Problem 1 is Exercise 7.1.36 from the book. Here is a graph showing the curve for

We rotate about the line the region bounded by the -axis and this curve.

To find its volume, a first natural attempt would be to use the washer method. We would then attempt to compute the volume as

where is the maximum of for and, for any given value of with and are the values of with such that

Unfortunately, this does not seem to be the best approach, as there does not seem to be a reasonable way of solving for the equation

(In fact, this equation cannot be solved in terms of elementary functions.

Similarly, there is no way of finding exactly what the value of is in terms of elementary functions.)

Since this approach seems to lead us nowhere, we now try to compute the volume using the shell method. Now the volume is expressed as

This expression looks approachable with the techniques we have studied. First, let’s rewrite the integral as

We compute both expressions using integration by parts:

To find we use and so and we can take Hence

We recognize from the graph of that the second expression is zero, and we have:

Similarly, for we have and so and and

The last expression is once more computed using parts, now with and so and This gives

Hence

Finally, the required volume is

Problem 2 asked to evaluate

A first attempt may go by using integration by parts, with and Unfortunately, this approach would not lead to simpler expressions, as both integrals and derivatives of and carry radicals.

If the expression inside the square root were of the form or we could use a trigonometric substitution. However, is not of this form. On the other hand, in Chapter 9 we saw that it is sometimes useful to complete squares, so we may want to try that here. We have:

This suggest trying the trigonometric substitution for We have and Also, when we have or and when we have or

In terms of the integral becomes

To evaluate expressions of this form, we use the identity and obtain

The second expression we recognize as For the first, we use either the reduction formula found in lecture, or integration by parts:

or

from which we get

Finally,

so the required integral equals

Typeset using LaTeX2WP. Here is a printable version of this post.

43.614000-116.202000

Advertisements

Like this:

LikeLoading...

Related

This entry was posted on Sunday, October 18th, 2009 at 1:27 pm and is filed under 175: Calculus II. You can follow any responses to this entry through the RSS 2.0 feed.
You can leave a response, or trackback from your own site.

It is not possible to provide an explicit expression for a non-linear solution. The reason is that (it is a folklore result that) an additive $f:{\mathbb R}\to{\mathbb R}$ is linear iff it is measurable. (This result can be found in a variety of places, it is a standard exercise in measure theory books. As of this writing, there is a short proof here (Intern […]

I learned of this problem through Su Gao, who heard of it years ago while a post-doc at Caltech. David Gale introduced this game in the 70s, I believe. I am only aware of two references in print: Richard K. Guy. Unsolved problems in combinatorial games. In Games of No Chance, (R. J. Nowakowski ed.) MSRI Publications 29, Cambridge University Press, 1996, pp. […]

Let $C$ be the standard Cantor middle-third set. As a consequence of the Baire category theorem, there are numbers $r$ such that $C+r$ consists solely of irrational numbers, see here. What would be an explicit example of a number $r$ with this property? Short of an explicit example, are there any references addressing this question? A natural approach would […]

Suppose $M$ is an inner model (of $\mathsf{ZF}$) with the same reals as $V$, and let $A\subseteq \mathbb R$ be a set of reals in $M$. Suppose further that $A$ is determined in $M$. Under these assumptions, $A$ is also determined in $V$. The point is that since winning strategies are coded by reals, and any possible run of the game for $A$ is coded by a real, […]

Yes. This is obvious if there are no such cardinals. (I assume that the natural numbers of the universe of sets are the true natural numbers. Otherwise, the answer is no, and there is not much else to do.) Assume now that there are such cardinals, and that "large cardinal axiom" is something reasonable (so, provably in $\mathsf{ZFC}$, the relevant […]

The two concepts are different. For example, $\omega$, the first infinite ordinal, is the standard example of an inductive set according to the first definition, but is not inductive in the second sense. In fact, no set can be inductive in both senses (any such putative set would contain all ordinals). In the context of set theory, the usual use of the term […]

I will show that for any positive integers $n,\ell,k$ there is an $M$ so large that for all positive integers $i$, if $i/M\le \ell$, then the difference $$ \left(\frac iM\right)^n-\left(\frac{i-1}M\right)^n $$ is less than $1/k$. Let's prove this first, and then argue that the result follows from it. Note that $$ (i+1)^n-i^n=\sum_{k=0}^{n-1}\binom nk i^ […]

I think it is cleaner to argue without induction. If $n$ is a positive integer and $n\ge 8$, then $7n$ is both less than $n^2$ and a multiple of $n$, so at most $n^2-n$ and therefore $7n+1$ is at most $n^2-n+1

Let PRA be the theory of Primitive recursive arithmetic. This is a subtheory of PA, and it suffices to prove the incompleteness theorem. It is perhaps not the easiest theory to work with, but the point is that a proof of incompleteness can be carried out in a significantly weaker system than the theories to which incompleteness actually applies. It is someti […]

Here is a silly thing; I am not sure it is an "advantage" (or, for that matter, a disadvantage), but it indicates a difference: Inside a model $M$ of $\mathsf{ZF}$ there may be "hidden'' models $N$ of $\mathsf{ZF}$. The situation I have in mind is something like the following, which uses the fact that $\mathsf{ZF}$ is not finitely ax […]