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Let $S$ be region enclosed by $y= 1/x$, $y=1$ and $y=2$. Goal is to find the volume of $S$ when it is revolved about $x= -1$

try:

Best to use washer method wrt y coordinate. A volume element is

$$ dV = ( \pi (1 + 1/y)^2 - \pi (-1)^2 ) dy $$ Thus,

$$ V = \int_1^2 (1 + 2/y + 1/y^2 - 1 ) dy = \int_1^2 (2/y + 1/y^2) dy $$

is this a correct setting for the required volume?

  • 0
    Also, in the integral, $(1+\frac {2}{y}+\frac {1}{y^2}-1) = (\frac {2}{y}+\frac {1}{y^2}) \neq (\frac {2}{y}-\frac {1}{y^2})$ – 2017-02-21
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    It is not a bounded region unless you include the line $x=0$ – 2017-02-21
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    yes, line x=0 is also included – 2017-02-21

1 Answers 1

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The shaded area below is the region $S$. enter image description here

Let us write $f(y)=\frac{1}{y}$. The outer radius is $f(y)+1$ and the inner radius is $1$. Thus, by using the washer method, $$dV=\pi\bigg(\big[f(y)+1\big]^2-1^2\bigg)dy$$ so that the volume is given by $$V=\int_{1}^2 dV=\int_{1}^2\pi\bigg(\big[f(y)+1\big]^2-1^2\bigg)dy=\pi\int_1^2 \left(\frac{2}{y} + \frac{1}{y^2}\right) dy.$$ Your only mistake is you introduced $(-1)^2$ and that you don't have $\pi$ in your integral.