20
![f(x): y=(x^2 /5) ⇒ f^(−1) (y): x=(√(5y)) x′=((√5)/(2(√y))) surface=2π∫_a ^b f^(−1) (y)(√(1+(f^(−1) )′^2 ))dy π(√5)∫_5 ^(10) (√(4y+5))dy=π(√5)[(((4y+5)^(3/2) )/6)]_5 ^(10) =((25πw)/6)(27−5(√5)) +the bottom circle which is 25π ⇒ surface is ((25)/6)π(33−5(√5))=285.618... cost is surface×.02≈5.71](https://www.tinkutara.com/question/Q80406.png)
$${f}\left({x}\right):\:{y}=\frac{{x}^{\mathrm{2}} }{\mathrm{5}}\:\Rightarrow\:{f}^{−\mathrm{1}} \left({y}\right):\:{x}=\sqrt{\mathrm{5}{y}} \\ $$$${x}'=\frac{\sqrt{\mathrm{5}}}{\mathrm{2}\sqrt{{y}}} \\ $$$$\mathrm{surface}=\mathrm{2}\pi\underset{{a}} {\overset{{b}} {\int}}{f}^{−\mathrm{1}} \left({y}\right)\sqrt{\mathrm{1}+\left({f}^{−\mathrm{1}} \right)'^{\mathrm{2}} }{dy} \\ $$$$\pi\sqrt{\mathrm{5}}\underset{\mathrm{5}} {\overset{\mathrm{10}} {\int}}\sqrt{\mathrm{4}{y}+\mathrm{5}}{dy}=\pi\sqrt{\mathrm{5}}\left[\frac{\left(\mathrm{4}{y}+\mathrm{5}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }{\mathrm{6}}\right]_{\mathrm{5}} ^{\mathrm{10}} =\frac{\mathrm{25}\pi{w}}{\mathrm{6}}\left(\mathrm{27}−\mathrm{5}\sqrt{\mathrm{5}}\right) \\ $$$$+\mathrm{the}\:\mathrm{bottom}\:\mathrm{circle}\:\mathrm{which}\:\mathrm{is}\:\mathrm{25}\pi \\ $$$$\Rightarrow \\ $$$$\mathrm{surface}\:\mathrm{is} \\ $$$$\frac{\mathrm{25}}{\mathrm{6}}\pi\left(\mathrm{33}−\mathrm{5}\sqrt{\mathrm{5}}\right)=\mathrm{285}.\mathrm{618}… \\ $$$$\mathrm{cost}\:\mathrm{is}\:\mathrm{surface}×.\mathrm{02}\approx\mathrm{5}.\mathrm{71} \\ $$
Commented by peter frank last updated on 03/Feb/20
$${thank}\:{you}\:{sir} \\ $$