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?
Best Answer
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 $$ \left(\frac{dx}{dy}\right)^2=\frac{1}{-4y^2\ln y} $$