Volume by cylindrical shells: y =1/x^2+1, x=0, x=1, y=0
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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.
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.
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Re: Volume by cylindrical shells
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 said:
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
BigGlenntheHeavy
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Re: Volume by cylindrical shells
\(\displaystyle 2\pi \int_{0}^{1}\frac{x \ dx}{x^{2}+1}\)
\(\displaystyle Let \ u \ = \ x^{2}+1, \ then \ du \ = \ 2xdx \ and \ index \ of \ integration \ goes \ from \ 1 \ to \ 2.\)
\(\displaystyle Ergo \ x \ dx \ = \ \frac{du}{2}, \ so \ \pi \int_{1}^{2} \frac{du}{u} \ = \pi ln|u|]_{1}^{2}.\)
\(\displaystyle Hence, \ V \ = \ \pi \ ln|2|.\)
\(\displaystyle 2\pi \int_{0}^{1}\frac{x \ dx}{x^{2}+1}\)
\(\displaystyle Let \ u \ = \ x^{2}+1, \ then \ du \ = \ 2xdx \ and \ index \ of \ integration \ goes \ from \ 1 \ to \ 2.\)
\(\displaystyle Ergo \ x \ dx \ = \ \frac{du}{2}, \ so \ \pi \int_{1}^{2} \frac{du}{u} \ = \pi ln|u|]_{1}^{2}.\)
\(\displaystyle Hence, \ V \ = \ \pi \ ln|2|.\)
D
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Re: Volume by cylindrical shells
hgaon001 said:Thank you as wellgalactus said: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!
BigGlenntheHeavy
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Re: Volume by cylindrical shells
galactus, your above definite integral is one of the few that I can do in my head.
galactus, your above definite integral is one of the few that I can do in my head.
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:
\(\displaystyle 2{\pi}\int_{\frac{1}{2}}^{1}\left[y\sqrt{\frac{1-y}{y}}+\frac{1}{4}\right]dy=\frac{{\pi}({\pi}+2)}{8}\)
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:
\(\displaystyle 2{\pi}\int_{\frac{1}{2}}^{1}\left[y\sqrt{\frac{1-y}{y}}+\frac{1}{4}\right]dy=\frac{{\pi}({\pi}+2)}{8}\)
