Disk vs Shells Volume Strategy

Question

In response to the question "When do you use the shell method vs the washer method?," a former math professor gave the answer "If the region is bounded by functions of x, put the element in vertically and if by functions of y, put the element in horizontally. Once you have put in the element, look at whether it forms a washer (disk) or shell when it rotates about the axis of revolution." Let's say my task is to find the volume enclosed rotating the area enclosed by $y=\sqrt{x}$, $y=0$, and $x=4$ around the $y$-axis. The definition of a function of $x$ is that $y$ depends upon the value of $x$. $y=\sqrt{x}$ is surely a function of $x$ but as for $y=0$, I am not sure if it is a function of $x$ or $y$. The same goes with $x=4$. Thus, I am not sure whether to put in the element vertically or horizontally for this problem.

Answer 1

For the purpose of applying the professor's advice,

• $y = 0$ gives $y$ as a function of $x$ (which happens to be a constant function).
• $x = 4$ gives $x$ as a function of $y$ (which happens to be a constant function).
• $y = \sqrt x$ gives $y$ as a function of $x,$ but the same curve is described by $x = y^2$ where $y \geq 0,$ which gives $x$ as a function of $y$.

It turns out that either shells or washers will work for this problem, and with approximately equal complexity. (The amount of extra complexity you get by making the "wrong" choice seems not worth spending any time worrying about in this case; you'll waste more time trying to decide than you would have spent doing the more "complicated" integral.)

What I like to do is actually plot the curves and identify the region of the $x,y$ plane that they bound. Then look at whether I would like to take vertical elements or horizontal elements. There are all kinds of situations that can come up that will make one way more complicated that the other, for example:

• Horizontal elements near the top of the region are bounded by the same curve at both ends, so I would have to cut the curve into a "left" part and a "right" part with a different function for each.
• Horizontal elements near the top of the region are bounded by a pair of curves, but horizontal elements near the bottom are bounded by a different pair of curves, so I'll need to write two integrals, one for the top of the region and one for the bottom.
• The same situation with horizontal elements near the bottom and top, but some elements between the top and bottom are bounded by yet other pairs of curves, so I need three or more integrals to get the volume using horizontal elements.

Vertical elements can have similar complications, just consider the ones at the left or right ends of the region rather than the bottom or top.

In your example, all horizontal elements are bounded by the same pair of curves and the third curve is just the last element (so it gives you one end of one of the integrals). A similar thing is true for the vertical elements. So both methods are straightforward to write as a single integral.

And there may be times when you look at the region and one way seems obviously best, but then when you have set up the integrals they look awful, so you try the other way instead, which might be easier (e.g. it has two integrals instead of one but the integrals are much easier to do) or it turns out even worse so you go back to the first way and slog through it.

If there were a simple, quick-to-apply rule that always gave the simplest answer, we wouldn't need to do so many exercises to get good at this!

Answer 2

Note that $y=\sqrt{x}$ and $x=y^2$ are equivalent. Whether you view it as a function of $x$ or $y$ is relative. In many cases, you could either use the shell or the washer/disk method to integrate volumes and you would get the same result.

For your specific case, you could integrate elther way.

1) Washer/disk method:

$$V=\int_0^2 \pi (4^2 - x^2) dy= \pi\int_0^2(16-y^4)dy= \frac{128}5\pi$$

2) Shell method,

$$V=\int_0^4 2\pi x y \>dx = 2\pi \int_0^4 x\sqrt x dx = \frac{128}5\pi$$

Answer 3

There is not exactly a fixed rule as when to use disks and when to use washers.

Generally, whey $y$ is expressed as a function of $x.$ (which is most common) And you are rotating around the x axis, use disks/ washers. And when rotating around the y axis use shells.

But often it can be done both ways just as easily. And sometimes, it is, in fact, better to switch up from the established rule.

First step is to make a picture of the region.

At this point we pick, which seems easier? Is there anything that might force me to break-up the region if I integrate with respect to x or with respect to y?

In this case, no. Both methods are about equally complicated. Shells looks a little bit easier.

Using shells, $y = 0$ forms the bottom of our verticals, $y= \sqrt x$ forms the top.

$\sqrt x - 0 = \sqrt x$ make the height of each cylinder wall.

Rotating around the y axis, $x$ is the radius of each cylinder

$V = 2\pi \int_0^4 x\sqrt x \ dx$

If you want to use washers you can.

$x = y^2$ makes the inner radius of each washer $x = 4$ is the outer radius.

The limit of integration -- we will integrate to $y = 2$ as that marks the point where $y = \sqrt x$ intersects $x = 4$

$V = \pi \int_0^2 4^2 - y^4 \ dy$