Volume formed by revolving the region between y = 1/x and y = 1, y = 3 about the y-axis

Key takeaway: This region is easier to see if you switch viewpoints and treat x as a function of y. That turns the y-axis rotation into a straightforward disk integral.

We revolve the region bounded by $y=\frac1x$, $x=0$, $y=1$, and $y=3$ about the y-axis.

1) Rewrite the curve as $x$ in terms of $y$

From

$$y=\frac1x,$$

we get

$$x=\frac1y.$$

For each $y$ between 1 and 3, the region runs from $x=0$ to $x=1/y$.

2) Use the disk method

Rotating around the y-axis gives disks of radius

$$R(y)=\frac1y.$$

So

$$V=\pi\int_1^3 \left(\frac1y\right)^2 dy$$

3) Integrate

$$V=\pi\int_1^3 y^{-2}dy =\pi\left[-\frac1y\right]_1^3$$ $$V=\pi\left(1-\frac13\right)=\frac{2\pi}{3}$$

Correct choice: C.

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Pitfalls the pros know 👇 A common mistake is using $2\pi\int x\,dy$ here even though the region touches the axis of rotation, so the disk method is cleaner. Another slip is forgetting to rewrite $y=1/x$ as $x=1/y$ before integrating with respect to $y$.

What if the problem changes? If the axis of rotation were the x-axis, you would need to express the boundaries differently and the integral would not look the same. If the top bound were $y=b$ instead of $3$, the volume would become $\pi\int_1^b y^{-2}dy = \pi\left(1-\frac1b\right)$.

Tags: disk method, inverse function, y-axis rotation