Find the surface area of the solid of revolution over the interval
1.$\displaystyle y=x^3, [0,2]$
Answer: $\displaystyle \pi(145\sqrt{145}-1)/27$
What I tried:
$\displaystyle \frac{dy}{dx}=3x^2$
Plug into formula for Surface Area of a solid of revolution about the x-axis:
$\displaystyle \int_a^b 2\pi y \sqrt{1+(\frac{dy}{dx})^2}dx$
$\displaystyle 2\pi\int_0^2 x^3\sqrt{1+(3x^2)^2}dx$
$\displaystyle 2\pi\int_0^2\sqrt{1+9x^4}x^3dx$
let $\displaystyle u=1+9x^4, \frac{du}{dx}=36x^3$
Convert limits: when$\displaystyle x=0, u=1$, when $\displaystyle x=2, u=289$
$\displaystyle \frac{2\pi}{36}\int_1^{289} \sqrt{u} du$
Integrate:
$\displaystyle \frac{\pi}{18}(\frac{2}{3}u^{\frac{3}{2}})\bigg|^{ 289}_1$
Plug in $\displaystyle u$, evaluate at limits, simplify.
I don't seem to be getting the answer that's in my book.