MHB Surface Area of Revolution with Double Integration

[Edward2022]

Please Help Me

View attachment 9079

 

[skeeter]

ok ... one more time

$\displaystyle S = 2\pi \int_a^b f(x) \sqrt{1+[f(x)]^2} \, dx$

rotating $f(x) = \sin{x}$ over the interval $[-\pi, 5\pi/4]$ is equivalent to rotating twice over the interval $[0,\pi]$ plus once over the interval $[0, \pi/4]$

so ...

$\displaystyle S = 4\pi \int_0^\pi \sin{x} \sqrt{1+\cos^2{x}} \, dx + 2\pi \int_0^{\pi/4} \sin{x} \sqrt{1+\cos^2{x}} \, dx$

let $u = \cos{x} \implies du = -\sin{x} \, dx$

$\displaystyle S = 4\pi \int_{-1}^1 \sqrt{1+u^2} \, du + 2\pi \int_{1/\sqrt{2}}^1 \sqrt{1+u^2} \, du$

let $u = \tan{t} \implies du = \sec^2{t} \, dt$

$\displaystyle S = 4\pi \int_{-\pi/4}^{\pi/4} \sqrt{1+\tan^2{t}} \sec^2{t} \, dt + 2\pi \int_{\arctan(1/\sqrt{2})}^{\pi/4} \sqrt{1+\tan^2{t}} \sec^2{t} \, dt$

$\displaystyle S = 4\pi \int_{-\pi/4}^{\pi/4} \sec^3{t} \, dt + 2\pi \int_{\arctan(1/\sqrt{2})}^{\pi/4} \sec^3{t} \, dt$

since the secant function is even ...

$\displaystyle S = 8\pi \int_0^{\pi/4} \sec^3{t} \, dt + 2\pi \int_{\arctan(1/\sqrt{2})}^{\pi/4} \sec^3{t} \, dt$

note integrating $\sec^3{t}$ by parts ...

$\displaystyle \int \sec^3{t} \, dt = \dfrac{1}{2} \bigg[\sec{t}\tan{t} + \ln|\sec{t}+\tan{t}| \bigg]$

so ...

$S = 4\pi \bigg[ \sec{t}\tan{t} + \ln|\sec{t}+\tan{t}| \bigg]_0^{\pi/4} + \pi \bigg[ \sec{t}\tan{t} + \ln|\sec{t}+\tan{t}| \bigg]_{\arctan(1/\sqrt{2})}^{\pi/4}$

$S = 5\pi\bigg[\sqrt{2}+\ln(\sqrt{2}+1)\bigg] - \pi \bigg[\dfrac{\sqrt{3}}{2} + \ln\left(\dfrac{\sqrt{3}+1}{\sqrt{2}}\right) \bigg] \approx 31.27$

sorry for the previous deleted posts ... too many opportunities to make careless errors, especially after consuming a few beers.