Finding the volume of a function rotating around the x-axis. y = xe^(-x^2) I was working on this question for a little while and I finally got it "done". However, I am not quite sure if what I did was correct or not. Here is the details:


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*Consider $y = xe^{-x^2}$ being rotated around the x-axis. The question says:


Does the resulting solid have a finite volume? If so, determine its volume.


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*My answer was yes, and here is what I've done:
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 A: Yep, you are correct that the volume is finite. The question is kinda ambiguous regarding the bounds, but since the range of the function is from negative infinity to positive infinity, my guess is that it is asking about the figure 
$$\int_{-\infty}^{\infty} \pi (xe^{-x^2})^2 dx $$
Which is equal to twice the answer you got from your work (but still finite). 
A: Starting from
$$I=\pi \int _0^{\infty} x^2e^{-2x^2}dx$$
we make a substitution $2x^2=t$ which gives $4xdx=dt$, or $dx=\dfrac{d t}{4x}$. From $2x^2=t$ we get $x^2=\dfrac{t}{2}$ and $x=\dfrac{t^{1/2}}{\sqrt{2}}$ and therefore $dx=\dfrac{\sqrt{2}}{4}t^{-1/2}dt$. Substituting gives
$$I=\pi \int _0^{\infty} x^2e^{-2x^2}dx=\pi \int _0^{\infty} \frac{t}{2}e^{-t}\dfrac{\sqrt{2}}{4}t^{-1/2}dt=\frac{\sqrt{2}\pi}{8}\int _0^{\infty }t^{1/2}e^{-t}dt=\frac{\sqrt{2}\pi}{8}\Gamma\left(\frac{3}{2}\right)$$
Where $\Gamma $ is the Euler gamma function. Using the property $\Gamma (n)=(n-1)\Gamma (n-1)$ we get $\Gamma \left(\dfrac{3}{2}\right)=\dfrac{1}{2}\Gamma \left(\dfrac{1}{2}\right)$ and the fact that $\Gamma \left(\dfrac{1}{2}\right)=\sqrt{\pi }$ we have $\Gamma \left(\dfrac{3}{2}\right)=\dfrac{\sqrt{\pi }}{2}$. Therefore
$$I=\frac{\sqrt{2}\pi}{8}\Gamma\left(\frac{3}{2}\right)=\frac{\sqrt{2}\pi}{8}\cdot \dfrac{\sqrt{\pi }}{2}=\frac{\sqrt{2}\pi ^{3/2}}{16}$$
