Rotate $y=e^{-x^2}$ about the y-axis to find the volume. 
Since we are rotating around the y-axis, my intuition is that I need to change my original function in terms of y instead of $x$. So to change $y=e^{-x^2}$, I should end with $x=(-ln(y))^{(1/2)}$. At first I thought this could cause a problem, but then I realized that my original function only has y values going form $0$ to $1$, so I thought I should be okay. Then to find volume, I should do the $\pi(Int[0,1])((-ln(y))^{(1/2)})^2 dy$. Which just ends up being $-\pi(Int[0,1])(ln(y))dy$. However, you cannot evaluate the integral of $ln(y)$ from 0 to 1 because there ends up being a part that is undefined. Therefore, I feel that I must have done something wrong along the way. I am teaching a calculus recitation but it has been a few years since I've dealt with these types of problems. I know that this may be solved using the shell method, however, the class requires specifically that they use the disk method for this problem
Thanks in advance.
 A: We do indeed want 
$$\int_0^1 -\pi \ln y\,dy.\tag{1}$$
The logarithm blows up as $y$ approaches $0$ from the right, so we want to find (if it exists)
$$\lim_{\epsilon\to 0^+} \int_\epsilon^1-\pi\ln y\,dy.\tag{2}$$
An antiderivative of $\ln y$ is $y\ln y-y$ (integration by parts). The integral (2) is therefore equal to 
$$-\pi\left[((1)\ln 1-1)-(\epsilon\ln \epsilon-\epsilon)                       \right].$$ 
Finally, take the limit as $\epsilon$ approaches $0$ from the right. We need to know that $\lim_{\epsilon\to 0} \epsilon\ln \epsilon=0$. This can be proved using L'Hospital's Rule, or in various other ways. 
A: Note that the highest part the curve is $(0, e^{-1})$, so our desired volume is:
$$
-\pi\lim_{k \to 0^+}\int_k^{1/e} \ln y \, dy
= -\pi\lim_{k \to 0^+}\left[y\ln y - y \right]_k^{1/e}
= \frac{2\pi}{e}
$$
A: Let $$I = \int_{-\infty}^\infty e^{-x^2}dx$$ Notice that our function is even, so $$\int_{-\infty}^\infty e^{-x^2}dx = 2\int_0^\infty e^{-x^2}dx$$ Next, consider $$2\int_0^\infty e^{-y^2}dy$$ Convince yourself that $$2\int_0^\infty e^{-x^2}dx=2\int_0^\infty e^{-y^2}dy$$ which means that $$I^2 = \left(2\int_0^\infty e^{-x^2}dx\right) \left(2\int_0^\infty e^{-x^2}dx\right) \\ =\left(2\int_0^\infty e^{-x^2}dx\right) \left(2\int_0^\infty e^{-y^2}dd\right) \\ =4\int_0^\infty \int_0^\infty e^{-(x^2+y^2)}dxdy $$ where we are guaranteed equality in the last line because $I$ is always positive. Lastly, we will now switch to polar coordinates. We have $x^2+y^2=r^2$ with $dxdy = rd\theta dr$ and our integral transforms to $$=4\int_0^\infty \int_0^{\frac{\pi}{2}} e^{-r^2}rd\theta dr$$ At this point the integral can be completed via a simple $u$-sub. I recommend integrating with respect to $\theta$ first, and using symmetry properties the integral will only require one improper integration instead of two.
You should get a volume of $\pi$ at the end.
A: Note that you get the full solid from just rotating the curve $e^{-x^2}$, $0\le x<\infty$ about the $x$-axis.
You can now find the volume using the shell method:
$V=\int_0^\infty 2\pi xe^{-x^2}dx$.  This integral can be done reasonably routinely, perhaps by a $u$-substitution. 
