Calculus problem involving the disk method
September 29, 2010 3:47 PM   Subscribe

The area between f(x), the x-axis and the lines x=a and x=b is revolved around the x-axis. The volume of this solid of revolution is b^3-a*b^2 for any a,b. What is f(x)?

First of all, this is a (starred) problem in a popular calculus text. I haven't been a student in years. This isn't homework.

I've been staring at this problem for longer than I care to admit.

The problem is from a set of end-of-chapter review problems. It's from a group of problems within that set which are all related to the section of the chapter devoted to the "disk method." The only integration technique we've already seen is substitution. The only functions we can integrate are powers of x, sin and cos.

I included the next three paragraphs to show that I've done some work on the problem. Unfortunately, I've spent far more time than this work would suggest. I have no reason to believe that any of the ideas contained therein will be helpful in any way. Frankly, I doubt it. Hopefully, if you know what's going on here you'll also be able to explain to me what's wrong down in the very last paragraph. It's easy to check and see that the function doesn't work, but I don't see where I made a mistake.

I've tried to reverse engineer the problem by guessing f's and looking for a pattern.

I fiddled with the function f(b)=int_a^b f(x)^2 dx. My brain is so baked by this problem, I don't really remember what went wrong with that.

I defined the function F(x)=int f(x)^2 dx. Then
int_a^(a+h) f(x)^2= F(a+h)-F(a).
But, from the formulation of the problem, it also equals (a+h)^3-(a+h)^2 * a
Set those equal, divide on both sides by h. Take the limit and as h->0 and you should get F'(x)=x^2. But apparently not, because that would mean that f(x)=x and I wouldn't be typing this if that was actually the answer.
posted by stuart_s to Science & Nature (27 answers total) 1 user marked this as a favorite
 
Your function F(x) isn't well defined. (You use x as the dummy variable of integration as well as the argument for the function F; this doesn't really make sense.) If you're looking for the volume of the solid using the disk method, you need a factor of pi in there as well. (The area of a circle is pi times the radius squared, the radius at the point x will be the value of f at x.)

I'm thinking you're going to want to use the fundamental theorem of calculus somehow, so give me a few to work on it. I have to go teach in about a half hour, this will give me something to do until then.
posted by King Bee at 4:01 PM on September 29, 2010


OK, here we go.

Set F(t) = \int_a^t \pi [f(x)]^2 dx. We are given that F(t) = t^3 - at^2.

Using the first fundamental theorem of calculus, we have that

F'(t) = \pi [f(t)]^2

on the one hand, and on the other, we have

F'(t) = 3t^2 - 2at.

Set the two equal to each other and solve for f(t).
posted by King Bee at 4:06 PM on September 29, 2010


Response by poster: I don't think that works. If I'm understanding you correctly, setting those two things equal will yield a solution for f(t) that involves a. a and b are both variables in this situation. They can't appear in the definition of the function f(t).
posted by stuart_s at 4:18 PM on September 29, 2010


Response by poster: Also, I don't agree with your analysis of my not-a-solution. Work through it again using a different variable if you don't like the argument and the dummy variable matching. The dummy variable is just that, a dummy variable, so I don't think it affects the logic. There's a mistake in there, but I don't think that's it.
posted by stuart_s at 4:25 PM on September 29, 2010


Setting b to 0 implies that the volume of the solid of revolution is 0 for any choice of a. This means that the function must be zero everywhere, which doesn't satisfy the requirements. No such function can exist; I'd go check the textbook really does say this.
posted by 7-7 at 4:52 PM on September 29, 2010 [1 favorite]


Well, it's cheating, but it may be a typo and it's supposed to be b^3-a^3, in which case f(x)=\sqrt(3)x/\sqrt(\pi) works (I think). It doesn't seem to me that there is a function in terms of only x that will yield the volume given in the statement of the problem, but I'm happy to be shown wrong. Have you checked the website of the publisher to see if there's a listing of errata?
posted by monkeymadness at 4:54 PM on September 29, 2010


7-7: It could have the added assumption of b>=a>=0, but without it stated that way you're right.
posted by monkeymadness at 4:56 PM on September 29, 2010


Best answer: Even with b>=a>=0 there are problems. Pick a point c between a and b and notice that the whole solid of revolution with volume b^3-a*b^2 is the sum of the 2 solids of revolution with volumes b^3-c*b^2 and c^3-a*c^2 and that this must hold for all a, b and c.
posted by Obscure Reference at 5:20 PM on September 29, 2010


I am think that there is an error in this as well. I'm trying to figure out the best way to show it, but let's say we have a function f such that it solves the problem. f(x)^2*pi, integrated from a to b, equals b^3-a*b^2 Now, if you flip the bounds of an integral, it's equal to the negative of the integral, which means that f(x)^2*pi, integrated from b to a, should equal -(a^3-b*a^2). This and the earlier should be equal. Let's analyze that.

b^3-a*b^2=a^3-b*a^2
factoring
b^2*(b-a)=a^2*(a-b)

Given b^2 and a^2 are both positive, this can only ever be true if a and b are equal.

Something is wrong with the answer as given. I mean if you think about it, imagine if you integrate from -a to b. Your volume will always be 0! so your function is 0 until you have b>0. Doing little tricks like that you can see why it breaks down.
posted by wooh at 5:56 PM on September 29, 2010


Response by poster: Crap! Sorry. Yes, b>a. I think that does mean that f(x)=0 for x<0>
I'm also pretty sure that this is not a typo. The typo monkeymadness hypothesizes makes the problem too easy. This is a starred problem. Other starred problems have been difficult. That leaves the possibility that the author has some fundamental misunderstanding of the theory involved, but my money is on the proposition that there's a meaningful answer to the question as stated above and amended here.

wooh, if you switch the limits of integration then you won't satisfy the new criteria. Sorry. Sorry. Sorry.

Please come back, everyone!
posted by stuart_s at 6:59 PM on September 29, 2010


Ok, here is another reason why I don't think it works.

We know that for x<0>
Consider the integral from 0 to b. The area is b^3. Now consider the integral from -b to b. The area is 2b^3...and yet, we know that it should be b^3 as well, because f(x)=0 for x<0>
It also doesn't make sense that b has to be greater than a, because it's just an ordinary integral like any other. It would violate rules about integration.
posted by wooh at 7:12 PM on September 29, 2010


We know that for x<0, f(x)=0. sorry for the typo there.
posted by wooh at 7:12 PM on September 29, 2010


Response by poster: Double crap!

Obscure Reference's objection to the problem as stated seems to be pretty solid.

That still leaves me wondering what the problem was intended to say. It can't be something as simple as monkeymadness suggested.

Any thoughts?
posted by stuart_s at 7:16 PM on September 29, 2010


The volume of any solid of revolution defined in this way, as a function of a and b, should have the form g(b) - g(a) for some function g. This is because f(x) is the integral from a to b of π f(x)2 dx.

But your claimed volume doesn't have this form. This is "obvious", but to give a proof that there's not just some clever representation I'm missing: the volume between 0 and 1 is, according to your formula, is 1. The volume between 1 and 2 is 4. So the volume between 0 and 2 should be 1 + 4 = 5 -- but your formula gives 8!

Monkeymadness's guess that it should be b^3 - a^3 seems reasonable.
posted by madcaptenor at 7:29 PM on September 29, 2010


Response by poster: Please come back for my question next week: What is George Simmons' home address and is he seriously allergic to anything?

Thanks, everyone.
posted by stuart_s at 7:30 PM on September 29, 2010


a and b are both variables in this situation. They can't appear in the definition of the function f(t).

This is not quite correct. You could view b as a variable with respect to your function F, where you integrate from a to b, but with respect to f(x) and the solid of rotation, a and b are just numbers. There is no reason why f(x) can't depend explicitly on a and or b, and the mixed terms of the volume hint that it must be so.

King Bee has the correct approach. I expect the problem is starred due to the direct use of the fundamental theorem (which is surprisingly uncommon in lower level calculus classes) and possibly the trickiness in dealing with numbers vs. variables.
posted by abc123xyzinfinity at 7:31 PM on September 29, 2010


Aha! If f(x) can depend on a and b, then this works:

f(x) = -b/(2√π) x-1/2

F(x) = -b/(√π) x1/2

F(x)2 = (b2/π)x

Volume of rotation from a to b

= π[F(b)2- F(a)2] = b2(b-a)
posted by 7-7 at 8:27 PM on September 29, 2010


Another option: f(x) does not exist, but f(x,b) does. If f(x,b)=b / sqrt{\pi} (a constant), then

\int_a^b \pi f^2(x,b) dx=b^2\int_a^b dx = b^2(a-b) = b^3-a b^2.

I agree with everyone that the question is ill posed, or there is a typo.
posted by bessel functions seem unnecessarily complicated at 8:39 PM on September 29, 2010


Should have previewed, 7-7 and I had the same idea.
posted by bessel functions seem unnecessarily complicated at 8:41 PM on September 29, 2010


BFSUC: I believe you win this one. I mixed up the order of squaring and integration. Your solution is correct, mine is wrong.
posted by 7-7 at 8:44 PM on September 29, 2010


You've probably already checked, but I have to ask: does a (star) by the problem mean there's an answer in the back of the book?
posted by chndrcks at 8:55 PM on September 29, 2010


My intuitive approach— which ends up at the same place as everyone else's— was to consider the change in volume as a function of either a or b. d(volume)/db is obviously equal to the area of the disk at the 'b' end of the solid of revolution, π(f(b))2 (visualize the infinitesimal extension of the volume as b→b+Δb); set these two expressions equal to each other and solve for f(b). You end up working through the same algebra as the more formal statements above.
posted by hattifattener at 10:11 PM on September 29, 2010


You should tell us which popular calculus text this comes from. I probably have it, and can take a closer look at what's going on in the chapter where it appears.

In hindsight though, there is unequivocally an error in the statement of the exercise. madcaptenor has shown above that it is not possible to have any such function.
posted by King Bee at 5:43 AM on September 30, 2010


Response by poster: The book is Calculus with Analytic Geometry by George Simmons, Second Edition. It's problem 26 in the review problem set for chapter 7.

Even problems don't have solutions in the answer key. Or in the Student Solutions Manual.

Incidentally, it's a good book (so far - I think), even though this isn't the first error that I've found. In fact, I think that I've recommended it in other ask.me questions.

Is madcaptenor's statement about volumes in general also true? Any volume of revolution must have the form g(b)-g(a)?
posted by stuart_s at 7:06 AM on September 30, 2010


Alas, I don't have that one. =(

I'm referring to madcaptenor's statement concerning how integrals work. The integral from 0 to 2 is the sum of the integrals from 0 to 1 and from 1 to 2. His example above shows that this integral does not obey that rule. The conclusion is that there is an error in the text.
posted by King Bee at 7:38 AM on September 30, 2010


Coming back to my guess on a typo, if we just fix a at 0 then we get some solid of revolution with volume equal to its length cubed. I can't think of anything else that works other than a cone. Not to say there isn't anything, but I can't think of one.
posted by monkeymadness at 12:51 PM on September 30, 2010


the whole solid of revolution with volume b^3-a*b^2 is the sum of the 2 solids of revolution with volumes b^3-c*b^2 and c^3-a*c^2 and that this must hold for all a, b and c.

I think this is only true for a single, given function f. If f depends on the choice of the limits of integration, this is not necessarily true. Integration from a to c plus integration from c to b are over completely different functions than integration from a to b.

posted by abc123xyzinfinity at 2:27 PM on September 30, 2010


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