Area of a Surface of Revolution over Y-Axis

Question

I need to set up an integral for the area of a surface rotated over both the x-axis and the y-axis. Given $y = e^{-x^2}$ over the interval $-1 \leq x \leq 1$.

I'm having difficulty, however, finding this equation when rotating over the $y$-axis, particularly in terms of the limits of integration. Here's what I have so far.

The operative formula is $\int_a^b 2 \pi x \ ds$ with $ds = \sqrt{1 + \Big(\frac{dy}{dx}\Big)^2} \ dx$. Based on $y = e^{x^2}$, we can get \begin{align*} & f(x) = e^{x^2} \Rightarrow f'(x) -2x e^{-x^2} \Rightarrow \Big(\frac{dy}{dx}\Big)^2 = \Big(-2x e^{-x^2}\Big)^2 = 4x^2 e^{-2x^2} \\ \end{align*} which means that $ds$ is \begin{align*} ds = \sqrt{1 + \Big(\frac{dy}{dx}\Big)^2} = \sqrt{1 + 4x^2 e^{-2x^2}} \end{align*} Putting together our integral then gives us \begin{align*} S = \int_{-1}^1 2 \pi x \sqrt{1 + 4x^2 e^{-2x^2}} \end{align*} This answer, however, is incorrect; the correct limits of integration are from $0$ to $1$, and I cannot understand why this is the case. If anyone has any insight on this, I would greatly appreciate it.

Answer 1

Notice that the graph of $y = e^{-x^2}$ for $-1 \leq x \leq 0$ is a mirror image (over the $y$ axis) of the graph of $y = e^{-x^2}$ for $0 \leq x \leq 1.$

So if you just take the right-hand half of the curve, $y = e^{-x^2}$ for $0 \leq x \leq 1,$ and rotate it by $\pi$ radians around the $y$ axis, you end up with the left-hand half of the curve. In the meantime you have swept over half of the desired surface of rotation. If you then continue to rotate another $\pi$ radians, you return to the original position of the right-hand half, while sweeping through the other half of the surface of rotation.

If you take both sides of the curve $y = e^{-x^2}$ for $-1 \leq x \leq 1$ and rotate them through $2\pi$ radians around the $y$ axis, they will each sweep through parts of the surface that the other side of the curve swept through (or will sweep through) $\pi$ radians earlier or $\pi$ radians later. You'll end up with twice the correct answer because you counted every part of the surface twice.

To fix this, you could take the entire curve and rotate it only $\pi$ radians, which would sweep through the entire surface exactly once. You would then evaluate $$ S = \int_{-1}^1 \pi x \sqrt{1 + 4x^2 e^{-2x^2}} dx. $$ (Notice the factor of $2$ that occurs in the formula in the question but not in the formula above.) But I think most people think it's simpler to take just one half of the curve and rotate it through a full revolution of $2\pi$ radians so that the calculation is more like the calculation for other surfaces of rotation. In particular, notice that $x$ is the radius of the circle traced by a point $(x,y)$ when rotating that point around the $y$ axis, and integrating over negative values of the radius is a bit weird.