How do I find the surface area of this function $y = e^{-x^2}$ when it's rotated around the y-axis? So I have this question:

Set up an integral for the area of the surface obtained by rotating the curve about the y axis.

So I have the x-axis solution:
$y = e^{-x^2}$ from $ -1 \leq x \leq 1$
I'm going to use the formula:
$$\int_{-1}^1 2 \pi e^{-x^2} \sqrt{1 + \frac{dy}{dx}^2} dx$$
$\frac{dy}{dx} = e^{-x^2} * -2x$ and so $\frac{dy^2}{dx} = 4x^2e^{-2x^2}$
Is that right? Did I do the squaring right?
So surface area = $$\int_{-1}^1 2 \pi e^{-x^2} \sqrt{1 + 4x^2e^{-2x^2}}dx$$ 
But what about the y-axis?
 A: When rotating the curve defined by the function $y=e^{-x^2}$ (with $-1\leq x\leq 1$) about the $y$-axis, you need to restrict the values of $x$ so that you can write (one half piece of) the curve as $x=g(y)$ for some function $g$ and find the corresponding range for $y$. 
Then you use the formula
$$
\int_a^b2\pi y\sqrt{1+\left(\frac{dx}{dy}\right)^2}\ dy
$$
Now, to get the function $g$, you either consider $x\in[0,1]$ or $x\in[-1,0]$ for the function $y=f(x)$ since if you consider all $x\in[-1,1]$, then you can not write $x$ as a function of $y$. Let us consider $x\in[0,1]$. Then $y\in [e^{-1}, 1]$.
Then 
$$
\ln y=-x^2
$$
implies by the chain rule that
$$
\frac{1}{y}=-2x\cdot\frac{dx}{dy}=-2 (\sqrt{-\ln y}) \cdot\frac{dx}{dy}
$$
which tells you that
$$
\left(\frac{dx}{dy}\right)^2=\frac{1}{-4y^2\ln y}
$$
A: Yes, it seems about right for the rotation around the x-axis.
As to the y-axis, I assume you rotate the positive part of your curve ($0 \leq x \leq 1$) around the axis? 
If so, try to express $x$ as a function of $y$ (which is valid, because $x\mapsto e^{-x^2}$ is bijective on this interval). And then, it is going to be the same formula, except you now integrate $x$ as a function of $y$ instead of the opposite :)
