Improper integrals and solids of revolution

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
2 replies · 2K views
phosgene
Messages
145
Reaction score
1

Homework Statement



Let n>1/2 and consider the function

[itex]f(x)=x^{-n}[/itex] for [itex]x\in[1,∞)[/itex]

Calculate the volume of the solid generated by rorating f(x) about the x-axis, showing all details of your working.

Homework Equations



Since it is rotated about the x-axis, its axis of symmetry is the x-axis, and by slicing up the solid of revolution into circles perpendicular to the x-axis, the volume is given by

[itex]V=lim_{t→∞}∫^{t}_{1}A(x)dx[/itex]
[itex]A= \pi r^{2}=\pi(x^{-n})^{2}[/itex]

So

[itex]V=lim_{t→∞}∫^{t}_{1}\pi(x^{-n})^{2}dx[/itex]

The Attempt at a Solution



[itex]V= \pi lim_{t→∞}∫^{t}_{1}(x^{-n})^{2}dx[/itex]


[itex]= \pi lim_{t→∞}∫^{t}_{1}(x^{-2n})dx[/itex]

[itex]= \pi lim_{t→∞} \frac{x^{-2n+1}}{-2n+1}|^{t}_{1}[/itex]

[itex]= \pi lim_{t→∞} \frac{1}{-2n+1}(t^{-2n+1}-1)[/itex]

[itex]= \frac{\pi}{-2n+1}[/itex]

So a few questions. First, am I right? Second, should I do another substitution with the limit and turn it into the limit as s goes to 0 where t=1/s? Appreciate any help :)
 
Physics news on Phys.org
phosgene said:

Homework Statement



Let n>1/2 and consider the function

[itex]f(x)=x^{-n}[/itex] for [itex]x\in[1,∞)[/itex]

Calculate the volume of the solid generated by rorating f(x) about the x-axis, showing all details of your working.

Homework Equations



Since it is rotated about the x-axis, its axis of symmetry is the x-axis, and by slicing up the solid of revolution into circles perpendicular to the x-axis, the volume is given by

[itex]V=lim_{t→∞}∫^{t}_{1}A(x)dx[/itex]
[itex]A= \pi r^{2}=\pi(x^{-n})^{2}[/itex]

So

[itex]V=lim_{t→∞}∫^{t}_{1}\pi(x^{-n})^{2}dx[/itex]

The Attempt at a Solution



[itex]V= \pi lim_{t→∞}∫^{t}_{1}(x^{-n})^{2}dx[/itex]


[itex]= \pi lim_{t→∞}∫^{t}_{1}(x^{-2n})dx[/itex]

[itex]= \pi lim_{t→∞} \frac{x^{-2n+1}}{-2n+1}|^{t}_{1}[/itex]

[itex]= \pi lim_{t→∞} \frac{1}{-2n+1}(t^{-2n+1}-1)[/itex]

[itex]= \frac{\pi}{-2n+1}[/itex]

So a few questions. First, am I right? Second, should I do another substitution with the limit and turn it into the limit as s goes to 0 where t=1/s? Appreciate any help :)

You are almost right. But if you put a number n>1/2 into that shouldn't you get something positive? And taking your limit works precisely because (-2n+1)<0, right? Just make sure you are clear on that. No need to do the limit another way.
 
Oops, you're right. I messed up the last step and forgot to multiply by the negative 1. Thanks :)