Which Volume Method?

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kneaiak

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"Find the volume of the solid of revolution. The region bounded by y=(1/x^2), y=0, x=2 and x=8."

I sketched the graph no problem but I'm confused as to which method to use. Is it as simply as the disk/washer method with respect to y? I started out with that but end up with 1/0 in a portion of my answer. I used 10 as my R and 2+(1/x^2) as my r. Any help would be great, I feel like a moron. My work so far........

int from 0 to 8 of (pi) ((10)^2 - ((2+(1/x^2))^2) dx

(pi) int from 0 to 8 of 100 - 4 - (2/x^2) - (1/x^4) dx

Integration:

I feel way off track
 
In that case, the shells method is more direct. It is possible to use the waster method, but you would need to split the region in two. For the shells method:

\(\displaystyle h(x) = x^{-2}\)
\(\displaystyle r(x) = x\)
\(\displaystyle 2\le x\le 8\)
 
ya. For some reason I'm stuck on wanting to use 0 to 8 and subtract 2 to 8 volume from it.
 
kneaiak said:
ya. For some reason I'm stuck on wanting to use 0 to 8 and subtract 2 to 8 volume from it.
Click to expand...
"0 to 8"? "2 to 8"? Do you mean "x between 0 and 8" and "x between 2 and 8"? You are told that the region is bounded by x= 2 and x= 8. The area you want is the area between x= 2 and x= 8. x= 0 to 2 has nothing to do with this.
 
I am just completely stuck on setting this thing up. I get that using the shell method my integral would be int from 2 to 8 of 2(pi)x(f(x) - g(x))dx but I can't seem to get what my f(x) and g(x) should be. Any help would be great.
 
kneaiak said:
I am just completely stuck on setting this thing up. I get that using the shell method my integral would be int from 2 to 8 of 2(pi)x(f(x) - g(x))dx but I can't seem to get what my f(x) and g(x) should be. Any help would be great.
Click to expand...

Could you not extract this information from my post?

\(\displaystyle h(x) = f(x)-g(x)\) where \(\displaystyle f(x)=x^{-2}\) and \(\displaystyle g(x)=0\).

You also understand that it won't always be \(\displaystyle 2\pi x (f(x)-g(x))\) right? This is only true for when the radius of revolution, \(\displaystyle r(x)\), is \(\displaystyle x\) and \(\displaystyle f(x)\ge g(x)\).
 
aaahhhh, that's what I'm getting stuck on with multiple problems, always thinking it's going to be f(x)-g(x). Sorry for all the confusion, been 9 years since I've taken this stuff. Thanks for all the help.
 
I come up with 2(pi) int from 2 to 8 x^-1 then when I integrate I get 2(pi) ln|x| evaluated from 2 to 8 gives me 2(pi) ln|8|-ln|2|, wrong. Where am I going wrong?!?!?!?!
 
kneaiak said:
I come up with 2(pi) int from 2 to 8 x^-1 then when I integrate I get 2(pi) ln|x| evaluated from 2 to 8 gives me 2(pi) ln|8|-ln|2|, wrong. Where am I going wrong?!?!?!?!
Click to expand...

Maybe it is your lack of parentheses? It can be simplified more also.

\(\displaystyle 2\pi(\ln|8|-\ln|2|) = 2\pi(\ln(8)-\ln(2)) = 2\pi(\ln(8/2)) = 2\pi\ln(4)\)
 
daon2 said:
Maybe it is your lack of parentheses? It can be simplified more also.

\(\displaystyle 2\pi(\ln|8|-\ln|2|) = 2\pi(\ln(8)-\ln(2)) = 2\pi(\ln(8/2)) = 2\pi\ln(4)\)
Click to expand...

And it can be taken further:

\(\displaystyle 2\pi\ln(4) \ = \ 2\pi\ln(2^2) \ = \ 4\pi\ln(2)\)
 
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