Answer to Question #154226 in Calculus for mafizur alam

Question

find the surface area of the object obtained by rotating y=sin(2x), 0<=x<=pi/8 about the x axis

Answer 1

$Solution: ~The ~formula ~that ~we’ll ~be ~using ~here ~is, \\Surface ~Area=~S= \int2\pi yds~~\\ where ~ds=\sqrt{1+ (\frac{dy}{dx})^2} dx~\\Here~y=sin(2x), ~0\leqslant x \leqslant \frac{\pi}{8}$

$\therefore \frac{dy}{dx}=2~cos(2x)\\ \Rightarrow ds=\sqrt{1+ (\frac{dy}{dx})^2} dx=\sqrt{1+[2~cos(2x)]^2}dx=\sqrt{1+4~cos^2 (2x)}dx$

$Therefore ~ the~ integral ~ for~the~surface~area~is,\\S=\int2\pi yds.........................................[Substitute~the~values ~ of~ y ~and~ds ]\\ \therefore ~S=\int_{0}^{\frac{\pi}{8}}2\pi~ sin(2x)~\sqrt{1+4~cos^2 (2x)}dx...................................................(1)$

$Now~we~have~ to~solve~this~integral.\\ We ~use ~the~ following~ trigonometric~ substitutions$

$cos(2x)=\frac{1}{2}tan(\theta).....................................................................................(2) \\ \Rightarrow -2 ~sin(2x)dx=\frac{1}{2}~sec^2(\theta)d\theta \Rightarrow sin(2x)dx=-\frac{1}{4}sec^2(\theta)d\theta~and \\ \sqrt{1+4~cos^2 (2x)}=\sqrt{1+tan^2(\theta)}=\sqrt{sec^2(\theta)}=|sec~\theta|$

$Now~we~decide~the ~ Plus~or~minus~sign~(+~ or~-)~of~sec~\theta,\\we~need~to~convert~the~x~limits~to~\theta~limits~$

$Here ~we ~use~ the~ equation ~(2)$

$x=0: cos(0)=1=\frac{1}{2}~tan(\theta)\Rightarrow2=tan(\theta)\Rightarrow\theta=tan^{-1}(2)=1.1071$

$x=\frac{\pi}{8}: cos(2(\frac{\pi}{8}))=cos(\frac{\pi}{4})=\frac{1}{\sqrt{2}}=\frac{1}{\sqrt{2}}(\frac{\sqrt{2}}{\sqrt{2}})=\frac{\sqrt{2}}{2} =\frac{1}{2}~tan(\theta)~and\\\Rightarrow ~\sqrt{2}=tan~(\theta) ~\\ \Rightarrow \theta=tan^{-1}(\sqrt{2})=0.9553$

$\therefore ~The~corresponding~range~of~\theta~is~0.9553\leqslant~\theta~\leqslant~1.1071.\\This~is~in~the~first~Quadrant ~and~secant~is~positive~there. ~\\Therefore, we~use~positive~secant~value.$

$Putting ~all ~the ~trigonometric~ substitutions ~ in~ equation~(1),$

$S=\int_{0}^{\frac{\pi}{8}}2\pi~ sin(2x)~\sqrt{1+4~cos^2 (2x)}dx\\ \Rightarrow S=\int_{1.1071}^{0.9553}2\pi~ sin(2x)~sec(\theta)dx~~~~~~~~~~~~[Since~ \sqrt{1+4~cos^2 (2x)}=sec(\theta)] \\ \Rightarrow S=\int_{.1071}^{0.9553}2\pi~ [-\frac{1}{4}sec^2(\theta)]~sec(\theta)d\theta~~~~~[Since~ sin(2x)dx=-\frac{1}{4}sec^2(\theta)d\theta] \\ \\\ \Rightarrow S=\int_{1.1071}^{0.9553}~ [-\frac{2\pi}{4}sec^2(\theta)]~sec(\theta)d\theta$

$\therefore S=-\frac{\pi}{2}\int_{1..1071}^{0.9553}~ sec^3(\theta)d\theta$

$Using~the~formula~for~the~integral~of~sec^3(\theta),$

$\therefore S=-\frac{\pi}{4}[sec(\theta)~tan(\theta)+ln|sec(\theta)+tan(\theta)|] \Biggr|_{1.1071}^{0.9553} ~~~~~~ \\ [since~\int sec^3x=\frac{1}{2}(sec(x)~tan(x)+ln|sec(x)+tan(x)|]$

$Now, we~ just~put~ upper~and~lower~limits~in ~above ~equation ~and ~solve,which ~gives$$S=1.8215$

$\therefore The ~Surface ~area ~of~ the~ object~ is~1.8215.$