Volume of revolution

[IssyAspen]

I'm having trouble with the following question:

Find the volume of the solid obtained by rotating the region under the curve

\(\displaystyle y = \dfrac{1}{x^2 + 1}\)

From 0 to \(\displaystyle \sqrt{e^3 - 1}\) about the y axis.

I assume it can be done using the washer method, because only one of my books has a description of the shell method, but all of the examples in the books have the area bounded by the graph, so I don't know how to set it up. All I know, I think, is that \(\displaystyle x^2 = \dfrac{1}{y} - 1\). It would be great If someone could point me in the right direction.

I did have a go at doing it using the shell method and got this: ( I can’t figure out how to show the limits, sorry).

\(\displaystyle 2\pi \displaystyle\int_{0}^{\sqrt{e^3 - 1}}\dfrac{x}{x^2 + 1} dx \)

\(\displaystyle 2\pi[\dfrac{1}{2}\ln(x^2 + 1)]_{0}^\sqrt{e^3 - 1}\)

\(\displaystyle \pi\ln\displaystyle[(\sqrt{e^3 - 1})^2 + 1] - 0\)

\(\displaystyle \pi\ln e^3\)

\(\displaystyle = 3\pi\)

Can someone please tell me where I have gone wrong? Thank you.

 

[Deleted member 4993]

I'm having trouble with the following question:

Find the volume of the solid obtained by rotating the region under the curve

\(\displaystyle y = \dfrac{1}{x^2 + 1}\)

From 0 to \(\displaystyle \sqrt{e^3 - 1}\) about the y axis.

I assume it can be done using the washer method, because only one of my books has a description of the shell method, but all of the examples in the books have the area bounded by the graph, so I don't know how to set it up. All I know, I think, is that \(\displaystyle x^2 = \dfrac{1}{y} + 1[/TEX] (that should be x² = 1/y - 1). It would be great If someone could point me in the right direction.

I did have a go at doing it using the washer method and got this: ( I can’t figure out how to show the limits, sorry).

\(\displaystyle 2\pi \displaystyle\int_{0}^{\sqrt{e^3 - 1}}\dfrac{x}{x^2 + 1} dx \)

\(\displaystyle 2\pi[\dfrac{1}{2}\ln(x^2 + 1)]_{0}^{e^3 - 1}\)

\(\displaystyle \pi\ln\displaystyle[(\sqrt{e^3 - 1})^2 + 1] - 0\)

\(\displaystyle \pi\ln e^3\)

\(\displaystyle = 3\pi\)

Can someone please tell me where I have gone wrong? Thank you.\)

\(\displaystyle

the elemental volume of an elemental washer is

\(\displaystyle dV = 2\ * \ \pi \ * \ x \ * \ y \ * dx\)

Then

\(\displaystyle V = 2\pi \displaystyle\int_{0}^{\sqrt{e^3 - 1}}\dfrac{x}{x^2 + 1} dx \)

Which is what you had finally - and as far as I can see your answer should be correct.\)

 

[IssyAspen]

the elemental volume of an elemental washer is

\(\displaystyle dV = 2\ * \ \pi \ * \ x \ * \ y \ * dx\)

Then

\(\displaystyle V = 2\pi \displaystyle\int_{0}^{\sqrt{e^3 - 1}}\dfrac{x}{x^2 + 1} dx \)

Which is what you had finally - and as far as I can see your answer should be correct.

Thank you, that +1 was a typo, but thanks for pointing it out, I've corrected it.

Thank you for checking the work.

As far as using the washer method, I think I would need the area bounded by:

the curve \(\displaystyle y = \dfrac{1}{x^2 + 1}\) the line \(\displaystyle x = \sqrt{e^3 - 1}\) and the y- axis, but how do I express that region so I can then rotate it round the y - axis, from 0 to 1?