# Thread: Integration help - Shell Method

1. ## Integration help - Shell Method

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

Area bound by the curves:

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

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

2. 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:

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

I just don't know where to begin integrating that first function. Thanks!
You want to know how to integrate $\displaystyle \int \frac{ e^{\frac{-x^2}{2}}}{\sqrt{2\pi}} \,\ dx$ ?

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

5. 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:

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

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) ?

6. Originally Posted by General
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.

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

7. 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.

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

8. 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.

$\displaystyle y=\frac{e^{{-x^2}/2}}{\sqrt{2\pi}}$, y=0, x=0, x=1
So you will not integrate what i asked you.
Do you know the steps of solving suck problem?

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

10. 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

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

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

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

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.

12. Originally Posted by xfriendsonfirex
$\displaystyle V=\int_0^1 x(\frac{e^{{-x^2}/2}}{\sqrt{2\pi}})dx$

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 $\displaystyle 2 \pi$

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

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

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

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

14. Originally Posted by xfriendsonfirex
Okay, so I've gotten as far as this:

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

Can I bring the $\displaystyle (2\pi)^{-1/2}$ 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

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

Now finish up

15. Well, after integrating, I am getting the following:

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

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

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

Any suggestions? Thanks for all the help so far.

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