Surface Area of Revolution

H

hingis

New member
Joined
Jul 11, 2006
Messages
6
Im working on some area of revolution problems and i think i have the main concepts right, but i might not be doing everything perfectly; im not quite sure. Here are some problems im trying to do to practice for my test:

#1Find the surface area; just set up integral:
y=x-x^3 on [0,1] about the x-axis

2 pi integral (from 0 to 1) (x-x^3)(2-6x^2+9x^4)^.5 dx

#2 Find the surface are
y=x^3 [1,2] about x-axis

My answer looks kinda weird.

2 pi integral x^3*(1+9x^4)^.5 dx
For this i used substitution is that where im going wrong?

pi/18 integral from 1 to 2 U^.5 = pi/18[2/3u^3/2] from 10 to 145?

#3 FInd surface area
x=1/8y^4 + 1/4y^(-2) [1,2] about x-axis

2pi integral (1/8y^4+1/4y^(-2))(1/2y^3+1/2y^(-3))

2pi[y^8/128 + 3y^2/32 + y^(-4)/-32) = 1179 pi/256

Thanks in advance for any help!
 
hingis said:
#2 Find the surface are
y=x^3 [1,2] about x-axis

My answer looks kinda weird.

2 pi integral x^3*(1+9x^4)^.5 dx
For this i used substitution is that where im going wrong?

pi/18 integral from 1 to 2 U^.5 = pi/18[2/3u^3/2] from 10 to 145?
Click to expand...
I'll help with this one, hingis.

You have:

\(\displaystyle \L\\2{\pi}\int_{1}^{2}x^{3}\sqrt{1+9x^{4}}dx\)

Use the substitution \(\displaystyle u=x^{4}\;\ du=4x^{3}dx\;\ \frac{du}{4}=x^{3}dx\)

Change limits of integration to:

\(\displaystyle (1)^{4}=1\;\ (2)^{4}=16\)

You now have:

\(\displaystyle \L\\\frac{\pi}{2}\int_{1}^{16}\sqrt{1+9u}du\)

Can you handle it now?.
 
hingis said:
i get why its pi/2 but why is the ^4 gone?
Click to expand...
Because "x<sup>4</sup>" was replaced with "u".

Are you not familiar with the method of u-substitution?

Eliz.
 
oh, sorry about that... i am familiar with it, i was just being dumb
 
Yes, the answers look OK, except, in #3 I believe you're revolving about the y-axis.
 
shoot. do i have to rearrange that equation and put it in terms of x? or can i leave it in y and still revolve about the x-axis?
 
aswimmer113 said:
do i have to rearrange that equation and put it in terms of x?
Click to expand...
Shouldn't "hingis" be asking these questions...?

Eliz.
 
aswimmer113 said:
shoot. do i have to rearrange that equation and put it in terms of x? or can i leave it in y and still revolve about the x-axis?
Click to expand...

No, I don't think so. It's just that your function is in terms of y and says revolve about the x-axis. The function should be in terms of x or left the way it is and revolved about the y-axis.

I doubt if they are expecting you to solve \(\displaystyle x=\frac{y^{4}}{8}+\frac{y^{-2}}{4}\) for y. That could prove to be quite daunting.
 
hmm.. this problem wasnt assigned, im just doing it to practice for my test, but the book does say revolved about the x-axis... :-/ ughhhhhhhh
 
You must log in or register to reply here.
Top