What substitution can be used to evaluate the integral giving the area of the surface of revolution of the curve $x = \sqrt[3]{y}$?

Question

The question is:

Find the area of the surface generated by revolving the given curve about the $x$-axis: $$x = \sqrt[3]{y}, \qquad 1 \leq y \leq 8.$$

Now, all is well, simple enough question, but to find the length by evaluating the integral of $2\pi \times (\text{radius})$ with respect to \text{length}$, we see the final form of the integrand is

$$\sqrt{{\frac{9 y ^ {4/3}+1}{9 y ^ {4/3}}}}$$

Therefore, surface area of the graph is

$$(\text{surface area}) = \int_1^8 2\pi y \sqrt{{\frac{9 y ^ {4/3}+1}{9 y ^ {4/3}}}}dy$$

What should the substitution be to solve this integral?

(A little query: How to add a space in latex instead of $\texttt{\hspace{.1cm}}$, as $\texttt{\hfill}$ doesn't work here?)

Answer 1

Hint:Use $\tan u = 3y^{2/3} \to \tan^3 u= 27y^2 \to 3\tan^2 u \cdot sec^2 u = 54ydy$ .Can you take it from here?

Answer 2

Hint Solving for $y$ shows that we can instead write the curve as $$y = x^3, \quad 1 \leq x \leq 2,$$ so in terms of $x$ we can write the area element more tractably (in particular without nonintegral exponents) as $$dS = 2 \pi y \sqrt{1 + \left(\frac{dy}{dx}\right)^2} dx = 2 \pi x^3 \sqrt{1 + 9x^4} dx,$$ the form of which suggests an easy substitution.

Answer 3

Consider $$I = \int2\pi y \sqrt{{\frac{9 y ^ {4/3}+1}{9 y ^ {4/3}}}}dy$$ and make a first change of variable to get rid of these fractional exponents : $y ^ {4/3}=t$, $y=t^{3/4}$, $dy=\frac{3}{4 \sqrt[4]{t}}$; this makes $$I=\frac{1}{2} \pi \int \sqrt{1+9 t}\,dt$$ Now, get rid of the radical and you will end with a very simple integrand.

I am sure that ou can take from here.