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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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$F[x_, y_, z_] = {x, y, (4 - z^2) (x^2 + y^2)}$

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${x, y, (x^2 + y^2) (4 - z^2)}$

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$Integrate[2 - 2 (x^2 + y^2) z, {x, -3, 3}, {y, -Sqrt[9 - x^2], Sqrt[9 - x^2]}, {z, 0, 2}]$

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In Cylindrical coordinates, the integral becomes

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$Integrate[(2 - 2 (r^2) z) r, {t, 0, 2Pi}, {z, 0, 2}, {r, 0, 3}]$

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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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$f[r_, t_] = {r Cos[t], r Sin[t], 0}$

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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:

$f[t_, h_] = {3Cos[t], 3 Sin[t], h}$

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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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${{r→ -2 3^(1/2)}, {r→2 3^(1/2)}}$

We want the positive square root, of course.

$S[r_, t_, z_] = {r Cos[t], r Sin[t], z}$

where

0 ≤ r ≤ 2
0 ≤ t ≤ 2π
/6 ≤ z ≤

(b)

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$F[x_, y_, z_] = {x + y z, Sin[x^9 z^6], Cos[x^7 y^8]}$

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${x + y z, Sin[x^9 z^6], Cos[x^7 y^8]}$

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$Integrate[1 * r, {r, 0, 2Sqrt[3]}, {t, 0, 2Pi}, {z, r^2/6, Sqrt[16 - r^2]}]$

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Problem 5

(a)

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$r[t_] = {3Sin[t], 3Cos[t], 0}$

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

(b)

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$F[x_, y_, z_] = {-y, x, Sin[x^7 y^8 z^9]}$

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${-y, x, Sin[x^7 y^8 z^9]}$

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${8 x^7 y^7 z^9 Cos[x^7 y^8 z^9], -7 x^6 y^8 z^9 Cos[x^7 y^8 z^9], 2}$

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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