# Volume by cyndrical shell

Gold Member
I think I've got it set up correctly, but I'm stuck on the integration.

Use the method of cylindrical shells to find the volume generated by rotating the region bounded by y=e^(-x^2), x=0, x=1, rotated about the y axis.

$$\begin{array}{l} \int\limits_{\rm{0}}^{\rm{1}} {\pi r^2 h\,\,dx} = \int\limits_{\rm{0}}^{\rm{1}} {\pi x^2 e^{ - x^2 } \,\,dx} = \\ \\ \pi \int\limits_{\rm{0}}^{\rm{1}} {x^2 e^{ - x^2 } \,\,dx} \\ \end{array}$$

I just don't know how to anti-deriv xe^(-x^2). I've tried substitution with u= -x2 with no joy. I'll bet this is easy, but it's got me stumped.

## The Attempt at a Solution

Wiemster
So if I understand correctly you rotate a Gaussian around the y-axis and want to know the volume enclosefd up to a distance 1 from the y-axis?

If that is the case I would reconsider the first integral you wrote down. keep in mind that you want to 'add' a lot of infinetesimally high cilinders (volume pi x^2 dy) and use the relation you have between x and y to change variables.

Good luck!

PS:
I just don't know how to anti-deriv xe^(-x^2).

Homework Helper
I think I've got it set up correctly, but I'm stuck on the integration.

Use the method of cylindrical shells to find the volume generated by rotating the region bounded by y=e^(-x^2), x=0, x=1, rotated about the y axis.

$$\begin{array}{l} \int\limits_{\rm{0}}^{\rm{1}} {\pi r^2 h\,\,dx} = \int\limits_{\rm{0}}^{\rm{1}} {\pi x^2 e^{ - x^2 } \,\,dx} = \\ \\ \pi \int\limits_{\rm{0}}^{\rm{1}} {x^2 e^{ - x^2 } \,\,dx} \\ \end{array}$$

I just don't know how to anti-deriv xe^(-x^2). I've tried substitution with u= -x2 with no joy. I'll bet this is easy, but it's got me stumped.
First, $xe^{-x^2}$ is simple to integrate, as Wiemster says, but your problem is integrating $x^2e^{-x^2}$ which is whole different matter.

Your real problem is that your formula, $\pi r^2h dx$, is wrong. It can't possibly be a differential of volume- it has units of distance4! A cylindrical shell has volume $2\pi r$, the circumference around the shell, times h, times the thickness,dy= dx. In this case, the thickness is dr, the height is the y value, $h= e^{-x^2}$, and the radius is r= x.
That does, in fact, reduce to
$$\pi \int_0^1 xe^{-x^2}dx$$
and the substitution $u= x^2$ works nicely.

Gold Member
I just don't know how to anti-deriv xe^(-x^2).

Sorry, I forgot to square x. I meant the anti-deriv of x^2*e^(-x^2).

AngeloG
radius = function; in this case e^(-x^2). But is it x*e^(-x^2) or e^(-x^2)?
h = dy or dx

If it's x*e^(-x^2), simple u substitution if it's the other it's just it's own derivative however you get ugly numbers.

$$\begin{array}{l} \int\limits_{\rm{0}}^{\rm{1}} {\left[ {{\rm{circumference}}} \right]\left[ {{\rm{height}}} \right]\left[ {{\rm{thickness}}} \right]} \\ \\ {\rm{circumference}} = 2\pi r,\,\,\,r = x \\ {\rm{height}} = e^{ - x^2 } \\ {\rm{thickness}} = \Delta x \\ \\ \int\limits_{\rm{0}}^{\rm{1}} {2\pi rh\,\,dx} = \int\limits_{\rm{0}}^{\rm{1}} {2\pi xe^{ - x^2 } \,\,dx} = \\ \\ 2\pi \int\limits_{\rm{0}}^{\rm{1}} {xe^{ - x^2 } \,\,dx} ,\,\,\,\,u = - x^2 ,\,\,\frac{{du}}{{dx}} = - 2x,\,\,dx = \frac{{du}}{{ - 2x}} \\ \\ 2\pi \int\limits_{\rm{0}}^{\rm{1}} {xe^{ - x^2 } \,\,dx} = 2\pi \int\limits_{\rm{0}}^{\rm{1}} {xe^u \,\,\frac{{du}}{{ - 2x}}} = \frac{{2\pi }}{{ - 2}}\int\limits_{\rm{0}}^{\rm{1}} {e^u \,\,du} = - \pi \int\limits_{\rm{0}}^{\rm{1}} {e^u \,\,du} = \\ \\ \left. { - \pi e^u } \right]_0^1 = \left. { - \pi e^{ - x^2 } } \right]_0^1 = \left( { - \pi e^{ - 1^2 } } \right) - \left( { - \pi e^{ - 0^2 } } \right) = - \pi \left( {e^{ - 1} - 1} \right) = \pi \left( { - e^{ - 1} + 1} \right) =\pi \left( {1 - 1/e} \right) \\ \end{array}$$
$$\begin{array}{l} \int_0^1 {} \\ and \\ \int\limits_0^1 {} \\ \end{array}$$