After watching Khan Academy, Org Chem tutor and a few others, my understanding is this: whether you solve in terms of $x$ and $y$ entirely depends on which axis you work with.

For shell - if you are working on $x$-axis, you draw your rectangle parallel to that axis. Because the width of this rectangle touches the $y$-axis, you thus work with $y$-terms. Similarly, if you are rotating against $y$-axis, then you are working with $x$-terms. For disc method, if you are rotating about the $x$ axis, then you work with $x$ terms - having drawn your disc perpendicular to that axis. Similarly, if rotating on $y$-axis, you draw your disc perpendicular to that and work with $y$-terms. Is this correct?

I also understand that both are interchangeable. One of the questions I attempted was this:

Find the volume of the solid generated by revolving the region bounded by $x= \frac{2}{y}$, and lines $y = 1$, $y= 4$, around the $y$-axis.

I first attempted with disc method and it worked:

$$\pi \int_1^4 \left(\frac{2}{y}\right)^2\, dy \quad \text{which was} \quad \frac{4}{y^2}$$

Then integrated this to get $-\frac{4}{y}$, and after substituting in the values I got $-1 - (-4)$ which got me $3$. Then multiplied by $\pi$ to get $3\pi$. This is correct.

Then I wanted to see if I could do it with shell method. I set up $2\pi$ and my integration limits as $2$ and $\frac{1}{2}$ (using $x= \frac{2}{y}$ to find out what my $x$ values would be when they intersected with $y= 1$ and $y= 4$ and I got $\frac{1}{2}$ and $2$ as my limits). So, my $R(x) H(x)$ was just $x\left(\frac{2}{x}\right)$, giving me $2$ to integrate. Then I get $2x$. Then substituting the values in I get $3$. However, I still need to multiply by the $2\pi$, giving me $6\pi$. How do I reconcile this?

  • $\begingroup$ I would encourage you to draw a picture to understand why you're getting such different results, as you're finding 2 very different volumes. In particular, do you know why the terms shell and disc (or washer) are used to describe the approach? $\endgroup$
    – Calvin Lin
    Commented Apr 29 at 16:08
  • $\begingroup$ Welcome to MSE! Please consider cleaning up your work with proper MathJax. I can gather that you found the volume via shells to be$$\int_\tfrac12^22\pi x \frac2x\,dx=\int_\tfrac12^24\pi x\,dx=6\pi$$(which is indeed not correct) but what about the part of the region where $0<x<\frac12$? $\endgroup$
    – user170231
    Commented Apr 29 at 16:43
  • $\begingroup$ Hi, I did draw a picture - my graph looked downward sloping, starting off quite high in the y axis and sloping down to the right with gradient decreases moving along the x-axis. Drawing it heklped me get 3pi, but not for shell. I'm not entirely sure what more to interpret with the graph as my integral limits check out. $\endgroup$
    – Anish Shah
    Commented Apr 30 at 1:47
  • $\begingroup$ @user170231 what do you mean by the part of the region where 0<x<1/2? Looking at my graph that's beyond the area for rotation as it becomes outside the bounds for the rotation, so I believe I just ignore that? $\endgroup$
    – Anish Shah
    Commented Apr 30 at 1:48

1 Answer 1


I strongly recommend that you read this answer first.

After reading it, your question has $x = f(y) = 2/y$, between $y = 1$ and $y = 4$, revolved about the $y$-axis.

  1. Is the interval of integration parallel or perpendicular to the axis of rotation?
  2. Based on your answer to (1), which method would you choose?
  3. What are your limits of integration?
  4. What is the differential volume of a representative shell, disk, or washer?

The correct integral setup for the disk method is:

$$V = \int_{y=1}^4 \pi (2/y)^2 \, dy.$$

The correct integral setup for the cylindrical shell method is:

$$V = 3\pi \left(\frac{1}{2}\right)^2 + \int_{x=1/2}^2 2 \pi x \left( \frac{2}{x} - 1 \right) \, dx.$$

Ask yourself why each method must be set up in the way shown. Draw a diagram to explain why.

  • $\begingroup$ Hi, thanks very much for the reply. I can understand fully with disc method. I already drew it out and it made sense - my intervals for disc being 4 and 1 on the y axis, and the corresponding x values being 2 and 0.5. I do want to ask, what is meant by the interval of integration? Does that just mean the limits of integration? I read the post you put and was surprised that they weren't interchangeable as I had believed they were. With perpendicularity I just draw the shells and discs when appropriate i.e. if shells, then draw your rectangle parallel to the y-axis, and for disc parallel to x $\endgroup$
    – Anish Shah
    Commented Apr 30 at 1:52
  • $\begingroup$ And if I could also inquire, I don't fully understand the integral set up for the shell method. I'm not entirely sure where 3pi(1/2)^2 comes from - we only get 3pi from disc method, but both are worked out independently of each other? I can understand most of the actual integration except for 2/x - 1. Surely if it was rotating against x axis then the -1 comes into play, because there is a constraint at y=1. Thus the need to subtract. But if we are rotating around y-axis, surely this doesn't matter? $\endgroup$
    – Anish Shah
    Commented Apr 30 at 1:57
  • $\begingroup$ @AnishShah "Interval of integration" refers to the interval $[a,b]$ over which a definite integral of the form $$\int_{x=a}^b f(x) \, dx$$ is evaluated, of an integrand $f(x)$ with respect to some variable $x$. As such, if $[a,b]$ is an interval defined on the $y$-axis (as in your case), this is parallel to the axis of rotation (in fact, it coincides with the axis of rotation). $\endgroup$
    – heropup
    Commented Apr 30 at 4:28
  • $\begingroup$ @AnishShah Regarding the shell method, the reason for $3\pi(1/2)^2$ has to do with the fact on the interval $x \in [0,1/2]$, the volume swept by rotating the region about the $y$-axis comprises a cylinder of radius $1/2$ and height $4 - 1 = 3$. Then on the interval $x \in [1/2, 2]$, the height of the cylindrical shell is equal to the difference of the function $y = 2/x$ and the line $y = 1$. This is why the shell has differential volume $dV = 2\pi x (2/x - 1) \, dx$. It has circumference $2\pi x$, and height $2/x - 1$. $\endgroup$
    – heropup
    Commented Apr 30 at 4:32
  • $\begingroup$ @AnishShah Failure to subtract $1$ will cause you to include the rectangular region $x \in [1/2, 2] \cap y \in [0,1]$. You cannot simply say "it doesn't matter." The function $y = 2/x$ represents the vertical distance from the $x$-axis to the curve. $\endgroup$
    – heropup
    Commented Apr 30 at 4:35

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