volume of solid

[mammothrob]

Im trying to graph the region bounded by y=e^x y=e^-x and x=2 about the x axis. my limits of intergration that im trying to are from e^2 and e^-2. I get these really ugly answers that are decimal? Am i using the wrong limits of integration? Should I be using exponential proerties rather than decimal ones? can anyone set up the integral for me.
Rob

 

[Unco]

You are rotating about the x-axis so you want x-limits (not y-limits). They are 0 and 2.

 

[soroban]

I get these really ugly answers that are decimal

Why are you surprised?
All "revolving" problems involve an integral with \(\displaystyle \pi\) in front.

 

[mammothrob]

okay. so I worked the problem out by revolving about the x axis using the washer type method. I went from 0 to 2. My answer was approx. 93.30. If someone else is interested in these types of problems and wants to verify this it would be cool. If not thanks for the previous help. Im going to try revolving about x with cylindricl shells, and that should get me the same answer.
Rob

 

[mammothrob]

by the way this is what my set up looked like

2
/
pi | [e^(2x) - e^(-2x)] dx
/
0

hope my notation is understandable
is there a tag for elongated s symbol used in intergration?

 

[Unco]

Your integral is spot on but your answer is a bit off. You should get 82.64...

Perhaps an integrating or calculator error?

Re. the integral symbol. You can use latex, with [ tex ] tags:

[tex]\pi \int^2_0 { \, e^{2x} \,  - \, e^{-2x} \, dx }[/tex]

gives

• \(\displaystyle \pi \int^2_0 { \, e^{2x} \, - \, e^{-2x} \, dx }\)

There may be a Forum Help menu at the top of the page (it doesn't show up on my screen, perhaps because of the skin I use). There should be some help on Latex there, or check out Karl's Notes. Your notation is legible, though.

 

[mammothrob]

so.....

from 0 to 2

pi * s(e^2x - e^-2x)dx

pi [ (e^2x+1/2x+1) - (e^-2x+1/(-2x+1)] 0 to 2

sub in X=2

pi [ (e^5/5) - (e^-3/-3) - (0) ] evaluated from 0 to 2

simplified I get somewhere in the neighborhood of 93 0r 94

I really have'nt spent much time with exponents... the only thing i can think of is that i made my own rules up somewhere, so i've busted put the pre-calc book from last year. uggg... i'm fireing up the coffee machine, its gonna be a long night here in san diego.

Thank you for the help so far, much appreciated.

 

[Unco]

\(\displaystyle \int \, e^{ax} \, dx \, = \, \frac{1}{a}e^{ax} \,\)!!!

Re. the coffee, that's how I've spent the last three weeks!

 

[mammothrob]

alright.... i got 82.649751.

Im stoked...

I'm kinda bummed at the same time though. My professor has never shown us that property S e^u = e^u. The property was right there in the front of my book though. we havent even intergrated (or differentiated for that matter) anything involving e or logs and such. Somewhat difficult for a weekend assignment I think, but who doesnt like a challange. Properties with e and logs are a chapter away.

Anways, thanks a bunch... my next goal is to revolve the same region about the y axis and about a line x=4 or something.... 1:30 am and counting.

 

[mammothrob]

I need to start reading the directions of my assignments.
I only had to set up the the intergral for x y and x=4 axis rotations for mondays lecture. I didnt have to solve the volumes yet. Im gonna finish the other two parts with solutions to the area anyways, who knows... mabey I score some extra credit.
This stuff rocks.