Surface area of revolution of curve

Question

I am wondering why this particular integration is being found difficult to solve. Would appreciate any help I can get. the graph is $y = x^3$ and the limits are $0 \leq y \leq 1$

Answer 1

If the problem is "Find the surface area of generated by $y=x^3$ , $0 \le y \le 1$ about the $y-$axis" then the setup is $\int_0^1 2 \pi x \sqrt{1+(\frac{d(x^3)}{dx} )^2}$ or $ \int_0^1 2 \pi y^\frac{1}{3} \sqrt{1+(\frac{d(y^\frac{1}{3})}{dy})^2} dy$. But if the problem is "Find the surface area of generated by $x=y^3$ , $0 \le y \le 1$ about the $y-$axis" then the setup is $\int_0^1 2 \pi y^3 \sqrt{1+(\frac{d(y^3)}{dy})^2} dy$ or $\int_0^1 2 \pi x \sqrt{1+\frac{d(x^\frac{1}{3})}{dx})^2} dx$. Though this setup $\int_0^1 2 \pi y^3 \sqrt{1+(\frac{d(y^3)}{dy})^2} dy$ is way more gorgeous. $(\frac{d(y^3)}{dy})^2=(3y^2)^2=9y^4$ and if $u=1+9y^4$ so $du=36y^3 dy$ which can be written as $\frac{1}{18} du=2y^3dy$ So we can write the integral as $\int_1^{10} \pi \sqrt{u} \frac{1}{18} du=\frac{\pi}{18} \int_1^{10} u^\frac{1}{2} du$

Answer 2

Since the OP has put effort into his work, I'll go ahead and put mines. I suppose you're accustom to functions of the form $y = f(x)$. Though it is not much different from graphing functions of the form $x = h(y)$.

Solving this problem, you have to solve for the line integral

\begin{align*} \int 2\pi x ds \tag{1} \end{align*}

(don't worry about it if you haven't heard of that terminology). But here's one way to do this. You have that

\begin{align*} ds & = \sqrt{\left(\frac{dx}{dy}\right)^2 + 1} \tag{2} \end{align*}

and you have the function $x = y^3$. Plug in these two. And you have your integral is

\begin{align*} \int 2\pi x ds & = \int_0^1 2\pi y^3 \sqrt{9y^4+1} dy \end{align*}

You can either do this in several ways, but I like $u$-substitution. So let $u = 9y^4 + 1$ and we have that $\frac{du}{18} = 2y^3dy$. So you'll get

\begin{align*} \frac{\pi}{18}\int_1^{10} \sqrt{u}du & = \frac{\pi}{27}[10\sqrt{10} - 1] \end{align*}