Question

find the surface area of the surface formed by rotation of the arc of the curve y=(x^5/5+1/(12x^3)), 1≤x≤2, around y axis.

Answer 1

I dont know how to integrate --- the last step

Answer 2

\(\Huge\color{yellow}{\bigstar}\Huge\color{Lime}{Hello}\Huge\color{yellow}{\bigstar}\)

Answer 3

well give me a sec to figure it out

Answer 4

that's a square inside the radical....

Answer 5

\(\color{#000000 }{ \displaystyle 2\pi \int\limits_{1}^{2}x\sqrt{x^8+\frac{1}{16x^8}+\frac{1}{2} }~dx }\)

\(\color{#000000 }{ \displaystyle 2\pi \int\limits_{1}^{2}x\sqrt{\frac{16x^{16} }{16x^8}+\frac{1}{16x^8}+\frac{8x^8}{16x^8} }~dx }\)

\(\color{#000000 }{ \displaystyle 2\pi \int\limits_{1}^{2}x\sqrt{\frac{16x^{16}+8x^8+1 }{16x^8} }~dx }\)

\(\color{#000000 }{ \displaystyle 2\pi \int\limits_{1}^{2}x\sqrt{\frac{16x^{16}+8x^8+1 }{16x^8} }~dx }\)

\(\color{#000000 }{ \displaystyle \frac{\pi}{2} \int\limits_{1}^{2}\frac{x\sqrt{(4x^8+1)^2} }{x^4} ~dx }\)

\(\color{#000000 }{ \displaystyle \frac{\pi}{2} \int\limits_{1}^{2}\frac{x(4x^8+1) }{x^4} ~dx }\)

\(\color{#000000 }{ \displaystyle p=x^2 }\)
\(\color{#000000 }{ \displaystyle (1/2)~dp=x~dx }\)
\(\color{#000000 }{ \displaystyle x=1\quad \Longrightarrow \quad p=1^2=1 }\)
\(\color{#000000 }{ \displaystyle x=2\quad \Longrightarrow \quad p=2^2=4 }\)

\(\color{#000000 }{ \displaystyle \frac{\pi}{4} \int\limits_{1}^{4}\frac{4p^4+1 }{p^2} ~dp }\)

\(\color{#000000 }{ \displaystyle \frac{\pi}{4} \int\limits_{1}^{4} 4p^2+p^{-2} ~dp }\)

\(\color{#000000 }{ \displaystyle \pi \int\limits_{1}^{4} p^2~dp+\frac{\pi}{4}\int\limits_{1}^{4}p^{-2} ~dp }\)