Answers to UMTYMP Calc III S07 Exam 3 Review

Problem 1

This is a fairly straightforward triple integral problem; to evaluate the integral by hand you'll have to switch to cylindrical coordinates.

In:= Out= In:= Out= In:= Out= In Cylindrical coordinates, the integral becomes

In:= Out= Problem 2

This problem is tedious, but doable.  You have to compute three separate flux integrals because the surface is in three pieces: the top of the cylinder, the bottom, and the sides.

Top

You can actually set up and evaluate the flux across the bottom of the cylinder, but without any calculations I know it has to be 0; the outward pointing normal vector will point straight up, and on the top of the cylinder z=2, so the vectors in the vector field will look like this:

In:= Out= In other words they won't move up or down at all; the movement is perpendicular to the normal vector (i.e. parallel to the surface) and hence there is no flux. Bottom

This one we actually have to compute.  Here's a parametrization; I'm using the letter f so that I can let r be a paremeter, namely the radius:

In:= Out= In:= Out= In:= Out= In:= Out= That points in the wrong direction, so we switch it:

In:= Out= In:= Out= In:= Out=  Sides

Quickly, using similar commands as above: Out= In:= Out= In:= Out= In:= Out= That's the correct vector, since it points out; as expected, it's the same as the vector from the center (axis) of the cylinder out to the point on the cylinder.

In:= Out= In:= Out=  Total Flux (Final Answer)

Adding the flux across the top, bottom and sides:

In:= Out= Problem 3

Look carefully at what you've computed in problems 1 and 2; they're the two sides of the Divergence Theorem!  According to the Divergence Theorem your answers should therefore be equal.

Problem 4

(a)

First find the shadow of the solid in the xy-plane; this is a circle, and we can find the radius by setting two equations equal.  For simplicity I'll replace + with .

In:= Out= We want the positive square root, of course. where

0 ≤ r ≤ 2 0 ≤ t ≤ 2π /6 ≤ z ≤ (b)

In:= Out= In:= Out= In:= Out= Problem 5

(a)

In:= Out= Here 0 ≤ t ≤ 2π.  Other answers are possible

(b)

In:= Out= In:= Out= Anything this ugly is a sign you shouldn't have to use curl F.

In the present problem we can use Stokes' Theorem to convert the surface integral to a line integral using the original vector field F.  F is ugly, but not as bad -- and on our curve it's even nicer:

In:= Out= Now let's do the line integral:

In:= Out= In:= Out= Problem 6

Here are two pictures of the region of integration, with one face left off and the other drawn as a wire mesh:

From In:=  Problem 7

There are, again, many different ways to do this.  The standard way is actually one of the longer ways, but probably the most familiar to all of you:

In:= Out= In:= Out= In:= Out= In:= Out= Remember that we computed G = Curl F above

In:= Out= In:= Out= For the first part of the problem I just need to set up te next integral (but I can have Mathematica evaluate it as long as I've typed it in):

In:= Out= Via Stokes Theorem this is equivalent to the integral of F along the standard unit circle:

In:= Out= In:= Out= In:= Out= Created by Mathematica  (May 8, 2007) 