# My Dark Calculus II Secret

December 28, 2007 8:49 AM Subscribe

Calculus II filter: I have trouble setting up the integral for the volume of a solid of revolution when it is being revolved around a horizontal or vertical line other than the x- or y- axis. Does anybody know a trick or mnemonic for these types of problems?

I don't care if you use disks, washers or shells. I will be teaching this topic in the Spring for the first time in many years, and I have always had kind of a block for these types of problems, ever since I first saw them in high school. I am hoping somebody out there has a really clever way of helping their students memorize how to set up these problems that will work for me and that I can share with my students.

(If you are wondering how I dealt with this the last time I taught Calc II, I only assigned problems that revolved the area around either the x-axis or y-axis. I could do that again, but I would rather be victorious over these problems once and for all.)

I don't care if you use disks, washers or shells. I will be teaching this topic in the Spring for the first time in many years, and I have always had kind of a block for these types of problems, ever since I first saw them in high school. I am hoping somebody out there has a really clever way of helping their students memorize how to set up these problems that will work for me and that I can share with my students.

(If you are wondering how I dealt with this the last time I taught Calc II, I only assigned problems that revolved the area around either the x-axis or y-axis. I could do that again, but I would rather be victorious over these problems once and for all.)

Dunno where you're falling down and I'm sure you've read this a million times, so maybe this won't help but what the hell. You're probably missing just one tiny step.

Using rings: You're integrating A = pi * (outer radius^2 - inner radius^2)

WLOG let's look at rotation around x-axis (y = 0 = constant.) First, you need to define your bounding functions so that they're functions of the axis you're rotating around. So write your bounding functions as f(x) & g(x).

Outer radius is (f(x) - y0). Since y = constant = 0 then outer radius just = f(x). That's the tiny step I suspect you're shortcutting -- note here we can skip it b/c axis is @ y=0. So you can just integrate ((f^2 - g^2) w.r.t. x over your limits and Bob's your uncle. Simple as a pimple and you own this move already.

So... Now, let's rotate around y = constant = 1. Same idea, and as long as your functions are the same, in terms of x, all is good. But now that tiny step becomes critical. This time, outer radius = (f(x) - constant) = (f(x) - 1). Similarly for inner radius.

So your integrand this time is (outer radius^2 - inner radius^2) = ((f(x) - 1)^2 - (g(x) - 1)^2)) w.r.t. x over the limits. Just that one tiny step -- getting the radius is all that you need.

Now, I have a hard time keeping all the x's and y's straight when doing this, so let's try to come up with some simple rules to memorize or keep on a cheat sheet.

First, remember the basic integral is pi * ((outer radius)^2) - (inner radius)^2) and the task is to put these radii in a form relevant to the problem.

Next -- and this is where I have to stop and think -- are we rotating around a horizontal line or vertical line. If we're integrating around a horizontal line (y = constant,) make sure the bounding functions are in terms of x and integrate pi * (f(x) - constant)^2 - (g(x) - constant)^2) w.r.t x over your x-limits.

If we are rotating around a vertical line (x = constant) then you need to make sure the functions are in terms of y and integrate ((h(y) - constant)^2 - (k(y) - constant)^2) times pi w.r.t. y over y-limits.

So...

Basic integrand is pi * (outer radius^2 - inner radius^2)

For a

Make sure inner and outer bounding function f and g are f(

Then inner and outer radii are (f(

And you integrate w.r.t.

Swap x for y in the above if you have to deal with a vertical line.

You can probably see this is the same as shifting the origin. If not, work it out.

Hope this helps. Look at Paul's examples keeping the above in mind and maybe it'll make more sense.

posted by Opposite George at 9:42 AM on December 28, 2007 [2 favorites]

Using rings: You're integrating A = pi * (outer radius^2 - inner radius^2)

WLOG let's look at rotation around x-axis (y = 0 = constant.) First, you need to define your bounding functions so that they're functions of the axis you're rotating around. So write your bounding functions as f(x) & g(x).

Outer radius is (f(x) - y0). Since y = constant = 0 then outer radius just = f(x). That's the tiny step I suspect you're shortcutting -- note here we can skip it b/c axis is @ y=0. So you can just integrate ((f^2 - g^2) w.r.t. x over your limits and Bob's your uncle. Simple as a pimple and you own this move already.

So... Now, let's rotate around y = constant = 1. Same idea, and as long as your functions are the same, in terms of x, all is good. But now that tiny step becomes critical. This time, outer radius = (f(x) - constant) = (f(x) - 1). Similarly for inner radius.

So your integrand this time is (outer radius^2 - inner radius^2) = ((f(x) - 1)^2 - (g(x) - 1)^2)) w.r.t. x over the limits. Just that one tiny step -- getting the radius is all that you need.

Now, I have a hard time keeping all the x's and y's straight when doing this, so let's try to come up with some simple rules to memorize or keep on a cheat sheet.

First, remember the basic integral is pi * ((outer radius)^2) - (inner radius)^2) and the task is to put these radii in a form relevant to the problem.

Next -- and this is where I have to stop and think -- are we rotating around a horizontal line or vertical line. If we're integrating around a horizontal line (y = constant,) make sure the bounding functions are in terms of x and integrate pi * (f(x) - constant)^2 - (g(x) - constant)^2) w.r.t x over your x-limits.

If we are rotating around a vertical line (x = constant) then you need to make sure the functions are in terms of y and integrate ((h(y) - constant)^2 - (k(y) - constant)^2) times pi w.r.t. y over y-limits.

So...

Basic integrand is pi * (outer radius^2 - inner radius^2)

For a

**line (of the form***horizontal**= constant), everything should be in terms of***y****:***x*Make sure inner and outer bounding function f and g are f(

**x**) and g(**x**)Then inner and outer radii are (f(

**x**) - constant) and (g(**x**) - constant)And you integrate w.r.t.

**x**over given**x**limits.Swap x for y in the above if you have to deal with a vertical line.

You can probably see this is the same as shifting the origin. If not, work it out.

Hope this helps. Look at Paul's examples keeping the above in mind and maybe it'll make more sense.

posted by Opposite George at 9:42 AM on December 28, 2007 [2 favorites]

Um, rotate/shift the axis to whatever is convenient since those transformations preserve volume. They have to learn how to change variables anyway, don't they? The bounds are probably messy either way, but this way they're clean in one variable.

posted by a robot made out of meat at 9:50 AM on December 28, 2007

posted by a robot made out of meat at 9:50 AM on December 28, 2007

I wouldn't shift anything, because that's guaranteed to confuse your students.

I too teach this stuff.

You need to be drawing very clear pictures. The right picture, and any problem like this is easy to set up. Draw the solid of revolution, having revolved your curve around whatever the line given is, and make sure your drawing is nice and big---say, at least half a page of paper per picture. Then draw in---as clearly as you can---a sample shell or washer, depending on what seems right. In your shell, figure out what the radius is and what the height is: this should be reasonably straightforward. In your washer, figure out the inner and outer radii.

Once you've got this info, setting up the integral is just like any other volume of revolution: for shells, you integrate from the center of the shells to the edge, integrating the quantity 2 Pi r h (where you now know r and h, because you indicated them on your drawing!), and for washers, integrating Pi R^2 - P r^2, where you now know R and r, again because you drew them!

The advantage to this method is your students (and you) don't need to learn any new techniques, and you don't need to memorize any other formulas. You just need to have the right drawing. Seriously. Unfortunately, I can't figure out how to post you a picture to illustrate. But it's absolutely the same process as doing volumes of revolution about the x- or y-axis, it's just that when you draw your pictures, you need to work a little harder at determining the radii/heights.

posted by leahwrenn at 2:01 PM on December 28, 2007

I too teach this stuff.

You need to be drawing very clear pictures. The right picture, and any problem like this is easy to set up. Draw the solid of revolution, having revolved your curve around whatever the line given is, and make sure your drawing is nice and big---say, at least half a page of paper per picture. Then draw in---as clearly as you can---a sample shell or washer, depending on what seems right. In your shell, figure out what the radius is and what the height is: this should be reasonably straightforward. In your washer, figure out the inner and outer radii.

Once you've got this info, setting up the integral is just like any other volume of revolution: for shells, you integrate from the center of the shells to the edge, integrating the quantity 2 Pi r h (where you now know r and h, because you indicated them on your drawing!), and for washers, integrating Pi R^2 - P r^2, where you now know R and r, again because you drew them!

The advantage to this method is your students (and you) don't need to learn any new techniques, and you don't need to memorize any other formulas. You just need to have the right drawing. Seriously. Unfortunately, I can't figure out how to post you a picture to illustrate. But it's absolutely the same process as doing volumes of revolution about the x- or y-axis, it's just that when you draw your pictures, you need to work a little harder at determining the radii/heights.

posted by leahwrenn at 2:01 PM on December 28, 2007

*I am hoping somebody out there has a really clever way of helping their students memorize how to set up these problems*

And, of course, the less you(r students) have to memorize, the better off everyone will be! It's better to memorize techniques to approach these problems than actual formulas.

posted by leahwrenn at 2:02 PM on December 28, 2007

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I can't remember what they taught us in school, but I'm pretty sure I used this method regardless what we were told to do.

posted by backseatpilot at 9:01 AM on December 28, 2007