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.

(As I pointed out in a comment) yes, partial Woodinness is common in arguments in inner model theory. Accordingly, you obtain determinacy results addressing specific pointclasses (typically, well beyond projective). To illustrate this, let me "randomly" highlight two examples: See here for $\Sigma^1_2$-Woodin cardinals and, more generally, the noti […]

I am not sure which statement you heard as the "Ultimate $L$ axiom," but I will assume it is the following version: There is a proper class of Woodin cardinals, and for all sentences $\varphi$ that hold in $V$, there is a universally Baire set $A\subseteq{\mathbb R}$ such that, letting $\theta=\Theta^{L(A,{\mathbb R})}$, we have that $HOD^{L(A,{\ma […]

A Wadge initial segment (of $\mathcal P(\mathbb R)$) is a subset $\Gamma$ of $\mathcal P(\mathbb R)$ such that whenever $A\in\Gamma$ and $B\le_W A$, where $\le_W$ denotes Wadge reducibility, then $B\in\Gamma$. Note that if $\Gamma\subseteq\mathcal P(\mathbb R)$ and $L(\Gamma,\mathbb R)\models \Gamma=\mathcal P(\mathbb R)$, then $\Gamma$ is a Wadge initial se […]

Craig: For a while, there was some research on improving bounds on the number of variables or degree of unsolvable Diophantine equations. Unfortunately, I never got around to cataloging the known results in any systematic way, so all I can offer is some pointers to relevant references, but I am not sure of what the current records are. Perhaps the first pape […]

Yes. Consider, for instance, Conway's base 13 function $c$, or any function that is everywhere discontinuous and has range $\mathbb R$ in every interval. Pick continuous bijections $f_n:\mathbb R\to(-1/n,1/n)$ for $n\in\mathbb N^+$. Pick a strictly decreasing sequence $(x_n)_{n\ge1}$ converging to $0$. Define $f$ by setting $f(x)=0$ if $x=0$ or $\pm x_n […]

All proofs of the Bernstein-Cantor-Schroeder theorem that I know either directly or with very little work produce an explicit bijection from any given pair of injections. There is an obvious injection from $[0,1]$ to $C[0,1]$ mapping each $t$ to the function constantly equal to $t$, so the question reduces to finding an explicit injection from $C[0,1]$ to $[ […]

One way we formalize this "limitation" idea is via interpretative power. John Steel describes this approach carefully in several places, so you may want to read what he says, in particular at Solomon Feferman, Harvey M. Friedman, Penelope Maddy, and John R. Steel. Does mathematics need new axioms?, The Bulletin of Symbolic Logic, 6 (4), (2000), 401 […]

"There are" examples of discontinuous homomorphisms between Banach algebras. However, the quotes are there because the question is independent of the usual axioms of set theory. I quote from the introduction to W. Hugh Woodin, "A discontinuous homomorphism from $C(X)$ without CH", J. London Math. Soc. (2) 48 (1993), no. 2, 299-315, MR1231 […]

This is Hausdorff's formula. Recall that $\tau^\lambda$ is the cardinality of the set ${}^\lambda\tau$ of functions $f\!:\lambda\to\tau$, and that $\kappa^+$ is regular for all $\kappa$. Now, there are two possibilities: If $\alpha\ge\tau$, then $2^\alpha\le\tau^\alpha\le(2^\alpha)^\alpha=2^\alpha$, so $\tau^\alpha=2^\alpha$. In particular, if $\alpha\g […]

Fix a model $M$ of a theory for which it makes sense to talk about $\omega$ ($M$ does not need to be a model of set theory, it could even be simply an ordered set with a minimum in which every element has an immediate successor and every element other than the minimum has an immediate predecessor; in this case we could identify $\omega^M$ with $M$ itself). W […]