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Can someone please check my work.

$R$ is the region in the first quadrant bounded by $y=1/x$, $y=1$ and $x=e$

Find the volume of the solid generated when $R$ is revolved about the line $y=1$

Disk:

$$V= \int_{1}^{e} \pi \left(1-\frac{1}{x}\right)^2 \,dx=\pi \left(e-\frac{1}{e}-2\right)$$

Shell:

\begin{align*} V&=-2\pi \int_{1/e}^{1} -\left(1-\frac{1}{y}\right)\left(e-\frac{1}{y}\right)\, dy\\[5pt] &=-2\pi\left[ey-(1+e)\ln(y)-\frac{1}{y}\right]\Bigg|_{1/e}^{1}\\[5pt] &=-2\pi\bigl[(e-(1+e)\cdot 0-1)-(1-(1+e)(-1)-e)\bigr]\\[5pt] &=-2\pi\bigl[(e-1)-2\bigr]\\[5pt] &=-2\pi(e-3) \end{align*}

DMcMor
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user130306
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1 Answers1

8

Your working for the disk method is correct. In shell method, the integral should be,

$\displaystyle 2 \pi \int_{1/e}^{1} \color {blue} {(1-y)} (e-\frac{1}{y}) \ dy$

The mistake that you made was in the radial distance of the shell from $y=1$. Instead of $(1 - \frac{1}{y})$, it should be $(1-y)$ as the shell is at distance $y$ from the x-axis.

Math Lover
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