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.

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In Cylindrical coordinates, the integral becomes

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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:

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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:

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

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Sides

Quickly, using similar commands as above:

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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.

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Total Flux (Final Answer)

Adding the flux across the top, bottom and sides:

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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 .

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We want the positive square root, of course.

where

0 ≤ r ≤ 2

0 ≤ t ≤ 2π

/6 ≤ z ≤

(b)

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Problem 5

(a)

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Here 0 ≤ t ≤ 2π. Other answers are possible

(b)

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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:

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Now let's do the line integral:

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Problem 6

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

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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:

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Remember that we computed G = Curl F above

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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):

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

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Created by Mathematica (May 8, 2007) |