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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 $\int_0^1 \left(e^y\right)^2 \, \mathrm{d}y = \frac{\pi}{2}\left(e^2 -1\right)$

Siminore
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yiyi
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    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. –  May 01 '13 at 11:29
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    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 01 '13 at 11:37

2 Answers2

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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

$$V=\int_{y=0}^1\pi(e^2-e^{2y})\,\mathrm{d}y=\pi\left[e^2y-\frac{1}{2}e^{2y}\right]_{y=0}^1=\pi\left(e^2-\frac{1}{2}e^2\right)+\frac{\pi}{2}=\frac{\pi}{2}(e^2+1)$$

Librecoin
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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

Hagen von Eitzen
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  • 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 01 '13 at 11:57
  • Hm, actually the line $x=1$ is redundant and led me to the wrong track. – Hagen von Eitzen May 01 '13 at 12:29