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I am trying to setup an integral to find the area of the surface of revolution of this function about the x-axis, but I am not sure of the steps to take .

$x = y + y^3$, $0 \leq y \leq 1$

I am not sure what steps to take to get the bounds for $x$ and to set the function up properly.

Bounds: Is it as simple as plugging in $0,1$ into y and solving for x such that $x = 0 + 0 = 0$ and $x = 1 + 1 = 2$?

Solving for y and dy:$$y = x - y^3$$

$$\begin{align} x &= y + y^3 \\ dx &= dy(1 + 3y^2) \\ \frac{dx}{(1+3y^2)} &= dy \\ \frac{1}{(1+3y^2)} &= \frac{dy}{dx} \end{align}$$

That leaves me with $$S = \int 2\pi y \sqrt{1+\bigg[ \frac{dy}{dx} \bigg]^2}dx \\ = \int_{0}^{2} 2\pi (x - y^3) \sqrt{1+\bigg[ \frac{1}{(1+3y^2)} \bigg]^2}dx$$

I feel that this integral isn't set up properly and that there shouldn't be a $y^3$ in the integral when integrating in relation to $x$.

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2 Answers

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Hint:

Note that if you use the already given parametric form of the rotated curve

• $(x(t),y(t)) = (t+t^3,t)$ for $t \in [0,1]$

you get immediately for rotation about the $x$-axis$$2\pi\int_0^1y(t)\sqrt{\dot x(t)^2 + \dot y(t)^2}\; dt = 2\pi\int_0^1 t\sqrt{(1+3t)^2 + 1}\; dt$$

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Integrate in terms of $y$ by using the transformation between $dx$ and $dy$.

$$\begin{align} S&=\int_0^1 2\pi y \sqrt{1+\dfrac{1}{(1+3y^2)^2}}(1+3y^2)dy \\ S&=\int_0^1 2\pi y \sqrt{(1+3y^2)^2+1}dy \end{align}$$

Let $u=1+3y^2$ such that $du=6ydy$ and by substitution

$$\begin{align} S&=\dfrac{\pi}{3}\int_1^4 \sqrt{u^2+1}du \end{align}$$

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