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.

The technique of almost disjoint forcing was introduced in MR0289291 (44 #6482). Jensen, R. B.; Solovay, R. M. Some applications of almost disjoint sets. In Mathematical Logic and Foundations of Set Theory (Proc. Internat. Colloq., Jerusalem, 1968), pp. 84–104, North-Holland, Amsterdam, 1970. Fix an almost disjoint family $X=(x_\alpha:\alpha

At the moment most of those decisions come from me, at least for computer science papers (those with a 68 class as primary). The practice of having proceedings and final versions of papers is not exclusive to computer science, but this is where it is most common. I've found more often than not that the journal version is significantly different from the […]

The answer is no in general. For instance, by what is essentially an argument of Sierpiński, if $(X,\Sigma,\nu)$ is a $\sigma$-finite continuous measure space, then no non-null subset of $X$ admits a $\nu\times\nu$-measurable well-ordering. The proof is almost verbatim the one here. It is consistent (assuming large cardinals) that there is an extension of Le […]

I assume by $\aleph$ you mean $\mathfrak c$, the cardinality of the continuum. You can build $D$ by transfinite recursion: Well-order the continuum in type $\mathfrak c$. At stage $\alpha$ you add a point of $A_\alpha$ to your set, and one to its complement. You can always do this because at each stage fewer than $\mathfrak c$ many points have been selected. […]

Stefan, "low" cardinalities do not change by passing from $L({\mathbb R})$ to $L({\mathbb R})[{\mathcal U}]$, so the answer to the second question is negative. More precisely: Assume determinacy in $L({\mathbb R})$. Then $2^\omega/E_0$ is a successor cardinal to ${\mathfrak c}$ (This doesn't matter, all we need is that it is strictly larger. T […]

Yes, the suggested rearrangement converges to 0. This is a particular case of a result of Martin Ohm: For $p$ and $q$ positive integers rearrange the sequence $$\left(\frac{(−1)^{n-1}} n\right)_{n\ge 1} $$ by taking the ﬁrst $p$ positive terms, then the ﬁrst $q$ negative terms, then the next $p$ positive terms, then the next $q$ negative terms, and so on. Th […]

Yes, by the incompleteness theorem. An easy argument is to enumerate the sentences in the language of arithmetic. Assign to each node $\sigma $ of the tree $2^{

A simple example is the permutation $\pi$ given by $\pi(n)=n+2$ if $n$ is even, $\pi(1)=0$, and otherwise $\pi(n)=n−2$. It should be clear that $\pi$ is computable and has the desired property. By the way, regarding the footnote: if a bijection is computable, so is its inverse, so $\pi^{-1}$ is computable as well. In general, given a computable bijection $\s […]

The question is asking to find all polynomials $f$ for which you can find $a,b\in\mathbb R$ with $a\ne b$ such that the displayed identity holds. The concrete numbers $a,b$ may very well depend on $f$. A priori, it may be that for some $f$ there is only one pair for which the identity holds, it may be that for some $f$ there are many such pairs, and it may a […]

The reflection principle is a theorem schema in ZFC, meaning that for each formula $\phi(\vec x)$ we can prove in ZFC a version of the principle for $\phi$. In particular, it gives us that if $\phi$ holds (in the universe of sets) then there is some ordinal $\alpha$ such that $V_\alpha\models \phi$. It follows from this that (assuming its consistency) $\math […]