Question Number 72789 by Learner-123 last updated on 02/Nov/19
$${Find}\:{the}\:{area}\:{of}\:{the}\:{surface}\:{generated} \\ $$$${by}\:{revolving}\:{the}\:{curve}\:{x}=\frac{{y}^{\mathrm{4}} }{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{8}{y}^{\mathrm{2}} }\: \\ $$$${about}\:{the}\:{x}−{axis}\:.\:\left({given}:\mathrm{1}\leqslant{y}\leqslant\mathrm{2}\right) \\ $$
Commented by MJS last updated on 02/Nov/19
$$\mathrm{about}\:\mathrm{the}\:{x}−\mathrm{axis}? \\ $$
Commented by Learner-123 last updated on 03/Nov/19
$${yes},{sir}! \\ $$$${In}\:{book}\:{the}\:{ques}.\:{says}\:{about}\:{the}\:{x}−{axis}.. \\ $$
Commented by MJS last updated on 03/Nov/19
$$\mathrm{it}'\mathrm{s}\:\mathrm{not}\:\mathrm{possible}\:\mathrm{about}\:\mathrm{the}\:{x}−\mathrm{axis} \\ $$$$\mathrm{we}\:\mathrm{would}\:\mathrm{need}\:\mathrm{to}\:\mathrm{transform}\:\mathrm{to}\:{y}={f}\left({x}\right) \\ $$$$\mathrm{this}\:\mathrm{can}\:\mathrm{be}\:\mathrm{done}\:\mathrm{but}\:\mathrm{within}\:\mathrm{the}\:\mathrm{borders} \\ $$$$\mathrm{given}\:\mathrm{for}\:{y}\:\mathrm{we}\:\mathrm{get} \\ $$$${y}=\mathrm{2}\sqrt[{\mathrm{4}}]{\frac{{x}}{\mathrm{3}}}\sqrt{\mathrm{cos}\:\left(\frac{\pi}{\mathrm{6}}+\frac{\mathrm{1}}{\mathrm{3}}\mathrm{arcsin}\:\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{32}\sqrt[{\mathrm{3}}]{{x}^{\mathrm{2}} }}\right)} \\ $$$$\mathrm{and}\:\mathrm{we}\:\mathrm{cannot}\:\mathrm{easily}\:\mathrm{go}\:\mathrm{on}\:\mathrm{with}\:\mathrm{this}…\:\mathrm{try} \\ $$$$\left.\mathrm{if}\:\mathrm{you}\:\mathrm{like}\:;−\right) \\ $$$$\mathrm{2}\pi\underset{\frac{\mathrm{3}}{\mathrm{8}}} {\overset{\frac{\mathrm{129}}{\mathrm{32}}} {\int}}{y}\sqrt{\mathrm{1}+\left({y}'\right)^{\mathrm{2}} }{dx} \\ $$
Commented by Learner-123 last updated on 03/Nov/19
$${yes},{i}\:{agree},\:{thanks}\:{sir}. \\ $$
Answered by MJS last updated on 02/Nov/19
![about the y−axis the formula is ∫_a ^b x(√(1+(x′)^2 ))dy x=((2y^6 +1)/(8y^2 )) ⇒ x′=((4y^6 −1)/(4y^3 )) 2π∫_1 ^2 ((2y^6 +1)/(8y^2 ))(√(1+(((4y^6 −1)/(4y^3 )))^2 ))dy= =2π∫_1 ^2 ((2y^6 +1)/(8y^2 ))(√((16y^(12) +8y^6 +1)/(16y^6 )))dy= =2π∫_1 ^2 ((2y^6 +1)/(8y^2 ))(√((((4y^6 +1)/(4y^3 )))^2 ))dy= =2π∫_1 ^2 ((2y^6 +1)/(8y^2 ))×((4y^6 +1)/(4y^3 ))dy= =(π/2)∫_1 ^2 y^7 dy+((3π)/8)∫_1 ^2 ydy+(π/(16))∫_1 ^2 (dy/y^5 )= =[(π/(16))y^8 +((3π)/(16))y^2 −(π/(64y^4 ))]_1 ^2 =((16911π)/(1024))](https://www.tinkutara.com/question/Q72793.png)
$$\mathrm{about}\:\mathrm{the}\:{y}−\mathrm{axis} \\ $$$$\mathrm{the}\:\mathrm{formula}\:\mathrm{is} \\ $$$$\underset{{a}} {\overset{{b}} {\int}}{x}\sqrt{\mathrm{1}+\left({x}'\right)^{\mathrm{2}} }{dy} \\ $$$${x}=\frac{\mathrm{2}{y}^{\mathrm{6}} +\mathrm{1}}{\mathrm{8}{y}^{\mathrm{2}} }\:\Rightarrow\:{x}'=\frac{\mathrm{4}{y}^{\mathrm{6}} −\mathrm{1}}{\mathrm{4}{y}^{\mathrm{3}} } \\ $$$$\mathrm{2}\pi\underset{\mathrm{1}} {\overset{\mathrm{2}} {\int}}\frac{\mathrm{2}{y}^{\mathrm{6}} +\mathrm{1}}{\mathrm{8}{y}^{\mathrm{2}} }\sqrt{\mathrm{1}+\left(\frac{\mathrm{4}{y}^{\mathrm{6}} −\mathrm{1}}{\mathrm{4}{y}^{\mathrm{3}} }\right)^{\mathrm{2}} }{dy}= \\ $$$$=\mathrm{2}\pi\underset{\mathrm{1}} {\overset{\mathrm{2}} {\int}}\frac{\mathrm{2}{y}^{\mathrm{6}} +\mathrm{1}}{\mathrm{8}{y}^{\mathrm{2}} }\sqrt{\frac{\mathrm{16}{y}^{\mathrm{12}} +\mathrm{8}{y}^{\mathrm{6}} +\mathrm{1}}{\mathrm{16}{y}^{\mathrm{6}} }}{dy}= \\ $$$$=\mathrm{2}\pi\underset{\mathrm{1}} {\overset{\mathrm{2}} {\int}}\frac{\mathrm{2}{y}^{\mathrm{6}} +\mathrm{1}}{\mathrm{8}{y}^{\mathrm{2}} }\sqrt{\left(\frac{\mathrm{4}{y}^{\mathrm{6}} +\mathrm{1}}{\mathrm{4}{y}^{\mathrm{3}} }\right)^{\mathrm{2}} }{dy}= \\ $$$$=\mathrm{2}\pi\underset{\mathrm{1}} {\overset{\mathrm{2}} {\int}}\frac{\mathrm{2}{y}^{\mathrm{6}} +\mathrm{1}}{\mathrm{8}{y}^{\mathrm{2}} }×\frac{\mathrm{4}{y}^{\mathrm{6}} +\mathrm{1}}{\mathrm{4}{y}^{\mathrm{3}} }{dy}= \\ $$$$=\frac{\pi}{\mathrm{2}}\underset{\mathrm{1}} {\overset{\mathrm{2}} {\int}}{y}^{\mathrm{7}} {dy}+\frac{\mathrm{3}\pi}{\mathrm{8}}\underset{\mathrm{1}} {\overset{\mathrm{2}} {\int}}{ydy}+\frac{\pi}{\mathrm{16}}\underset{\mathrm{1}} {\overset{\mathrm{2}} {\int}}\frac{{dy}}{{y}^{\mathrm{5}} }= \\ $$$$=\left[\frac{\pi}{\mathrm{16}}{y}^{\mathrm{8}} +\frac{\mathrm{3}\pi}{\mathrm{16}}{y}^{\mathrm{2}} −\frac{\pi}{\mathrm{64}{y}^{\mathrm{4}} }\right]_{\mathrm{1}} ^{\mathrm{2}} =\frac{\mathrm{16911}\pi}{\mathrm{1024}} \\ $$
Commented by Learner-123 last updated on 03/Nov/19
$${thanks}\:{sir}. \\ $$