Find the volume of the solid of revolution, or state that it does not exist.

[abc1]

Find the volume of the solid of revolution, or state that it does not exist. The region bounded by f(x)= the square root of ((x+3)/(x^3)) and the x-axis on the interval [1,infinity) is revolved around the x-axis.

I tried using the disk method: pi* (sqrt(((x+3)/(x^3)))^2
Then I think I have to take the limit as b is approaching infinity from 1 to b of pi* (sqrt(((x+3)/(x^3)))^2. But I don't know how to take the limit now. Am I doing this problem correctly? Can someone please help me solve it?

 

[MarkFL]

We are given:

\(\displaystyle f(x)=\sqrt{\frac{x+3}{x^3}}\)

The volume of an arbitrary disk is:

\(\displaystyle dV=\pi r^2\,dx\)

where:

\(\displaystyle r=f(x)=\sqrt{\frac{x+3}{x^3}}\)

Hence:

\(\displaystyle dV=\pi\frac{x+3}{x^3}\,dx\)

And so:

\(\displaystyle V=\pi\int_1^{\infty}\frac{x+3}{x^3}\,dx\)

Since this is an improper integral, we may write:

\(\displaystyle V=\pi\lim_{t\to\infty}\left(\int_1^{t}\frac{x+3}{x^3}\,dx \right)\)

I would suggest rewriting the integrand:

\(\displaystyle V=\pi\lim_{t\to\infty}\left(\int_1^{t}x^{-2}+3x^{-3}\,dx \right)\)

Now, apply the FTOC and then take the limit of the result. Can you proceed?

 

[abc1]

Thank you so much for replying! I was just wondering, would it be possible to use lhopital's rule to find the limit since V=πlimt→∞(∫t1x+3x3dx) would be infinity over infinity? I tried that and I got 1/(3x^2) and then tried to apply the fundamental theorem of calculus, but I got the wrong answer, and I don't understand why.

 

[abc1]

Also, I tried proceeding from where you left off, applying the FTOC and I got pi * (lim as b approches infinity of (b^-2 +3b^-3) - 4. So then wouldn't that equal pi * ( infinity + 4) so it would be infinity so it would diverge?

 

[MarkFL]

No, it's not an indeterminate form...I would write:

\(\displaystyle V=\pi\lim_{t\to\infty}\left(\int_1^{t}x^{-2}+3x^{-3}\,dx \right)=\pi\lim_{t\to\infty}\left(\left[\frac{x^{-1}}{-1}+\frac{3x^{-2}}{-2} \right]_1^t \right)\)

\(\displaystyle V=-\pi\lim_{t\to\infty}\left(\left[\frac{1}{x}+\frac{3}{2x^2} \right]_1^t \right)=-\pi\lim_{t\to\infty}\left(\frac{1}{t}+\frac{3}{2t^2}-\frac{1}{1}-\frac{3}{2} \right)\)

Can you take the limit now?

 

[abc1]

Thanks so much! I got 5/2!

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I have one more question though. Why wasn't it an indeterminate form? It looked like it would be infinity over infinity.

 

[MarkFL]

Thanks so much! I got 5/2!

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I have one more question though. Why wasn't it an indeterminate form? It looked like it would be infinity over infinity.

Don't forget the factor of $\pi$. :D

Do you mean the integrand in its original form? The following is not true in general:

\(\displaystyle \int_a^b\frac{f(x)}{g(x)}\,dx=\frac{\int_a^b f(x)\,dx}{\int_a^b g(x)\,dx}\)

 

[abc1]

Oh okay! Thanks so much again for your help! :)