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Question Number 72789 by Learner-123 last updated on 02/Nov/19

$$Findtheareaofthesurfacegenerated$$$$byrevolvingthecurvex=\frac{y^4}{4}+\frac{1}{8y^2}$$$$aboutthex-axis.(given:1⩽y⩽2)$$

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

$$yes,sir!$$$$Inbooktheques.saysaboutthex-axis..$$

Commented by MJS last updated on 03/Nov/19

$${\mathrm{it}}^{\prime }\mathrm{s}\mathrm{not}\mathrm{possible}\mathrm{about}\mathrm{the}x-\mathrm{axis}$$$$\mathrm{we}\mathrm{would}\mathrm{need}\mathrm{to}\mathrm{transform}\mathrm{to}y=f\left(x\right)$$$$\mathrm{this}\mathrm{can}\mathrm{be}\mathrm{done}\mathrm{but}\mathrm{within}\mathrm{the}\mathrm{borders}$$$$\mathrm{given}\mathrm{for}y\mathrm{we}\mathrm{get}$$$$y=2\sqrt[4]{\frac{x}{3}}\sqrt{\cos (\frac{\pi }{6}+\frac{1}{3}\arcsin \frac{3\sqrt{3}}{32\sqrt[3]{x^2}})}$$$$\mathrm{and}\mathrm{we}\mathrm{cannot}\mathrm{easily}\mathrm{go}\mathrm{on}\mathrm{with}\mathrm{this}...\mathrm{try}$$$$\mathrm{if}\mathrm{you}\mathrm{like};-)$$$$2\pi \underset{\frac{3}{8}}{\stackrel{\frac{129}{32}}{\int }}y\sqrt{1+{\left(y^{\prime }\right)}^2}dx$$

Commented by Learner-123 last updated on 03/Nov/19

$$yes,iagree,thankssir.$$

Answered by MJS last updated on 02/Nov/19

$$\mathrm{about}\mathrm{the}y-\mathrm{axis}$$$$\mathrm{the}\mathrm{formula}\mathrm{is}$$$$\underset{a}{\stackrel{b}{\int }}x\sqrt{1+{\left(x^{\prime }\right)}^2}dy$$$$x=\frac{2y^6+1}{8y^2}\Rightarrow x^{\prime }=\frac{4y^6-1}{4y^3}$$$$2\pi \underset{1}{\stackrel{2}{\int }}\frac{2y^6+1}{8y^2}\sqrt{1+{\left(\frac{4y^6-1}{4y^3}\right)}^2}dy=$$$$=2\pi \underset{1}{\stackrel{2}{\int }}\frac{2y^6+1}{8y^2}\sqrt{\frac{16y^{12}+8y^6+1}{16y^6}}dy=$$$$=2\pi \underset{1}{\stackrel{2}{\int }}\frac{2y^6+1}{8y^2}\sqrt{{\left(\frac{4y^6+1}{4y^3}\right)}^2}dy=$$$$=2\pi \underset{1}{\stackrel{2}{\int }}\frac{2y^6+1}{8y^2}\times \frac{4y^6+1}{4y^3}dy=$$$$=\frac{\pi }{2}\underset{1}{\stackrel{2}{\int }}{y}^{7}dy+\frac{3\pi }{8}\underset{1}{\stackrel{2}{\int }}ydy+\frac{\pi }{16}\underset{1}{\stackrel{2}{\int }}\frac{dy}{y^5}=$$$$={[\frac{\pi }{16}{y}^8+\frac{3\pi }{16}{y}^2-\frac{\pi }{64y^4}]}_1^2=\frac{16911\pi }{1024}$$

Commented by Learner-123 last updated on 03/Nov/19

$$thankssir.$$

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