Volume by cylindrical shells: y =1/x^2+1, x=0, x=1, y=0

[hgaon001]

I'm having trouble with the integration in this problem.

Use cylindrical shells to find the volume of the solid generated when the region enclosed by the given curves is revolved about the y-axis.

y =1/x^2+1, x=0, x=1, y=0

 

[galactus]

Re: Volume by cylindrical shells

I assume that is $\displaystyle \frac{1}{x^{2}+1}$?.

Or is it $\displaystyle \frac{1}{x^{2}}+1$?.

The way you have it written means the latter.

The last one diverges, so we will assume that is not it.

That is why grouping symbols are important.

So, assuming the former, $\displaystyle 2{\pi}\int_{0}^{1}\frac{x}{x^{2}+1}dx$

To integrate, Let $\displaystyle u=x^{2}+1, \;\ \frac{du}{2}=xdx$

Then, it turns into an easy one to integrate.

 

[hgaon001]

Re: Volume by cylindrical shells

I assume that is $\displaystyle \frac{1}{x^{2}+1}$?.

Or is it $\displaystyle \frac{1}{x^{2}}+1$?.

The way you have it written means the latter.

The last one diverges, so we will assume that is not it.

That is why grouping symbols are important.

So, assuming the former, $\displaystyle 2{\pi}\int_{0}^{1}\frac{x}{x^{2}+1}dx$

To integrate, Let $\displaystyle u=x^{2}+1, \;\ \frac{du}{2}=xdx$

Then, it turns into an easy one to integrate.

Yes i got up to what you wrote $\displaystyle 2{\pi}\int_{0}^{1}\frac{x}{x^{2}+1}dx$

but i dont understand why when you integrate it they get (pi)ln(x^2+1) but i dont know what happens to the x that was on the top

 

[galactus]

Re: Volume by cylindrical shells

I just showed you all that. Make the suggested substitutions and you should see it.

 

[BigGlenntheHeavy]

Re: Volume by cylindrical shells

$\displaystyle 2\pi \int_{0}^{1}\frac{x \ dx}{x^{2}+1}$

$\displaystyle Let \ u \ = \ x^{2}+1, \ then \ du \ = \ 2xdx \ and \ index \ of \ integration \ goes \ from \ 1 \ to \ 2.$

$\displaystyle Ergo \ x \ dx \ = \ \frac{du}{2}, \ so \ \pi \int_{1}^{2} \frac{du}{u} \ = \pi ln|u|]_{1}^{2}.$

$\displaystyle Hence, \ V \ = \ \pi \ ln|2|.$

 

[hgaon001]

Re: Volume by cylindrical shells

$\displaystyle 2\pi \int_{0}^{1}\frac{x \ dx}{x^{2}+1}$

$\displaystyle Let \ u \ = \ x^{2}+1, \ then \ du \ = \ 2xdx \ and \ index \ of \ integration \ goes \ from \ 1 \ to \ 2.$

$\displaystyle Ergo \ x \ dx \ = \ \frac{du}{2}, \ so \ \pi \int_{1}^{2} \frac{du}{u} \ = \pi ln|u|]_{1}^{2}.$

$\displaystyle Hence, \ V \ = \ \pi \ ln|2|.$

Thank you so much! yea i wasn't thinking of using substitution that's why it was confusing. Makes perfect sense.. thanx again!

 

[galactus]

Re: Volume by cylindrical shells

Apparently the same sub I mentioned was a waste of time then?. Did you read it?. :roll:

 

[hgaon001]

Re: Volume by cylindrical shells

Apparently the same sub I mentioned was a waste of time then?. Did you read it?. :roll:

Thank you as well it was just easier when i saw it.. but no it was not a waste of time.. thanx again!

 

[Deleted member 4993]

Re: Volume by cylindrical shells

Apparently the same sub I mentioned was a waste of time then?. Did you read it?. :roll:

Thank you as well it was just easier when i saw it.. <<< you mean Completly worked out or ready for cut-and-paste or spoon-fed - don't you
but no it was not a waste of time.. thanx again!

 

[galactus]

Re: Volume by cylindrical shells

Just for kicks, do it in washers as well.

$\displaystyle {\pi}\int_{\frac{1}{2}}^{1}\left(\frac{1-y}{y}+1\right)dy$

 

[BigGlenntheHeavy]

Re: Volume by cylindrical shells

galactus, your above definite integral is one of the few that I can do in my head.

 

[galactus]

Re: Volume by cylindrical shells

Yes, because of the $\displaystyle \frac{1}{y}+1-1$?. It is obviously ln.

Good Glenn.

 

[galactus]

Re: Volume by cylindrical shells

Hey, let's revolve it about the x-axis now.

Washers:

$\displaystyle {\pi}\int_{0}^{1}\left(\frac{1}{1+x^{2}}\right)^{2}dx=\frac{{\pi}({\pi}+2)}{8}$

Shells:

$\displaystyle 2{\pi}\int_{\frac{1}{2}}^{1}\left[y\sqrt{\frac{1-y}{y}}+\frac{1}{4}\right]dy=\frac{{\pi}({\pi}+2)}{8}$