Integration help - Shell Method

**xfriendsonfirex** · February 14th 2010, 10:25 AM

Hey guys. Any help integrating this function would be much appreciated, so I can apply the shell method.

Area bound by the curves:

$y=\frac{e^{{-x^2}/2}}{\sqrt{2\pi}}$
$y=0$
$x=0$
$x=1$

I just don't know where to begin integrating that first function. Thanks!

**General** · February 14th 2010, 10:28 AM

Originally Posted by xfriendsonfirex

Hey guys. Any help integrating this function would be much appreciated, so I can apply the shell method.

Area bound by the curves:

$y=\frac{e^{{-x^2}/2}}{\sqrt{2\pi}}$
$y=0$
$x=0$
$x=1$

I just don't know where to begin integrating that first function. Thanks!

You want to know how to integrate $\int \frac{ e^{\frac{-x^2}{2}}}{\sqrt{2\pi}} \,\ dx$ ?

**xfriendsonfirex** · February 14th 2010, 10:30 AM

Yes please.

**General** · February 14th 2010, 10:32 AM

You can not integrate it in terms of functions we know.
see :
integrate ( e^[ (-x^2)/2 ] ) - Wolfram|Alpha

Post the whole question, please.

**Archie Meade** · February 14th 2010, 10:33 AM

Originally Posted by xfriendsonfirex

Hey guys. Any help integrating this function would be much appreciated, so I can apply the shell method.

Area bound by the curves:

$y=\frac{e^{{-x^2}/2}}{\sqrt{2\pi}}$
$y=0$
$x=0$
$x=1$

I just don't know where to begin integrating that first function. Thanks!

Hi,
Do you want to find the area of the function between the 3 lines,
or do you want to rotate the area about an axis (shells) ?

**xfriendsonfirex** · February 14th 2010, 10:36 AM

Originally Posted by General

Post the whole question, please.

Use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving the plane region about the y-axis.

$y=\frac{e^{{-x^2}/2}}{\sqrt{2\pi}}$ , y=0, x=0, x=1

**Jhevon** · February 14th 2010, 10:39 AM

Originally Posted by xfriendsonfirex

Use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving the plane region about the y-axis.

$y=\frac{e^{{-x^2}/2}}{\sqrt{2\pi}}$ , y=0, x=0, x=1

ok, then that function is not what you would integrate. what would you need to integrate using the Shell formula?

**General** · February 14th 2010, 10:41 AM

Originally Posted by xfriendsonfirex

Use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving the plane region about the y-axis.

$y=\frac{e^{{-x^2}/2}}{\sqrt{2\pi}}$ , y=0, x=0, x=1

So you will not integrate what i asked you.
Do you know the steps of solving suck problem?

**xfriendsonfirex** · February 14th 2010, 10:43 AM

What would I need to integrate then? Maybe I can make more sense of that.

**Jhevon** · February 14th 2010, 10:45 AM

Originally Posted by xfriendsonfirex

What would I need to integrate then? Maybe I can make more sense of that.

volume by Shells is given by

$V = 2 \pi \int_a^b (\text{radius})(\text{height})~dx$

have you seen this formula before? have you practiced setting up other problems like this?

**xfriendsonfirex** · February 14th 2010, 10:57 AM

$V=\int_0^1 x(\frac{e^{{-x^2}/2}}{\sqrt{2\pi}})dx$

Is this what I would be integrating? The x as the width, and the function as the height?

Sorry, I'm really slow with LaTeX. Never used it before.

**Jhevon** · February 14th 2010, 11:00 AM

Originally Posted by xfriendsonfirex

$V=\int_0^1 x(\frac{e^{{-x^2}/2}}{\sqrt{2\pi}})dx$

Is this what I would be integrating? The x as the width, and the function as the height?

Sorry, I'm really slow with LaTeX. Never used it before.

actually, you need to multiply that by $2 \pi$

now, as for integrating. do a substitution, $u = \frac {-x^2}2$

**xfriendsonfirex** · February 14th 2010, 11:06 AM

Okay, so I've gotten as far as this:

$y=-2\pi\int_0^1(\frac{e^u}{\sqrt{2\pi}})du$

Can I bring the $(2\pi)^{-1/2}$ in front of the integral, as it is a constant? If so, I can integrate the rest easily.

**Jhevon** · February 14th 2010, 11:12 AM

Originally Posted by xfriendsonfirex

Okay, so I've gotten as far as this:

$y=-2\pi\int_0^1(\frac{e^u}{\sqrt{2\pi}})du$

Can I bring the $(2\pi)^{-1/2}$ in front of the integral, as it is a constant? If so, I can integrate the rest easily.

yes, you can bring it out. and your limits would change actually. your integral becomes

$V = - \sqrt {2 \pi} \int_0^{- \frac 12}e^u~du$

Now finish up

**xfriendsonfirex** · February 14th 2010, 11:21 AM

Well, after integrating, I am getting the following:

$-\sqrt{2\pi}(1-e^{-1/8})$

However, my book key says I should be getting this:

$\sqrt{2\pi}(1-\frac{1}{\sqrt{e}})$

Any suggestions? Thanks for all the help so far.

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