Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Give the volume of the solid generated by revolving the region bounded by the graph of $y = \ln(x)$, the $x$-axis, the lines $x = 1$ and $x = e$, about the $y$-axis.

Question 2 the solution guide gives $\frac{\pi}{2}\left(e^2+1\right)$ has the correct answer.

My work to a solution:

$y = \ln x \rightarrow e^y=x$ and changing the bounded values of $x$ to the respective $y$ values gives $e^y$ from 0 to 1.

Thus the integral of the area is $\int_0^1 \left(e^y\right)^2 \, \mathrm{d}y = \frac{\pi}{2}\left(e^2 -1\right)$

share|cite|improve this question
I got your answer $\frac{\pi}{2}(e^2-1)$ for $\int^1_0(e^y)^2dy$, I think something is wrong with your integral. – user67258 May 1 '13 at 11:29
You need to use the washer method (draw the picture; you're revolving the "triangular" region with "vertices" $(1,0)$, $(e,0)$, and $(e,1)$ about the $y$-axis). The integrand should be $\pi( e^2-(e^y)^2)$. – David Mitra May 1 '13 at 11:37
up vote 3 down vote accepted

Consider a volume element $\mathrm{d}V=A(y)\,\mathrm{d}y$. The area of the annulus generated by a revolving line segment from $x$ to $e$ is $\pi(e^2-x^2)$. Hence $\mathrm{d}V=\pi(e^2-x(y)^2)\,\mathrm{d}y=\pi(e^2-e^{2y})\,\mathrm{d}y$ and


share|cite|improve this answer

The correct integration is indeed $$ \int_0^1(\pi e^2- \pi x^2)\,\mathrm dy = \pi \int_0^1(e^2-e^{2y})\,\mathrm dy=\left.\pi\left(e^2y-\frac{e^{2y}}2\right)\right|_{y=0}^{y=1}=\frac{\pi}2(e^2+1).$$

(I got confused by the constraint $x\ge1$ as that is redundant by the constraints $y\le\ln x$ and $y\ge 0$; I should have read David Mitra's comment to the question).

The area to be rotated

share|cite|improve this answer
The region is bounded by the graph of $y=\ln x$, the $x$-axis, and the lines $x=1$ and $x=e$. It seems you're using the graph of $y=\ln x$, the line $y=1$, and the lines $x=1$ and $x=e$. Or, am I off? – David Mitra May 1 '13 at 11:57
Hm, actually the line $x=1$ is redundant and led me to the wrong track. – Hagen von Eitzen May 1 '13 at 12:29

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.